Jerry Qin

A Working List of Probability Questions: Week 3

Mon, 06 Jul 2020 22:40:32 GMT

Programming, Statistics, and Past Interview Questions
July 6, 2020 | Updated August 5, 2020

Welcome to the last week of my trading interview prep series! My focus this week is to review programming and statistics concepts, along with exploring more challenging problems in areas I’ve already done practice problems in. Aside from these topics, I’ll mostly be pulling past interview questions from Glassdoor pages and Leetcode. As always, I’ll be providing solutions to the problems I do and a list of resources I use for reviewing concepts, along with a list of extra problems that I work on after this week is over.

Day 13 | Day 14 | Day 15 | Day 16 | Day 17 | More

Day Thirteen: July 6

Most of today’s problems are past interview questions in trading!

Problem 1. You have a drawer with an infinite number of two colors of socks, which exist in equal probability. What is the expected number of attempts at taking out socks individually from the drawer before a matching pair is found?
Solution. This is a past Jane Street interview question that should be answered without pen/paper. Note that a matching pair of socks must be found by the third draw by the pidgeonhole principle. After our first draw, we have probability $\frac{1}{2}$ of drawing the same color our second round and $\frac{1}{2}$ of drawing a different color. Should we draw a third color, it is guaranteed that we will complete a pair of matching socks the third round so we have and equal probability of completing the pair in two or three rounds. Then then expected number of attempts is $2.5$ .

Problem 2. There is a raffle with 80 total tickets. Three tickets win a prize. I buy 5 tickets. What is the approximate probability that I win exactly one prize? Solution. We can approach this problem either combinatorially or by brute force (I prefer the combinatorial method but the brute force method is more intuitive). There are $\binom{3}{1}$ ways to choose one (and only one) of the three prizes. Of the remaining $77$ tickets, there are $\binom{77}{4}$ ways to choose four non-winning tickets. We can multiply these two together to get the number of outcomes in which we win only one prize. Finally, there are $\binom{80}{5}$ ways to choose $5$ tickets (winning or non-winning) so we divide our numerator by this to get the probability of getting a prize, or about $16.89\%$ . To brute force it, let’s consider each of the three prizes. Note that the probability of winning the first prize is $\frac{5}{80}$ . Then, considering that we can only win one of the three prizes,

\begin{aligned} P(\text{win only first prize}) &= \tfrac{5}{80} \cdot \tfrac{75}{79} \cdot \tfrac{74}{78} \\ P(\text{win only second prize}) &= \tfrac{75}{80} \cdot \tfrac{5}{79} \cdot \tfrac{74}{78} \\ P(\text{win only third prize}) &= \tfrac{75}{80} \cdot \tfrac{74}{79} \cdot \tfrac{5}{78} \end{aligned}

Combining these probabilities gives us the same answer as above with our combinatorial solution.

Problem 3. Consider a deck with $26$ black and $26$ red cards. You draw one card at a time and you can choose either guess on whether it is red beforehand or simply observe the result. If the card is red you get $\$1$ and the game ends whenever you decide to guess. What is your strategy to play this game and the expected earnings? Solution. I have no idea what the right answer is so solving this question was rather fun. The most pedestrian strategy is to guess one the first card with probability $\frac{1}{2}$ of guessing correctly, with EV $\$0.50$ . However, we can aim to slighly improve this by basing our future guesses on previously drawn cards. We can find the expected value guessing at any given round based on the number of red and black cards left in the deck. Let $f(r,b)$ be the expected payoff for the state where there are $r$ red and $b$ blue cards left. The probability of guessing correctly at any given state of red and black cards is $f(r,b) = \frac{r}{r+b}$ (the expected value of guessing at each round is also described by this function). This is where my strategy gets a bit more dicey: I’d first draw $5$ cards with the hope that I pick out more black cards than red cards which would increase the EV of the rest of the deck. Should I end up behind on this initial sequence of draws, I’m more likely to draw more black cards than red cards in the commin turns so I’ll continue to do so until I reacha bit over $50\%$ of drawing another red card. This is an inexact science and I don’t really have an answer to encapsulate every situation that could arise other than drawing more cards up to a point where there are few enough cards or high enough percentage of drawing a red card that I can maximize my probability of winning. This strategy introduces more variability into our expected payoff: instead of having a strictly $50\%$ chance of getting a dollar, we introduce a bit of rng into our draw. We can potentially go above or below $50\%$ of guessing a red card correctly. Doing this, I run the risk of never getting back up to $50\%$ should I draw $26$ red cards right off the bat (an extremely unlikely and unfortunate outcome). Update: I’ll be posting a revised answer to this question soon!

My calculation for the expected payoff is also a bit sketchy and relies on a symmetry argument: consider that we draw a sequence of cards such that we have probability $p$ of drawing a red card next time (EV $p$ ). However, we are just as likely to have draw the sequence of cards such that each $i^\text{th}$ card drawn in the second sequence is the opposite of its counterpart in our actual draw, giving EV $1-p$ . Applying this to any state of cards that we decide to guess at, we are equally likely to have a payoff $p$ as a payoff $1-p$ so our expected value of the overall game is $\$0.50$ regardless of our strategy. All in all, I’m not really sure if I improve anything for myself by taking on increase variability but I am inclined to be less risk averse given that I cannot lose money.

Problem 4. How would I design the elevators for a new 40 story office building that had an average of 100 people per floor to most efficiently fill and empty the building given a standard 9-5 workday and traffic conditions in my city? The answer should to be as detailed as possible, including expected passengers per car, time per stop, average floors stops per trip at various hours, etc. Solution. This is a Google PM question but the nature of this question feels very trading-esque. Note hat $4000$ people must enter and exit the building each day, presumably during the morning and afternoon close to $9$ and $5$ . Some people will enter in early and some will funnel in a bit later in the workday so I’ll asume we have some sort of normal distribution of people entering the office centered at $9:00$ , with sd of $30$ minutes. Then, we have approximately the following number of people using the elevator in various times during the morning:

7:30 - 8:00 - $80$
8:00 - 8:30 - $560$
8:30 - 9:00 - $1360$
9:00 - 9:30 - $1360$
9:30 - 10:00 - $80$
10:00 - 10:30 - $80$

During lunch some people will probably leave the office, but it’s quite unlikely that the elevators will be operating at full capacity since some people (perhaps even a majority) will stay in for lunch. Let’s say that there are $10$ elevators and each elevator can carry an average of $10$ people, disregarding weight, so we have $1360$ people being funnelled into the building in a $30$ minute interval. Then each elevator should make $\frac{1360}{100} = 13.6$ trips between the first floor and carrying people up to higher floors. Since this office complex is relatively large, we should consider splitting up the floors serviced by each elevator so the time spent travelling is minimized. We can break this up so that each elevator services $8$ floors: elevator 1 & 2 servicing floors $1-8$ , elevators 3 & 4 servicing floors $1, 9-16$ , etc. Then each elevator makes an average of $9$ floor stops per trip, with the outlier of elevators $1,2$ which average $8$ floors per trip. While this helps reduce inefficiency, the elevators that travel the most should still travel between te first floor and the last $8$ floors, making an average of one trip per $2.2$ minutes. It takes relatively little time for people to exit an elevator, so we’ll allocate an average of $5$ seconds per stop before moving on to the next floor. Then elevators $9,10$ will spend $40$ seconds travelling between top floors, which gives them about $90$ more seconds to travel between floor 1 and floor 40 and fill up another elevator full of people. Note that this answer can be a bit more detailed, focusing in on maximizing the carriage capacity of higher floor elevators by making use of the fact that people on lower floors will take the stairs, and other random variables that I haven’t considered.

Problem 5. Consider a card game with two black cards and two red cards. You play a game where you draw a card each round. If it’s black you lose a dollar and if its red you win a dollar. You can choose to stop at any time. What’s your winning strategy and its associated expected value? Solution. Note that by playing the game out completely we ensure breaking even. To cover the obvious cases, if we finish the third draw with one dollar ( $RBR$ ) or the second draw with two dollars ( $RR$ ), it makes sense to end the game since we are guaranteed to lose part of our winnings by continuing. Also, if we’re behind at any stage of the game we continue drawing since we can guarantee an even outcome and if we’re even at any point in the game we should keep drawing since we will either have positive gains or play out the entire game and end even. Let’s consider what happens when we draw $R$ first. Then we will be up one dollar and have probability $\frac{1}{3}$ of drawing another red (net $+2$ ), $\frac{2}{3}\cdot \frac{1}{3} = \frac{1}{3}$ of drawing two blacks and forcing us to finish the game ( $+0$ ), and $\frac{1}{3}$ of drawing $BR$ afterward and ending ( $+1$ ). So conditional of drawing a red first turn, we have ev $\frac{1}{3} \cdot 2 + \frac{1}{3} \cdot 1$ . Now consider rolling $B$ first: if you draw two consecutive reds afterwards ( $\frac{1}{3}$ ) you can end with $\$1$ and break even otherwise ( $\frac{2}{3}$ ). Then, our total EV is $\frac{1}{2} \cdot 1 + \frac{1}{2}(\frac{1}{3} \cdot 1 + \frac{2}{3} \cdot 0) = \frac{2}{3}$ using this strategy. This can be extended to any number of cards.

Day Fourteen: July 7

Todays problems are a mix of programming questions from Leetcode and some past trading interview questions.

Problem 1. Given a string, find the length of the longest substring without repeating characters. Solution. This is a leetcode medium question. We can solve this problem with a sliding window technique and minimize time complexity by using a hashmap, gicing an $O(n)$ solution. Starting from the first index of the string, we can insert each character into the hashmap and begin counting substrings sies through a counter variable. Should we happen to come across a non-unique character in our hashmap, we slide our “window” once index to the right and scan the next substring and update our counter variable each time we come across a unique character.

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int left(0), counter(0);
        unordered_map<char, int> myMap;

        for (int i = 0; i < s.size(); i++) {
            if (myMap.find(s[i]) != myMap.end()) {
                left = std::max(left, myMap[s[i]] + 1);
            }
            myMap[s[i]] = i;
            counter = std::max(counter, i - left + 1);
        }
        return counter;
    }
};

This solution runs faster than $54\%$ C++ submissions so it could definitely be streamlined/improved. I implemented a basic sliding window technique and didn’t dedicate much time to improving it.

Problem 2. There are $3$ balls distributed in $4$ boxes. What is the probability that the maximum number of balls in any given box is exactly $2$ ? Solution. This is an incredibly simple combinatorics question that I got wrong on a recent HackerRank. The number of ways to arrange $n$ balls into $k$ boxes is $\binom{n+k-1}{n}$ , so there are $\binom{6}{3} = 20$ ways to arrange the the balls into boxes. You can learn more about balls and boxes combinatorial methods here. To find the desired probability we will consider the case that $2$ balls are in one box. Then, the final ball is in one of the three remaining boxes. There are $\binom{4}{1}$ ways to pick a box to carry two balls and $\binom{3}{1}$ ways to pick which of the remaining boxes we place the last ball into for each $2$ -balled box. Then, we have probability $\frac{\binom{4}{1} \binom{3}{1}}{\binom{6}{3}} = \frac{3}{5}$ of the given scenario occuring.

If you didn’t know the formula for the total number of ways to arrange the balls into boxes we can still derive it with combinatorial methods. There are three cases that we must consider: all balls are contained in different boxes, two balls are contained in one box and the final one in another box, and all three balls in one box. There are $\binom{4}{3}$ ways to choose three boxes for which each contains one ball, $\binom{4}{1}\binom{3}{1}$ ways to choose one box to hold two balls and a second box to hold the final ball (as previously calculated), and $\binom{4}{1}$ ways to choose one box to contain all the balls. Then our probability of obtaining the desired outcome is $\frac{\binom{4}{1}\binom{3}{1}}{\binom{4}{3} + \binom{4}{1}\binom{3}{1} + \binom{4}{1}} = \frac{3}{5}$ .

Problem 3. You have a box filled with cash. Cash value is uniformly and randomly distributed from 0 to 1000. You are trying to win the box in an auction: you win the box if you bid at least the value of the cash in the box; you win nothing if you bid less (but you lose nothing). If you win the box, you can resell it for 150% of its value. How much should you bid to maximize the expected value of your profit (resale of box minus bid)? Solution. We would bid $\$0$ as the expected value of this game is always negative. I’ll assume that cash means there will not be fractions of a dollar in the box. Let $X$ be the payout of the box and $Y$ be the amount that we pay for the box after bidding. Then we have $P(X) = 1.5X - Y$ , $X\leq Y$ as ur profit function. Since $\mathbb{E}[P(X)] = 1.5\mathbb{E}[X] - \mathbb{Y}$ , we first need to find the expected value of $X$ and $Y$ . Each integer between $0$ and $1000$ is equally likely to be in the box ( $P(X=x)= \frac{1}{1001}$ ). Since we only realize the payout of the box if $X\leq Y$ we only need to sum up to $Y$ :

\mathbb{E}[X] = \sum_{x= 0}^Y \frac{1}{1001} = \frac{Y(Y+1)}{2002}

To find the expected value of $Y$ , note that we will pay $Y$ dollars for the box if $X\leq Y$ and $0$ otherwise. Given a $Y$ , the probability that $X\leq Y$ is $\frac{Y+1}{1001}$ since each potential value of $X$ shy of $Y$ has probability $\frac{1}{1001}$ . Then $\mathbb{E}[Y] = \frac{Y(Y+1)}{1001}$ . It follows that

\begin{aligned} \mathbb{E}[P(X)] &= 1.5\mathbb{E}[X] - E[Y] = \frac{1.5Y(Y+1)}{2002} - \frac{Y(Y+1)}{1001} \\ &= \frac{-3Y(Y+1)}{4004} \end{aligned}

which is always negative so we should not bid any amount for the box. Our optimal bid then is $\$0$ , with expected payoff $\$0$ . Note that my original assumption was the impetus for using a summation; should I have assumed that any value was possible in the box, then we would have to approach $X$ as though it were a continuous variable.

Problem 4. Imagine I flip 100 coins in a sequence, and you need to guess the sequence in one try. You are allowed to ask one yes/no question. What do you ask to maximize the probability of guessing the sequence? Solution. This question is a bit open-ended as there isn’t really a definitive correct answer. Without information, we have probability $(\frac{1}{2})^{100}$ of guessing the entire sequence correctly. We can double our probablility to $(\frac{1}{2})^{99}$ by asking a question about any specific index in the sequence (we should ask for the first one since it is the only index that we are guaranteed to reach). We could potentially ask whether there are more heads than tails or vice versa, but the probability that one type of flip occurs more frequently than the other in the sequence is $\frac{1}{4}$ (convince yourself of this) and a yes to the question provides minimal benefit in choosing the values of any specific index until very late in the sequence. A no to the question, which is far more likely, means that the question is esentially wasted as we don’t know whther there are more tails than heads or if they are equal in number. Potentially we could ask about the second or the third index, which would give us probability $\frac{1}{2}, \frac{1}{4}$ of guessing the first two and three indicies respectively. However, this still leaves the first index up to chance and runs the risk of guessing no index correctly should we guess the first one wrong and the game ends. I can’t really think of any other ways to gather information that would be readily pertinent to guessing the the sequence and it seems like asking for the first index is the best option since we can guarantee at least one index is correct. Were I more confident in my understanding of the coin flips or the nature of the sequence I potentially would employ the more risky strategy of asking the identity of some non-first index, or even if the number of heads is greater than the number of tails if I were confident enough in my ability to guess more than $50\%$ of the sequence. As it stands however, asking about anything but the first index seems to be extremely high risk low reward.

Day Fifteen: July 8

Here are some more betting games from previous interviews.

Problem 1. You have all the clubs from a deck, 13 cards, and you can choose 2 from the deck and get paid their product, where all face cards are considered to be 0. You can pay $1$ to reveal the difference of any two cards you choose, how much would you pay to play this game?
Solution. Our probability of getting a nontrivial outcome (not picking a $J,Q,K$ ) just guessing off the top is $\frac{\binom{10}{2}}{\binom{13}{2}} = \frac{15}{26}$ . We’ll aim to improve this and maximize our payout from this game. Let’s index our cards and call them $C_1, C_2, \dots, C_13$ , with $C_i$ describing the $i^\text{th}$ card from the left. WOLOG lets make $C_1$ the card that we compare all other ones to. If $C_1$ is a face card, then there are $2$ more face cards within the next $12$ cards so it will take $11$ comparisons for us to encounter a second face card, at which we’ll see the difference between $C_1$ and the new face card is $0$ . From there, we know that all of the previous comparisons we’ve done have yielded the exact value of each card, which allows us to select the $9$ and $10$ cards for a $\$90$ payout. Note that this is the worse case scenario and there is a chance we come across the second zero much earlier and it’s simply our imperative to find the $9,10$ cards through direct comparison.

If the first card is not a face card, then there are $3$ face cards in the next $12$ cards that we’ll compare to $C_1$ . Then, within $11$ cards, we’ll come across at least $2$ face cards, at which we’ll realize that two of the comparisons that we’ve done yield the same difference between cards. The two cards that have the same difference when compared to $C_1$ are then face cards and their difference with $C_1$ is simply the value of $C_1$ . From there, you can reverse engineer the values of the rest of the cards you went through in order to find the $9,10$ cards. Again, we might come across the case where this occurs much earlier than the $11^\text{th}$ comparsion which would mean that we’d simply have to find the $9,10$ cards again through comparisons. From these two cases, we’ve confirmed that spending $\$11$ can guarantee a payout of $\$90$ . I’d then pay up to $\$79$ to play this game, which ensures that I’ll always have a nonnegative expectation. Note that $79$ is not the expected value of the game, as we’d have to consider the multitude of cases that can arise with the cards, but that’s relatively tedious to calculate so I’m inclined to pay up to, or maybe a bit more than, $\$79$ . The reason I’m willing to go beyong $\$79$ is because paying that amount guarantees a nonnegative payout, but the expected profit of doing this is positive, as we’ve essentially done a worst case analysis and haven’t considered the game on average.

Problem 2. We randomly sample integers $a,b,c$ from a uniform distribution of $[0,1]$ . What is the probability that $ax^2 + 2bx + c = 0$ has real roots? Solution. In order for the given quadratic equation to have real roots, we must have $4b^2 - 4ac \geq 0$ , or $b^2 \geq ac$ . This can be solved by finding the probability that $b^2 \geq ac$ on the unit cube, but I’m not particularly good at multivariable calculus so I opted for a solution using random variables. For a uniformly distributed r.v. $X$ on $[0,1]$ , we have that $-\ln(X)$ follows an exponential distribution with $\lambda = 1$ . Then, let $A,B,C$ be uniform independent random variables on $[0,1]$ and

P(B^2 \geq AC) = P(-\ln(B) \leq \frac{-\ln(A) - \ln(C)}{2}) = P(X \leq \frac{Y}{2})

for $X$ exponential and $Y$ with the pdf $xe^{-x}$ for $x>0$ . Since $X$ and $Y$ are independent, we have

\int_0^\infty P(X \leq \frac{Y}{2} \, | \, Y = x)f_Y(x) \, dx = \int_0^\infty [1 - e^{-x/2}]xe^{-x} \, dx = \frac{5}{9}

Then the probability that our quadratic equation has real roots is $\frac{5}{9}$ . This solution relies on drawing the connection between the negative log of a uniform distribution on $[0,1]$ being exponential and the sum of exponential distributions with $\lambda$ parameter having a pdf of $\lambda^2 xe^{-\lambda x}$ . A more basic case of this problem with $ax^2 + bx + c = 0$ can be found here.

Problem 3. Let’s play a game - I give you a 12 sided die and will pay you whatever the die lands on. If you are unhappy with the roll, you can choose to roll another two 6-sided dice and I will pay you the sum of the two dice. How much are you willing to pay to play this game? Solution. This is a Jane Street interview question that I came across on their Glassdoor page. Note that the expected value of the second dice is $7$ (just the sum of expected value of two fair dice) so if we roll less than a $7$ on the $12$ -sided die, we’ll be incentivised to roll again. This occurs with probability $\frac{1}{2}$ so we have $\frac{1}{2}$ chance of getting a $\$7$ payoff. On the other hand, we also have $\frac{1}{2}$ probability that we roll at least a $7$ , giving us expected payoff of $\frac{1}{6}(7 + 8 + 9 + 10 + 11 + 12) = 9.5$ for the scenario that we roll at least a $7$ of the $12$ -sided die. Then, the total expected value is $\frac{1}{2} \cdot 7 + \frac{1}{2} \cdot 9.5 = 8.25$ . Then it follows that I’d pay up to $\$8.25$ to play this game, assuming that I’m using the optimal strategy that I described above.

Problem 4. Given n non-negative integers $a_1, a_2, \dots, a_n$ , where each represents a point at coordinate $(i, a_i)$ . n vertical lines are drawn such that the two endpoints of line $i$ is at $(i, a_i)$ and $(i, 0)$ . Find two lines, which together with x-axis forms a container, such that the container contains the most water. Solution. This is another Leetcode medium question. The brute force solution involves comparing every possible pair of lines and the volumes of containers that they build, which run in $O(n)$ time. On the other hand, we can approach the question with the following intution: given two lines, the area of the container formed is limited by the height of the lower stick. In hopes of getting a taller line, we would shift the lower line one unit to the right (or left depending which is shorter), which will decrease the width but add the potential of increasing the height of the next line. After each iteration of line moving, we’ll check the area again and make updates to the maximum area should that be applicable. This is done as opposed to moving the taller line, which would decrease the width of the container and add no possibility of improved height. This method won’t always guarantee us a larger area, but it encompasses all the possibilites of higher sticks adding area as we “traverse” the array. This gives us the following solution:

class Solution {
public:
    int maxArea(vector<int>& height) {
        int size = height.size();
        int left = 0;  
        int right = size - 1;
        int maxArea = 0;
        
        while(left != right){
            maxArea = max(maxArea, containerArea(height[left], height[right], right - left));
            
            if(height[right] > height[left]){
                left++;
            }else{
                right--;
            }
        }
        return maxArea;
    }
    
    int containerArea(int lHeight, int rHeight, int width){
        return width * min(lHeight, rHeight);
    }
};

This solution runs in $O(n)$ time, but does waste some resources. When we shift over from one line to another, we’re doing it in hopes that we’ll compensate for the smaller width by getting to a higher line. We can guarantee that the next area we consider is with a stick of higher height by using a while loop to continue incrementing our line index until we reach a line that is taller than the original one we deemed was too short. This would be relatively simple to implement, simply changing left++ to while(height[left] <= min(height[left], height[right]) {left++;} (do this for right as well), which would make up for the wasted time we iterate through our origina while loop.

Day Sixteen: July 9

Here are a couple more solutions to some leetcode and past interview questions at trading firms.

Problem 1. What is the expected value of the largest of four dice rolls? Solution. This is a pretty straightforward order statistic question. Let $X_i$ be the random variable denoting the outcome of the $i^\text{th}$ dice rolls. Our goal is to find the pmf of $Z = \max\{X_1, X_2, X_3, X_4\}$ , which is the $4^\text{th}$ . One thing to keep in mind is that each of the four dice rolls is independent so we can multiply their probabilies together when combining cdf’s. Then,

\begin{aligned} P(Z \leq z) &= P(X_1, X_2, X_3, X_4 \leq z) = P(X_1 \leq z) \cdots P(X_4 \leq z)\\ &= P(X_1 \leq z)^4 = \left(\frac{x}{6} \right)^4 \end{aligned}

Since each $X_i$ is discretely distributed, we have that $P(X_i = x) = P(X \leq x) - P(X \leq x-1)$ so $p_{Z}(z) = \frac{z^4 - (z-1)^4}{6^4}$ . We can directly use this pmf to calculate the expected value of $Z$ :

\mathbb{E}[Z] = \sum_{\mathbb{R}} p_Z(z)z = \sum_{z=1}^6 z \cdot \frac{z^4 - (z-1)^4}{6^4} = \frac{6797}{1296} \approx 5.24

Then, the expected value of the largest of four independent dice rolls is $5.24$ . Note that we can extend this problem to $n$ dice rolls by employing a similar strategy of using the definition of expected value and calculating the cdf of $Z = \max\{X_1 , \dots, X_n\}$ , where $X_1, \dots X_n$ signify the outcomes of $n$ independent dice rolls.

Problem 2. If I toss 3 Tic Tacs on a tic-tac-toe board at random, and I pay you $\$20$ if the pieces create tic-tac-toe, and you pay me $\$1$ if they dont, do you want to play the game? Solution. This is a past interview question at SIG. My intuition led me to solve it combinatorially since it’s a varient of the ball and boxes question from a couple days ago. Assuming that Tic Tacs don’t have to land on unique boxes on the board, we have $\binom{9}{3} + \binom{9}{1}\binom{8}{1} + \binom{9}{1} = 165$ ways to arrange the Tic Tacs on the board (this can also be done by $\binom{9+3-1}{3}$ ). Additionally, there are $8$ ways to arrange the Tic Tacs in order to create a tic-tac-toe with the board (6 ways across the rows and columns and 2 additional for the diagonals). Then, our expected payout to this game is

\mathbb{E}[X] = \frac{8}{165} \cdot 20 + \frac{157}{165} \cdot -1 \approx 0.018

Our expected payout from playing this game is about $2$ cents per turn so I would play this game. The original question on SIG’s Glassdoor page can be found here, and documents the scenario in which all the Tic Tacs land on unique boxes and the payout for a tic-tac-toe is only $9$ dollars instead of $20$ .

Problem 3. Flip a fair coin many times until either a pattern of $HHT$ or $HTT$ occurs. What’s the probability that HHT occurs before HTT? Solution. This is a classic question that has a relatively elegant solution. We could potentially model this with a Markov chain but this approach is rather extensive and unnecessary. Note that since both sequences begin with $H$ , sequences with leading tails can be ignored up until the appearance of the first $H$ . Then within the first three flips, the following sequences each appear with probability $\frac{1}{4}$ :

$HHH$ : We will have $HHT$ appear first as soon as the next heads shows up $HHT$ : $HHT$ appears first
$HTT$ : $HTT$ appears first
$HTH$ : The sequence is basically reset as $H$ is no longer within the most recent $3$ -flip sequence

Note that we dont consider the $4$ outcomes that begin with a tail flipped first due to our earlier observation. The first two cases guarantee that $HHT$ will be flipped first, the third guarantees $HHT$ appears first, and the final case is a reset. Then $HHT$ appears first with probablity $\frac{2}{3}$ , twice that of $HTT$ appearing first. You can find a similar question dealing with consecutive tennis wins here.

Problem 4. You roll a dice until you get a $5$ . What is the expected value of the minimum value rolled? Solution. This is a past interview at Jane Street. It can be solved with the tail sum formula, $\mathbb{E}[x] = \sum_{x=0}^\infty P(X\geq x)$ but I approached it combinatorially. Let $X$ be the minimum value of the dice rolls until the first $5$ is rolled. The maximum value of $X$ is $5$ as rolling only $6$ ’s gets cancelled out when the first $5$ is rolled so $\mathbb{E}[X] = \sum_{x=1}^5 P(X = x)$ . Consider $P(X=1)$ : since $1$ is the smallest possible value for $X$ we don’t need to consider the values $2,3,4$ being rolled before the first $5$ . We only need a $1$ to be rolled before the first $5$ , which happens with probability $\frac{1}{2}$ . For $P(X=2)$ we need $2$ to be rolled before $5$ and $1$ to not be rolled before $5$ . If we consider the $3!$ unique permuations of $\{1,2,5\}$ , there is only one of them ( $251$ ) that satisfies this condition. Then, there is $\frac{1}{3!} = \frac{1}{6}$ chance that $P(X=2)$ . For $P(X=3)$ we need $3$ to be rolled before $5$ and $2$ and $1$ to be rolled after $5$ . Only $3512$ and $3521$ satisfy this giving probability $\frac{2}{4!} = \frac{1}{12}$ . For $P(X=4)$ we need $4$ rolled before $5$ and $1,2,3$ rolled afterward, which has $\frac{3!}{5!}$ probability. Finally, for $P(X=5)$ , we need to roll $5$ first, followed by $1,2,3,4$ in any order. This has $\frac{4!}{5!} = \frac{1}{5}$ probability, giving

\mathbb{E}[X] = \sum_{x=1}^5 P(X=x) = \frac{1}{2} + 2 \cdot \frac{1!}{3!} + 3 \cdot \frac{2!}{4!} + 4 \cdot \frac{3!}{5!} + 5 \cdot \frac{4!}{5!} = \frac{137}{60}

Then the expected value of the minimum value of dice rolled is about $2.28$ .

Day Seventeen: July 10

Today is the last day I’ll be consistently updating my website with new problems but stay tuned below for more resources in the future!

Problem 1. Suppose there are $n$ points on a circle. How many regions can be formed by connecting all of these points to one another? Solution. This is a problem that can solved for by studying the first few cases. Lets called $R(n)$ the number of regions that can be formed with $n$ points. Then $R(1)=1$ and $R(2)=2$ . Let $R'(n)$ be our approximation of $R(n)$ . On first glance, it seems like the recurring pattern is $R'(n) = 2^{n-1}$ . However, this breaks down when $n=6$ as $R(6) = 31 \neq 32$ . We can approach this problem instead by considering the number of regions that each new chord between points creates. For $n$ points, connecting each of them gives us $\binom{n}{2}$ chords. However, in the cases of $n=1, 2$ we have $\binom{n}{2} = 1, 2$ respectively while $R(1) = 1$ and $R(2) =2$ . In both of these cases, there is an additional region present that doesn’t seem to be dependent on the number of line segments drawn. This can be confirmed by the fact that $R(0) = 1$ and there is always going to one region present before we consider any points on the circle. Then we potentially have $R'(n) = \binom{n}{2} + 1$ . However when $n >= 3$ this pattern seems to break down. $R(4) = 8$ , $R(5) = 16$ , and $R(6) = 31$ but we have the estimated values $R'(4) = 7$ , $R'(5) = 11$ and $R'(6) = 16$ . This difference that we observe between $R(n)$ and $R'(n)$ can be explained through the number of intersection between chords in the circle (convince yourself of this). We have an intersection between chords for every two lines, which require $4$ points so there are a total of $\binom{n}{4}$ intersections for $n$ points. Then $R(n) = \binom{n}{2} + \binom{n}{4} + 1$ (you can check that adding $\binom{n}{4}$ doesn’t break any of the smaller cases). For a proof of why $\binom{n}{2} + \binom{n}{4} + 1 = 2^{n-1}$ for $n \in [1,5]$ , see the top comment on this Stack Exchange post.

Problem 2. You are given a 100-sided die. After you roll once, you can choose to either get paid the dollar amount of that roll OR pay one dollar for one more roll. What is the expected value of the game? Solution. Let $X$ be the payout from playing this game. For this approach I’m assuming that I’ll want to stop playing the game once I reach some value $x\in [1,100]$ and find the value of $x$ that maximizes $\mathbb{E}[X]$ . In my opinion, this approach is flawed as we aren’t keeping track of the amount of money we’ve already paid which prevents us from maximizing profit, but this might be subverted by the fact that we’re calculating the EV of the game overall. So our strategy is to keep our winnings if we roll $X > x$ and and keep rerolling if we get $X\leq x$ . An interesting thing to note is that since each roll of the dice is independent, each $n^\text{th}$ roll of the dice has exactly the same conditions and distribution as the first roll (except we’ve spent $n$ dollars). Then on our first roll we have one of two cases:
$1$ : we roll $X> x$ with probability $\frac{100 - x}{100}$ with expected winning $\frac{101 + x}{2}$
$2$ : we roll $X\leq x$ with probability $\frac{x}{100}$ and we pay $\$1$ to reroll for payout $\mathbb{E}[X] - 1$
Combining these two, we have

\begin{aligned} \mathbb{E}[X] &= \frac{100 - x}{100} \cdot \frac{101 + x }{2} + \frac{x(\mathbb{E}[X] - 1)}{100}\\ &= \frac{101 + x}{2} - \frac{x}{100-x}\\ \end{aligned}

We can then differentiate to find the maximum:

\begin{aligned} \frac{dE}{dx} &= \frac{-x}{(100-x)^2} - \frac{1}{100-x} + \frac{1}{2}\\ &= \frac{x^2 - 200x + 9800}{2(x-100)^2} = 0 \end{aligned}

Solving for $x$ gives us $x\approx 85.86$ . Since $x$ can only take on discrete values, we check $85$ and $86$ to find that $\mathbb{E}[85] = 87.33$ and $\mathbb{E}[86] = 87.36$ . Then, the expected value of the game optimized by using this strategy is $\$87.36$ .

Problem 3. Suppose two players are playing a game. The first player chose an integer between 1-30. The second player picks a different one. After that, a random integer (x) between 1-30 will be generated. The player whose number is closer to x get paid x dollars. Would you go first or second? What number should you pick? Solution. Let $m$ and $n$ be Player 1 and Player 2’s choices respectively. Player 2 can maximize his payout by choosing $n$ to be close to $m$ but he must decide whether it is optimal to choose above or below Player 1’s choice. In order for Player 2 to choose above Player 1, then expected value of choosing higher than $n$ must be greater than the expected value of choosing lower than $n$ , that is

\begin{aligned} \frac{1}{30} \sum_{i = n + 1}^{30} i &> \frac{1}{30} \sum_{i = 1}^{n-1} i \\ \frac{30 \cdot 31}{2} &> 2\frac{n(n-1)}{2} + (n+1) + n\\ 21.56 &> n \end{aligned}

Then, should Player 1 choose $n \leq 21$ , then Player 2 would choose $m= n+1$ and if Player 1 chose $n\geq 22$ Player 2 would choose $m=n-1$ . The expected value is clearly maximized at either $n=21$ or $n=22$ for Player 1, and a quick calculation finds that $\mathbb{E}[21] = \frac{77}{10}$ and $\mathbb{E}[22] = \frac{78}{10}$ . Then Player $1$ would choose $22$ and Player 2 would choose $21$ . Since Player 1’s expected payoff is slightly higher than that of Player $2$ , I would choose to go first and pick $n=22$ .

Problem 4. A survey is given to all passengers on a number of different planes. The survey asks each person how full their plane was. The people answer honestly. If 50% of people claim that their plane was 80% full, while the other 50% claim that their plane was 20% full, how full was the average plane? Solution. This is a relatively simple question that’s intuitively misleading. An incorrect approach to the question would be to average out the $20\%$ and the $80\%$ passengers for a total of $50\%$ fullness on plans on average. Instead note that since the total number of people that claim $80\%$ of the plane is full is equal to the total number of people who claim their plane is $20\%$ full, there must be more planes carrying people at $20\%$ capacity than at $80\%$ . In fact, there are $4$ times as many people in planes $20\%$ full as there are peope in planes $80\%$ full. Then, the average capacity of each plane is

\frac{1}{5}\cdot 80\% + \frac{4}{5} \cdot 20\% = 32\%

Note that the first approach is wrong since averaging out the percentages assumes that the total number of people flying in each type of plane is distributed uniformly across the proportions of plane capacity. You can find this question on Jane Street’s Glassdoor.

Extra Problems

Problem. Given an array of length $n$ find the largest possible sum of consecutive elements of the array. Solution. This question is one of the classic dp questions from Brian Dean’s academic site. Clearly, the max sum will be along a subarray that is completely positive, so we need only keep track of the positive elements. We can then use a sliding window technique: keep track of the maximum sum up to a certain index $j$ and decide whether to add the next element based on how big it is compared to the existing sub-array. Along the way, we can update the overall maxSum by comparing the current maxSum to the sum of the subarray being considered.

int maxContiguousSum(int arr[], int n){
    int maxSum = arr[0];
    int maxSub = arr[0];

    for(int i = 1; i < n; i++){
        maxSub = max(arr[i], maxSub + arr[i]);
        maxSum = max(maxSum, maxSub);
    }

    return maxSum;
}

This solution runs in $O(n)$ time.

Problem. We randomly sample from a uniform distribution on $[0,1]$ until the sum of all sampled values exceeds $1$ . What is the expected number of samples we take? Solution. Let $N$ be the number of samples we take and sum together to get a number greater than $1$ . We would like to find $P(N = n) \geq 0$ since $N$ is a discrete random variable. We can find this by taking $P(N \leq n) - P(N < n) = P(N \leq n) - P(N \leq n-1)$ , or equivalently $P(N > n - 1) - P(N > n)$ . Note that $P(N > n)$ means that it takes more than $n$ samples to sum up to $1$ , or equivalently that the first $n$ uniform r.v.’s that we draw sum up to less than $1$ . The probability of this occuring is simply $\frac{1}{n!}$ , as shown on Day 5 Problem 5. Then, we have

P(N= n) = \frac{1}{(n-1)!} - \frac{1}{n!} = \frac{n-1}{n!}

where $n = 1, 2, 3, \dots$ We can use the definition of expected value to solve for the expected number of samples we need to take:

\begin{aligned} \mathbb{E}[N] &= \sum_{n=1}^\infty nP(N=n) = \sum_{n=1}^\infty \frac{n(n-1)}{n!} = \sum_{n=0}^\infty \frac{(n+1)n}{(n+1)!}\\ &= \sum_{n=0}^\infty \frac{1}{n!} = e \end{aligned}

Then we expect to draw $e$ times in order for the sum to exceed $1$ .

A Working List of Probability Questions: Week 2

Sun, 28 Jun 2020 22:40:32 GMT

Market Making, Expected Value, and General Topics
June 28, 2020 | Updated July 6, 2020

Welcome to week two of my trading prep series! This week I’m going to focus most of my attention on learning and practicing market making, expected value, and other expanded topics. I’ll also post resources and links to additional problems as I go along for my documentation purposes. Thanks for tuning in!

Day 8 | Day 9 | Day 10 | Day 11 | Day 12

Update: Instead of doing $5$ problems every day consecutively, I’ll be doing problems on weekdays and spending weekends reviewing problems and topics I’ve already covered.

Day Eight: June 29

A friend linked me to a blog post by Messy Matters on confidence intervals/overconfidence that you can find here. I’ve been spending some time with market making questions so I found the excercise and results of the “study” pretty interesting. I also found this powerpoint on recuiting tips when I was looking for new problems to do.

Problem 1. Make me a market on the temperature in an hour. Solution. I was thinking through this problem after taking a walk this morning. At the time, I estimated the temperature to be about $60^\circ - 65^\circ$ (the actual temperature was $64^\circ$ ). Given it was $10\colon 35AM$ , it’s still relatively early in the day and there’s quite a bit of room for the temperature to increase to before it plateaus in the afternoon. From my experience, the temperature in a typical summer day in San Francisco usually tops off around $75^\circ$ to $85^\circ$ . This morning was relatively cool and so I’d expect that today would be one of the cooler days, toping off between $75^\circ$ to $78 ^\circ$ . Then, just assuming that the temperature increases somewhat linearly, I’d set my bid at $63^\circ$ and offer at $68^\circ$ .

After setting my initial spread, I’d adjust my interval accordingly to how the second party/market makers chooses to buy from or sell to me. To do this, I sampled from a normal distribution meaned at $65.5$ with $\sigma = 0.83$ for the price and flipped a coin to determine whether I was being bought or sold to at that price.

Bid at $66.9^\circ$ : buying from me at a value leaning so far to the right in my range indicates that I severely underestimated the temperature in an hour, as the willingness to buy higher might be an attempt to execute an unfavorable trade with me to then later sell at a higher price with an acceptable margin. However, I also know the temperature cannot change too much so I’d adjust my spread to account for this new information by shifting my position to $65.5-71$ . I widened my range a bit since I’m pretty skeptical that my original range reflected the actual temperature well and I want to make sure to cover all my bases.
Offer at $68.4^\circ$ : A willingness to sell indicates that, taking into account my decision-making, the second party believes that $68.4^\circ$ optimizes both margins and chance of trade execution. I wouldn’t expect the exact temperature to be $68.4^\circ$ since not all buyers with information on the weather would want to buy the exact of the instrument as it’s harder to later sell for profit. I’d adjust my spread down to $66.5 - 69.5$ .

Unfortunately, this simulation is flawed as I didn’t really have a second party with better information on the specific product being exchanged. I tried to adjust for that by using the sampler and randomizing but since the real value of the temperature at $11\colon35$ wasn’t known to either me or my fictitious party, I had to operate under the assumption that my initial spread wasn’t completely wrong, which could have been the case. Next time, I’ll probably try this with a partner. The actual temperature at $11:35$ was about $66^\circ$ so you can see how the spread deviated from that without this information, but this exercise still helps me practice adjusting my spread to new information.

Problem 2. Smith is in jail and had $3$ dollars and can get out on bail if he has $8$ dollars. A guard agrees to make a series of bets with him, with Smith’s probability of winning $p = 0.4$ and probability of losting $p = 0.6$ . Find the probability that he can leave on bail before losing all his money if he bets $1$ every time versus if he bets as much as possible each round to get to $8$ dollars but nothing more than that. Which is the optimal strategy? Solution. This problem is from Aldous’s STAT 150 class and is a spinoff of the gambler’s ruin problem. If Smith bets one dollar at a time, the Markov chain that describes his money at each state $i$ follows:

Let $p_i$ , $i\in[0,8]$ be the probability that Smith’s money reaches $8$ before $0$ when starting from state $i$ . Then we have

\begin{aligned} &p_i = 0.4p_{i+1} + 0.6p_{i-1} \text{ for } i \in [1,7]\\ &p_0 = 0\\ &p_8 = 1\\ \end{aligned}

You can solve this extremely convoluted system of linear equations to find that $p_3 = 0.0964$ . Alternately, the transition matrix for this Markov chain is:

P = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0.6 & 0 & 0.4 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0.6 & 0 & 0.4 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0.6 & 0 & 0.4 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0.6 & 0 & 0.4 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0.6 & 0 & 0.4 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0.6 & 0 & 0.4 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0.6 & 0 & 0.4\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}

Taking $P^n$ with $n$ sufficiently large gives us the steady state of this Markov chain, and $P^n_{4,9}$ gives us $p_3$ as well!

For the case that Smith bets as much as possible each round as needed to reach $8$ dollars, we have the Markov chain:

This in turn gives us the system of equations

\begin{aligned} &p_3 = 0.4p_6 + 0.6p_0 = 0.4p_6\\ &p_4 = 0.4p_8 + 0.6p_0 = 0.4\\ &p_6 = 0.4p_8 + 0.6p_6 = 0.4 + 0.6p_4 \end{aligned}

Solving for this gives us $p_3 = 0.256$ . This is considerably higher than betting one dollar each time, so Smith should choose this strategy when betting.

Problem 3. What is the expected number of tosses for a us to toss a $HTH$ ? Solution. This is another Markov chain problem. Note that we start all the way over whenever we roll a tails before our first heads so let’s call this state $0$ . We can also account for all the other possible states: $1 = H$ , $2 = HT$ , and $3 = HTH$ . Assuming a fair coin we have the following Markov chain:

Let’s denote $\mathbb{E}[i]$ as the expected number of turns to get $HTH$ from each state $i$ . Then we have the following system of linear equations:

\begin{aligned} &\mathbb{E}[0] = 1 + 0.5\mathbb{E}[0] + 0.5\mathbb{E}[0]\\ &\mathbb{E}[1] = 1 + 0.5\mathbb{E}[1] + 0.5\mathbb{E}[2]\\ &\mathbb{E}[2] = 1 + \mathbb{E}[3] + 0.5\mathbb{E}[0]\\ &\mathbb{E}[3] = 0 \end{aligned}

Solving these gives $\mathbb{E}[0] = 10$ so we expect it to take 10 turns to flip our first $HTH$ .

Problem 4. Players $A$ and $B$ play a marble game. Each player has a red and a blue marble in a bag and draws one marble from their bag each round. If both marbles are blue $A$ wins $\$3$ and if both are red $A$ wins $\$1$ . If the two colors are different, $B$ wins $\$2$ . If you can play this game 100 times, who would you play as?

Solution. Since there is an equal chance of picking marbles with different colors as picking two marbles that are the same color, the expected value of playing player $A$ versus player $B$ is the same (EV $= \$1$ ). Then, minimizing risk, we can calculate the variance of the payoff for each player:

\begin{aligned} \sigma^2_A &= (1-1)^2 \cdot \frac{1}{4} + (3-1)^2 \cdot \frac{1}{4} + (0-1)^2 \cdot \frac{1}{2} = \frac{3}{2}\\ \sigma^2_B &= (0-1)^2 \cdot \frac{1}{2} + (2-1)^2 \cdot \frac{1}{2} = 1 \end{aligned}

$A$ ’s payoff has larger variance and he holds more risk so it’s advantageous to play as player $B$ when risk-averse. However, since $n$ is relatively large, the actual payoff approaches the expected payoff of playing, minimizing the risk of playing as player $A$ .

Problem 5. You are trapped in a dark cave with three indistinguishable exits on the walls. One of the exits takes you 3 hours to travel and takes you outside. One of the other exits takes 1 hour to travel and the other takes 2 hours, but both drop you back in the original cave through the ceiling, which is unreachable from the floor of the cave. You have no way of marking which exits you have attempted. What is the expected time it takes for you to get outside? Solution. This is a simple EV problem. Since the doors are indistinguishable and we can’t makr our escape attempts, the probability of selecting each of the exists is identical and independent from the others. Then let $X$ be a random variable measuring the time it takes to make it out of the cave. There is $\frac{1}{3}$ chance we make it outside in time $3$ , $\frac{1}{3}$ chance we waste an hour and end up back in the cave, and $\frac{1}{3}$ chance we waste two hours and end up back in the cave. Then,

\mathbb{E}[X] = \frac{1}{3} \cdot 3 + \frac{1}{3}(1 + \mathbb{E}[X]) + \frac{1}{3}(2 + \mathbb{E}[X])

Solving this equation gives $\mathbb{E}[X] = 6$ so we expect to spent $6$ hours attempting to escape before we’re finally outside.

Day Nine: June 30

For today’s problems I took a step back and reviewed some general problems and conditional probability.

Problem 1. A deck of cards has 3 Red and 3 Blue cards. At each stage, a card is selected at random. If it is Red, it is removed from the deck. If it is Blue then the card is not removed and we move to the next stage. Find the average number of steps till the process ends. Solution. Note that the proces ends when there are no red cards left. To get started, I’ll draw the Markov chain for this problem:

Now let’s index the states: State $0 = R^3B^3$ , State $1 = R^2B^3$ , State $2 = RB^3$ , State $3 = B^3$ . Let $\mathbb{E}[i]$ denote the expected time until the process ends. Then we have the following equations:

\begin{aligned} \mathbb{E}[0] &= \tfrac{1}{2}(1+ \mathbb{E}[0]) + \tfrac{1}{2}(1+ \mathbb{E}[1])\\ \mathbb{E}[1] &= \tfrac{2}{5}(1 + \mathbb{E}[2]) + \tfrac{3}{5}(1 + \mathbb{E}[1])\\ \mathbb{E}[2] & = \tfrac{1}{4} \cdot 1 + \frac{3}{4}(1 + \mathbb{E}[2]) \end{aligned}

Solving this system gives $\mathbb{E}[0] = 8.5$ .

Problem 2. A process moves on the integers 1, 2, 3, 4, and 5. It starts at 1 and, on each successive step, moves to an integer greater than its present position, moving with equal probability to each of the remaining larger integers. State five is an absorbing state. Find the expected number of steps to reach state five. Solution. The Markov chain for this problem is

This gives us the transition matrix

\begin{pmatrix} 0 & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4}\\ 0 & 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3}\\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2}\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}

We’re interested in finding the expected number of steps to reach state five from state one so let $\mathbb{E}[i]$ denote the expected number of steps to reach step $5$ from step $i$ . Then we have the equations

\begin{aligned} \mathbb{E}[1] &= 1 + \tfrac{1}{4}\mathbb{E}[2] + \tfrac{1}{4}\mathbb{E}[3] + \tfrac{1}{4}\mathbb{E}[3] + \tfrac{1}{4}\mathbb{E}[5]\\ \mathbb{E}[2] &= 1 + \tfrac{1}{3}\mathbb{E}[3] + \tfrac{1}{3}\mathbb{E}[4] + \tfrac{1}{3}\mathbb{E}[5]\\ \mathbb{E}[3] &= 1 + \tfrac{1}{2}\mathbb{E}[4] + \tfrac{1}{2}\mathbb{E}[5]\\ \mathbb{E}[4] &= 1\\ \mathbb{E}[5] &= 0 \end{aligned}

Solving this system of linear equations gives us the set of expected steps from each state $\{\mathbb{E}[1], \mathbb{E}[2], \mathbb{E}[3], \mathbb{E}[4], \mathbb{E}[5]\} = \{\frac{25}{12}, \frac{11}{6},\tfrac{3}{2}, 1, 0\}$ so we expect about $2.0833$ steps to reach $5$ .

Problem 3. An unfair coin has probability $p$ of tossing heads and $q = 1-p$ of tossing tails. How do you make this coin fair? Solution. Note that $p_H = p$ and $p_T = p$ . The only way for us to make the coin fair is to somehow combine the $H,T$ outcomes as rolling consecutive heads vs tails has different probability of success. Note that $HT$ is rolled with a probability of $pq$ and $TH$ is rolled with a probability of $qp$ (note these are equal). Also $HH$ has $p = p^2$ and $TT$ has $p = q^2$ . We can modify the typical rules of coin flipping in in the following way to make it fair: Let heads be tossing an $HT$ and tails be tossing $TH$ , which both have equal probability and is fair. Then, to account for the remaining two possible outcomes, we just start over when we get $HH$ or $TT$ , which gives us an automatic reset when we toss two of the same face. Technically, $HHT$ would be rolling a $HT$ , but note that $P(HHT) = p^2q \neq q^2p = P(TTH)$ so we can’t simply take the first occurance of $HT$ or $TH$ as a fair outcome is not guaranteed.

Problem 4. This is a variation of the gambler’s ruin problem. You have $9$ and want to reach $10$ . Your probability of success is $p = 0.4$ and failure is $q=0.6$ . You may only place bets in multiples of $10p$ . How would you maximize the probability of reaching $10$ ? Solution. I would bet $\$4$ each time I play. Note that from $\$9$ we can bet either $4$ or $8$ . We have $p= 0.4$ of reaching $\$10$ from each of these bets, but $0.6$ chance of reaching $\$1$ if we bet $8$ dollars and $q$ chance of reaching $\$5$ if we bet $4$ dollars. Once we reach $1$ , we can no longer bet as all of our bets must be a multiple of $4$ . Then, if we bet $\$8$ , we have a total of probability $0.4$ to reach our goal. If we bet $\$4$ , then we have probability $0.4$ of getting to $\$10$ the first round. If this doesnt happen and we lose the first bet instead, then we’re at $\$5$ and we have the opportunity to go down to $\$1$ (probability $0.4$ ) in which we’re done and can’t bet anymore or go back up of $\$9$ and play over our original scenario over again. Clearly, betting $\$4$ per round is optimal as we have probability $0.4$ of winning the first round (matching the probability of winning when betting $\$8$ ) and there also is some additional probability $p_0 >0$ that we can reach $\$10$ in later rounds.

Problem 5. Two players, $A$ and $B$ , toss a fair coin in that order. The sequence of heads and tails is recorded. If there is a $HT$ sequence in the rolls, then whoever rolled the tails wins and the game ends. What is the probability that $A$ wins? Solution. Let $P(A)$ and $P(B)$ denote the probabilities that $A$ and $B$ win respectively. Note that $P(B) = 1 - P(A)$ . This next part is pretty cool—we can condition $P(A)$ on $A$ ’s first coin toss, with has probability $0.5$ to roll $H$ and probability $0.5$ to roll $T$ . Then we have $P(A) = \tfrac{1}{2}P(A \, | \, H) + \frac{1}{2}P(A \, | \, T)$ . If $A$ tosses heads, then $B$ essentially becomes the first to toss and so $P(A , |, T) = P(B) = 1-P(A)$ . If $A$ tosses heads however, we can then condition the probability further on $B$ ’s first toss. $B$ has probability $0.5$ of rollng $T$ , in which $A$ loses, and probability $0.5$ of tossing $H$ , which essentially brings us to the scenario that $B$ is the first person to toss an $H$ so $A$ would have probability $1-P(A \, | \, H)$ of winning. Then we have

P(A \, | \,H) = \tfrac{1}{2} \cdot 0 + \tfrac{1}{2}(1 - P(A\, | \, H)) \to P(A \, | \,H)= \tfrac{1}{3}

We can then plug this back into the original expresion to find that $P(A) = \frac{4}{9}$ .

Day Ten: July 1

Come back later for the rest of today’s problems! My friend recommended me take a look at Fifty Challenging Problems in Probability, and I’m taking a couple of todays problems from here.

Problem 1. You play a game with an urn that contains $75$ clue balls $25$ red balls and $1$ yellow ball. For each red ball you draw you gain a dollar and if you draw the yellow ball you lose everything and the game ends. What is your strategy for this game given that you can decide to stop or redraw after each turn? Solution. Note the payoff for each blue ball draw is $0$ and the payoff for the yellow ball is $-r$ where $r$ is the number of red ball already drawn. If we let $b$ be the number of blue balls already drawn, we can derive the marginal expected value (not sure if this is a term):

\begin{aligned} \mathbb{E}[X] &= \frac{75 - b}{(75 - b) + (25 - r) + 1}\cdot 0 + \frac{25 - r}{(75 - b) + (25 - r) + 1} \cdot 1 \\ &+ \frac{1}{(75 - b) + (25 - r) + 1}\cdot (-r) = \frac{25- 2r}{(75 - b) + (25 - r) + 1} \end{aligned}

Our marginal $EV$ should be greater than $0$ for us to keep playing. Then, setting the numerator to $0$ and solving gives us $r= \frac{25}{2}$ . This is the number of balls that have already been drawn, so after drawing $12$ balls our $EV$ is still positive so we’ll draw a $13$ th one. Then we draw $13$ balls and call it a day (of course this will sometimes be cut short if we draw a yellow ball before then).

Problem 2. A drawer contains red socks and and black socks. When two socks are chose at random, the probability that both are red is $\frac{1}{2}$ . How small can the number of socks in the drawer be? Solution. Let $r$ and $b$ denote the number of red and black socks respecively. Since we are selecting socks without replacement, we have

\frac{r}{b+r} \cdot \frac{r-1}{b+r-1} = \frac{1}{2}

Then,

2r(r-1) = (b+r)(b+r -1)\\ 2r^2 = b^2 + 2br + r^2 - b - r\\ 2r^2 - 2r - b^2 - 2br - r^2 + b + r = 0\\ r^2 -(1= 2b)r - (b^2 - b) = 0

Note that $b\neq 0$ since $P(r \, | \, b = 0) = 1$ . So if we choose $b= 1$ and plug it into the quadratic equation to solve for $r$ , in which case we get $r = 3$ . Then, the minimum number of socks in the drawer is $\$4$ .

Problem 3. Mr. Brown always bets a dollar on the number $13$ at roulette agains the advice of his friend. To help cure Mr. Brown of playing roulette , his friend bets Brown $\$20$ at even money that Mr. Brown will be behind at the end of $36$ plays. How is the cure working? Mr. Brown is piad $\$35$ of his bet amount in roulette when his number comes up and roulette has $38$ numbers to bet on. Solution. In order to go even in $36$ rounds, Mr. Brown needs only win one turn. So he’ll be behind if he loses all $36$ turns, which has probability $(\frac{37}{38})^36 = 0.383$ , which is the probability that he loses his bet with his friend. Note also that his ev after 36 turns is

36(\tfrac{1}{38} \cdot 35 - \tfrac{37}{38} \cdot 1) \approx -1.89

Then Mr. Brown’s EV of betting with his friend is

20(1- 0.383) - 20(0.383) - 1.89 = 2.79

So this cure doesn’t seem to be working too well. Mr. Brown’s EV of eahc cycle of $36$ turns is $\$2.79$ whenever he makes the $\$20$ bet with his friend.

Problem 4. We throw three dice one by one. What is the probability that we roll three nunbers in strictly increasing order? Solution. Note that in order for the three numbers to be in strictly increasing order, they have to be unique. Then $P(\text{unique and strictly increasing}) = P(\text{unique})$ . We have probability $\frac{6}{6} \cdot \frac{5}{6} \cdot \frac{4}{6} = \frac{5}{9}$ . When conditioned on us rolling three numbers, the probability of rolling strictly increasing numbers is simply based on how we want them arranged. There is only one correct arragement, but there are $6$ ways to choose how to arrange the three numbers, so we have $P(\text{stringtly increasing} \, | \, \text{unique}) = \frac{1}{6}$ . Then we have

\begin{aligned} P(\text{strictly increasing}) &= P(\text{strictly increasing and unique})\\ &= P(\text{unique}) \cdot P(\text{stringtly increasing} \, | \, \text{unique})\\ &= \frac{5}{9} \cdot \frac{1}{6} = \frac{5}{54} \end{aligned}

Then we have probability $0.093$ of rolling such an outcome.

Problem 5. The game of craps, played with two dice, is one of America’s fastest and most popular gambling games. These are the rules: we throw two dice. If we roll $7$ or $11$ on the first throw, we win immediately and if we roll $2, 3$ or $12$ we lose immediatly. And other throw is called a “point”. If the first toss is a point, then we keep rolling until we roll our point, or roll a $7$ and lose. Wht is the probability of winning? I thought this was a Markov chain problem at first but a closer look made me realize it was just a bit of conditional probability. Note how I got the numbers: there are $36$ outcomes for two dice. There is one way for me to roll a $2$ , on way to roll a $12$ , abd $2$ ways to roll a $3$ . Then the probabiity of losing the first round is $\frac{1 + 1+ 2}{36} = \frac{1}{9}$ . Additionally, there are $6$ ways to roll a $7$ and $2$ ways to roll an $11$ so our probability of winning the first round is $\frac{2}{9}$ and there is a $\frac{2}{3}$ chance of rolling a point. Now, we just need to find the probability of winning conditioned on rolling a point and adding that to the probability of winning the first round to arrive at the total probability of winning. Note that there is probability $\frac{1}{12}$ that we roll $4$ or $10$ , $\frac{1}{9}$ that we roll a $5$ or a $9$ , and $\frac{5}{36}$ of rolling a $6$ or an $8$ .

Note that in the event that we roll a point in our first round, we need only consider the probabilities of rerolling the point (to win) or a $7$ (to lose) as rolling a nonpoint/seven will bring us back to our original scenario of wanting to roll the point. Then, there is probability $\frac{3}{3 + 6}$ of rolling a $4$ or $10$ , $\frac{4}{6}$ of rolling of a $5$ or $9$ , $\frac{5}{5+6}$ of rolling a $6$ or an $8$ (these can all be calculated by the number of ways to roll the point vs. the $7$ ). Then, the probability of winning conditional on rolling a point the first turn is

\begin{aligned} P(W \, | \, \text{point}) &= \tfrac{1}{12} \cdot \tfrac{3}{9} + \tfrac{1}{9} \cdot \tfrac{4}{10} + \tfrac{5}{36} \cdot \tfrac{5}{11} + \tfrac{5}{36} \cdot \tfrac{5}{11} + \tfrac{1}{9} \cdot \tfrac{4}{10} + \tfrac{1}{12} \cdot \tfrac{3}{9}\\ &\approx 0.27071 \end{aligned}

Adding this to $\frac{2}{9}$ , which is our probability of winning round $1$ , we have a total probability of winning $p = 0.493$ , which is relatively fair for a casino game.

Day Eleven: July 2

Dynamic programming is also relatively fair game at some trading firms so I’ve included some dp questions for future problems.

Problem 1. A basketball player is taking $100$ free throws. She scores one point if the ball passes through the hoop and zero point if she misses. She has scored on her first throw and missed on her second. For each of the following throw the probability of her scoring is the fraction of throws she has made so far. For example, if she has scored $23$ points after the $40^\text{th}$ throw, the probability that she will score in the $41^\text{th}$ throw is $23/40$ . After $100$ throws (including the first and the second), what is the probability that she scores exactly $50$ baskets? Solution. This is another problem from A Practical Guide. I approached this problem with the mindset of Markov chains at first but then I realized that I could build it up iductively with conditional probability. Let’s define $(n,k)$ to be the event that the player scores $k$ baskets after $n$ shots and $p_{n,k} = P((n,k))$ . Since we make the first shot and miss the second, the third shot has success probability $\frac{1}{2}$ ( $p_{3,1} = \frac{1}{2} = p_{3,2})$ . For $n=4$ , we can condition it on $(3,1)$ and $(3,2)$ having taken place. Then,

$p_{4,1} = P((4,1)\, | \, P(3,1)) + P((4,1)\, | \, P(3,2)) = \frac{2}{3}\cdot \frac{1}{2} + 0 \cdot \frac{1}{2} = \frac{1}{3}$
$p_{4,2} = P((4,2)\, | \, P(3,1)) + P((4,2)\, | \, P(3,2)) = \frac{1}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{3}$
$p_{4,3} = P((4,3)\, | \, P(3,1)) + P((4,3)\, | \, P(3,2)) = 0 \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$

The pattern seems to be emerging that $p_{n,k} = \frac{1}{n-1}$ for $k \in [1, n]$ regardless of what $k$ is (obviously $k < n$ since we have missed the second shot), which we can prove inductively.
Base Case: $p_3$ is our base case
Inductive Step: Suppose that $P_{n,k} = \frac{1}{n}$ holds for all $k \in [1, n]$ . Then for $n+1$ :

\begin{aligned} p_{n+1, k} &= P(\text{score} \, | \, (n, k -1))p_{n,k-1} + P(\text{miss} \, | \, (n, k))p_{n,k}\\ &= \frac{k-1}{n} \cdot \frac{1}{n-1} + \left( 1-\frac{k}{n} \right) \cdot \frac{1}{n-1} = \frac{1}{n} \end{aligned}

Then we have probability $\frac{1}{99}$ of making exactly $50$ shots out of $100$ .

Problem 2. Dynamic Programming: Given two arrays of numbers, find the length of the largest subsequence present in both arrays. For example the length of largest subsequence present in $[1,2,3,4]$ and $[2,3,5,7]$ is $2$ . Solution. Our goal is to build a method that returns the largest common subsequence (lcs). Let’s look at the brute force solution: we can approach the problem by finding each of the substrings in one of the arrays and seeing if it is present in the second array. There are $\binom{n}{k}$ combinations for substrings of length $k$ given an array of length $n$ . Then we have $\binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n} = 2^n$ operations for our brute force solution, indicating $O(2^n)$ time. Now let’s approach it with dynamic programming: assuming our first array has length $m$ and our second array has length $n$ , we can build an $m\times n$ matrix in which the $(i,j)$ entry holds the size of largest common multiple of $\text{array1}[0, 1, \dots i]$ and $\text{array2}[0, 1, \dots j]$ .Then we have the following solution in C++:

int largestCommonSubsequence(int *array1, int *array2){
	int m = array1.size();
	int n = array2.size();
	int lcs[m+1][n+1]; // our "matrix"

	// constructs m by n matrix of lcs's using dp
	for(int i = 0; i < m; i++){
		for(int j = 0; j < n; j++){
			if(i == 0 || j == 0){
				lcs[i][j] = 0;
			} else if (array1[i-1]= array2[j-1]){
				lcs[i][j] = lcs[i - 1][j - 1] + 1;
			} else{
				lcs[i][j] = max(lcs[i-1][j], lcs[i][j-1]);
			}
		}
		return lcs[m][n];
	}
}

Walking through the for loop: when either $i$ or $j$ is $0$ , at least one of the two subarrays we’re looking at are empty. Then, the lcs is just $0$ . Otherwise, if the previous number we looked at in the first array is the same as the previous number we looked at in the second array, then we simply need to add one to the value of our previous $(i,j)$ entry since we’ve increased the common subseqeunce by one number. Finally, if neither of these are the case, the lcs of the current $i,j$ matrix is the same as the max of the previous $i-1 \times j$ or $i \times j-1$ matrix. This solution has an $O(mn)$ time, where $m$ and $n$ are the lengths of the two arrays.

Problem 3. $500$ ants are randomly put on a $1$ -foot string (independent uniform distribution for each ant between $0$ and $1$ ). Each ant randomly moves toward one end of the string (equal probability to the left or right) at constant speed of $1$ foot/minute until it falls off at one end of the string. Also assume that the size of the ant is infinitely small. When two ants collide head-on, they both immediately change directions and keep on moving at $1$ foot/min. What is the expected time for all ants to fall off the string? Solution. This problem is relatively simple when you start thinking about it. The most troublesome part seems to be the collision as each collision complicates the process and each any can collide with another at any time. Interesting collisions don’t seem to be relevant to solving this question. When an ant moving to the right collides with an ant moving to the left, the result is the two ants switching directions, but there is still one ant moving left and one ant moving right (at the same speed as well). So its’ as if collisions don’t really take place. Then, the expected time is just the max time for an ant to move across the string given a uniform distribution. This is identical to asking what the expected value of the maximum of $500$ i.i.d random variables with uniform distribution on $[0,1]$ , aka an order statistic problem. Lets find the CDF of $X_{500}$ , the maximum of the independently distributed ants.

\begin{aligned} P(X_{500} \leq x) &= P(X_i \leq x \, | \, i \in [1,500]) \\ &= P(X_1 \leq x)P(X_2 \leq x) \cdots P(X_{500} \leq x) = x^{500} \end{aligned}

Then, we can find the PDF by differentiating to get $p_X(x)= 500x^{499}$ . The expected value of the max of the $500$ i.i.d random variables is

\int_{-\infty}^\infty xp_X(x) \, dx = \int^1_0 x\cdot 500x^{499} \, dx = \frac{500}{501}

There is an alternate solution to this problem on the Berkeley EECS 126 site, where it’s listed as a test review problem.

Problem 4. In a common carnival game a player tosses a penny from a distance of about five feet onto the surface of a table ruled in $1$ -inch squares. If the penny (diameter $\frac{3}{4}$ inch) falls entirely inside a square, the player received five cents but does not get his penny back and he loses his penny otherwise. If the penny lands on the table, what is the expected payoff of the game. Solution. We’re given that that the penny lands on the table so we know that the penny will fall on a square. Then we only need to find the regions where our penny can land on for it to be fully encapsulated. I’ve drawn two diagrams to model two different appraoches to this problem. We cant first visualize the penny landing on any four-square unit since any four adjacent squares are guaranteed to fully encompass a penny. Then, the part of the penny that associates with each corner must be squeezed into one of the four $\frac{1}{4} \cdot \frac{1}{4}$ squares, which has a total area of $\frac{1}{4}$ and probability $\frac{1}{16}$ of being landed on.

Regarding the second visualization, this is much more accurate to the possible outcomes of the penny toss since we cannot have an edge of the penny touch a specific corner of the square. Then, we can visualize the locations that the center of the penny must reside in. The radius of the penny is $\frac{3}{8}$ so the center of the penny must reside at least $\frac{3}{8}$ inch from each side of the square, giving a smaller square of potential center point wuth area $\frac{1}{16}$ . Then, the expected value of the game is $-1 \cdot \frac{15}{16} + 4 \cdot \frac{1}{16} = \frac{-11}{16}$ .

Problem 5. Eight eligible men and seven women happen randomly to have purchased single seats in a $15$ seat row of a theater. On average, how many pairds of adjacent seats, are ticketed for marriageable couples? Solution. The book that this problem is from is a bit old so the wording is a bit old-fashioned. Essentially we’re looking to find the average number of potential male-female couples who sit next to one another. The best case and worst case scenario follow

\text{MWMWMWMWMWMWMWM}\\ \text{MMMMMMMMWWWWWWW}\\

In order to build a marriageable couple we need to have $\text{MW}$ or $\text{WM}$ , or unlike adjacent seatings. If they are unlike, then we have $1$ marriageable couple and if not then we have $0$ . The probability of a marriageable couple in the first two seats is $2 \cdot\frac{8}{15} \cdot \frac{7}{14}$ . Note that $\frac{8}{15}$ is also the expected number of marriageable couples in the first two seats. Our sam calculation can applies to any adjacent pair so we can get the average number of marriageable adjacent pairs by taking the number of adjacent pairs and multiplying by the expected value of each adjacent pair being marriageable: $14 \cdot \frac{8}{15} = \frac{112}{15} \approx 7.47$ . Then the expected number of marriageable couples is just under half of the people present.

Moving on to a more general case, when we can $b$ elements of one type and $m$ of another randomly arranged in a line, the expected number of unlike adjacent elements is

(m+b+1) \left[2\cdot \frac{bm}{(m+b)(m+b-1)}\right]

which can be shown easily by going through the same process we used for our specific case.

Day Twelve: July 3

Come back later for today’s problems!

Problem 1. Two players bet on the number of rolls of the total of two standard six faced dice. Player $A$ bets tha a sum of $12$ will occur first and player $B$ bets that two consecutive $7$ ’s will be rolled before that. What is the probability that $A$ will win? Solution. This question can be solved either through conditional probability or a Markov chain approach. I’ve opted for a Markov chain since it’s much more straightforward and easily illustrated. Below is the Markov diagram for this problem. I’ve chosen to denote $\empty$ the event of rolling for the first time. Since every time we roll something other than a $7$ or a $12$ we essentially reset, each state gives us a certain probability of returning to our original state. It’s worth noting that after rolling the first $7$ there is still a nontrivial probability for $B$ to win as $12$ can still be rolled in the next round.

Then we have the following system of equations to solve:

\begin{aligned} p_\empty &= \tfrac{1}{36} + \tfrac{29}{36}p_\empty + \tfrac{1}{6}p_7\\ p_7 &= \tfrac{1}{36} + \tfrac{29}{36}p_\empty \end{aligned}

Noting that $p_{7,7} = 0$ and $p_{12} = 1$ , solving gives us that $p_\empty = \frac{7}{13}$ . Then, player $A$ is slightly favored to win over player $B$ in their bet.

Problem 2. There are 51 ants on a square with side length of 1. If you have a glass with a radius of 1/7, can you put your glass at a position on the square to guarantee that the glass encompasses at least 3 ants? Solution. We can do this by creating $25$ sub-squares and using the pidgeonhole principle to conclude that ones of these sub-squares must have at least $3$ ants. In creating these $25$ sub-squares, we want them to cover the entirety of the original $1\times 1$ square so each of these smaller squares will have a $\frac{1}{5}$ side length. Note that a sub-square with side length $\frac{1}{5}$ has diagonal length $\sqrt{2}\frac{1}{5} \approx 0.282$ . Meanwhile, the radius $\frac{1}{7}$ glass cup has a diameter of $\frac{2}{7} \approx 0.285$ , so any sub-square can be inscribed within the glass. Then, we simply need to choose the sub-square with the most ants and place our glass carefully to cover at least three ants.

Problem 3. The probability you’ll see a falling star in the sky over the course of one hour is 0.44. What’s the probability you’ll see one over half an hour? Solution. This is a past Citadel interview question that I’ve seen variants of at other companies. We can assume that each half hour is independent from the other regarding falling star appearances. Then, the probability of not seeing a shooting star in either half-hour intervals is $0.56$ . Let $p$ be the probability that we see a shooting star in a thirty minute interval. Then we have

(1-p)(1-p) = 0.56

Solving gives us $p \approx 0.25$ . This question is meant to be solved mentally so I didn’t include and pen or paper for it. For taking the square root of $0.56$ , I did a linear approximation based off of $\sqrt{0.49}$ and $\sqrt{0.64}$ .

Problem 4. You are given two ropes that when lit burn in one hour. Which one of the following time periods CANNOT be measured with your ropes? a) 50 min b) 30 min c) 25 min d) 35 min. Solution. The only time period I can’t think of a solution to is $50$ min. The key to this problem is realizing that we can burn the rope wherever we want. It’s easy to finding $30$ minutes by burning a single rope on both ends. We can burn $35$ by first measuring $30$ minutes. Then, we measure $5$ additional minutes by burning the second rope on both ends and $5$ additional times in the middle. This process will create two ropes, all of which burn on both of their ends. If any of these ropes burn out fully, all we need to do is light one of the other $5$ ropes in the middle. Keep doing this proccess and you’ll have an additional $5$ minutes measured. To measure $25$ minutes, we can burn one of the ropes on both ends and once in the middle. If one of the two ropes formed by the burning burns out, then light the other sub-rope in the middle, making sure the rope is still burning in $4$ places. This ensures that the first rope will finish burning in $10$ minutes. Then, we need only light the second rope on both ends and two times in the middle. This ensures the second rope burns at $6$ places, and so we’ll measure $10$ additional minutes that way.

Problem 5. You have three cards, each labeled $n$ , $n+1$ , $n+2$ , and you don’t know $n$ . All cards start facing down so you can’t see them. You flip one card. If you choose to “stay”, you get that card’s value. If you don’t “stay”, then you flip another card. Again, choose to “stay” (and keep the 2nd card’s value) or flip the final card and keep the final card’s value. Design the optimal strategy to maximize the value of the card that you choose and find the expectation of that value. Solution. Note that your expected value will be at least $n$ . Making our decision on draw $1$ ensures we have no information and making our decision of draw $3$ presents us with no choice. Then it makes sense for us to make our decision on round $2$ . Since we have the ability to draw once again after round $2$ , our decision is simply to take what we drew or draw again. I will refer to each draw $i$ as $d_i$ . Then, if
$d_2 - d_1 = -2:$ we have the sequence $n+2, n, n+1$ so we choose to draw again for ev $n+1$
$d_2 - d_1 = 2:$ we have the sequence $n, n+2, n+1$ so we stay for ev $n+2$
$d_2 - d_1 = 1:$ we either go up $1$ ( $n,n+1, n+2$ ) or down $2$ ( $n+1, n+2, n$ ) so we choose to stay for ev $n + 1.5$ $d_2 - d_1 = -1:$ we either go up $2$ ( $n+1, n, n+2$ ) or down $1$ ( $n+2, n+1, n$ ) so we choose to draw again for ev $n+1$ Using this strategy, we combine our total expected value:

\mathbb{E}[X] = \tfrac{3}{6}(n+1) + \tfrac{1}{6}(n+2) + \tfrac{2}{6}(n+1.5) = n+ \tfrac{4}{4}

My optimal strategy is to draw the first two cards and determine whether to draw again with the thought process above with expected value $n + \frac{4}{3}$ . Putting it more simply, if the second draw is higher than the first, then stay, and if not, then draw again.

Week Two Review:

This week I focused on working through more of A Practical Guide, and starting to pull general probability question from resources. i’ve done a bit of market-making practice behind the scenes and am still figuring out the best way of going about that. I didn’t get to review as many topics as I had hoped to and next week I hope to work on some more programming/staticstics questions as recruiting season begins.

Here are this week’s list of resources:

A Practical Guide to Quantitative Finance Interviews (Ch 2, 4)
Fifty Challenging Problems in Probability
eFinancialCareers Past Interview Questions
Confidence Intervals
MIT Quant Finance Presentation
Citadel Glassdoor Questions

A Working List of Probability Questions: Week 1

Sun, 21 Jun 2020 22:40:32 GMT

In Preparation For Quantitative Trading Interviews
June 21, 2020 | Updated June 28, 2020

This series is mostly for my own self-documentation purposes as I prepare for recruiting at the end of summer. I’ll be updating this page with 5 new problems/brain teasers and solutions every day to keep myself accountable. These problems will come from a variety of sources, including A Practical Guide to Quantitative Finance Interviews, Math Stack Exchange, and UC Berkeley’s EECS 126 course notes. You can skip to my week one review here or navigate to a specific day using the legend below. Thanks for tuning in!

Day 1 | Day 2 | Day 3 | Day 4 | Day 5 | Day 6 | Day 7

Day One: June 21

Today’s problems are relatively simple as I spent a while figuring out the LaTeX formatting on websites.

Problem 1. What is the expected number of tosses to get three consecutive heads when flipping a fair coin? Solution. Let $\mathbb{E}[3]$ denote the expected number of rolls for $3$ consecutive heads. If we start rolling and roll a tail first, which has probability $\frac{1}{2}$ , then the expected number of tosses will be $\mathbb{E}[3] + 1$ . If we roll a head then a tail, which has probability $\frac{1}{4}$ , then the expected number will be $\mathbb{E}[3] + 2$ . If we roll 2 heads then a tail with probability $\frac{1}{8}$ then expected value is $\mathbb{E}[3] + 3$ . Finally if we roll $3$ heads consecutively, which has probability $\frac{1}{8}$ , then $\mathbb{E}[3] = 3$ . Combining these together, we have

\mathbb{E}[3] = \frac{1}{2}(\mathbb{E}[3] + 1) + \frac{1}{4}(\mathbb{E}[3] + 2) + \frac{1}{8}(\mathbb{E}[3] + 3) + \frac{1}{8}(3)

Solving this equation gives $\mathbb{E}[3] = 14$ . We can extend this to find the expected number of rolls for $n$ consecutive heads to get $\mathbb{E}[n]$ :

\mathbb{E}[n] = \frac{1}{2}(\mathbb{E}[n] + 1) + \frac{1}{4}(\mathbb{E}[n] + 2) + \cdots + \frac{1}{2^n}(\mathbb{E}[n] + n) + \frac{1}{2^n}(n)

Solving for this in general gives us the formula $\mathbb{E}[n] = 2(2^n - 1)$ . A number of other solutions of this type of problem can be found here.

Problem 2. Suppose there are five pirates with $100$ pieces of gold who are voting on splitting up the gold. The most senior pirate will propose a way to distribute the gold and requires $50\%$ of the votes in order to pass his proposal. If he receives less than $50\%$ of the votes then he will be killed and the process will begin with the next most senior pirate. How will the gold coins be divided? Solution. Let’s label each pirate in ascending order based on seniority, so pirate $5$ will be most senior and $1$ will be least senior. Building this up from the two pirate case: Pirate 2 can safely propose he take all $100$ gold pieces as he constitutes $50\%$ of the votes. Then Pirate 1 receives nothing.

Moving on to the three pirate case—Pirate 3 knows that if his plan is voted down, then Pirate 2 can take all $100$ coins and Pirate 1 will receive nothing. All he needs to do is give Pirate 1 one coin to secure his vote. Adding a fourth pirate, Pirate 4 knows that if his plan is voted down, then Pirate 2 will get nothing. Then all he needs to do is propose to offer Pirate 2 one coin and keep the remaining 99. Now considering all five pirates, Pirate 5 knows that if his plan is voted down, Pirate 1 and Pirate 3 will receive nothing. Then all he needs to do is offer 1 gold coin each to Pirate 1 and Pirate 3, keeping the 98 coins.

We can extend this to any number $n$ pirates (so long as $n<= 200$ ). If $n$ is odd, the most senior pirate will offer Pirates $1, 3, \dots, n - 2$ each one coin and keep the remainder for himself. If $n$ is even, then he will offer Pirates $2, 4, \dots, n-2$ one coin each and keep the remainder.

Problem 3. For a best of 3 set tennis match, would you bet on it finishing in two sets or three sets given that probabilty of winning a set is constant? Solution. Let $p = P(\text{player 1 wins})$ and $q = P(\text{player 2 wins})$ . To win in two sets, we can have player 1 or player 2 win two consecutive sets, giving a total probability of $p^2 + q^2$ . In order for the match to end in three sets, one player must win 2 sets (assume thats its player 1) and the other one set. This corresponds with a probability of $2pq$ . Note that $(p - q)^2 \geq 0$ , implying that $p^2 + q^2 \geq 2pq$ , so I would bet on the match finishing in two sets.

Problem 4. You break a stick of length 1 at two points uniformly. What is the probability that the three resulting pieces can form a triangle? Solution. A triangle is possible to form so long as one piece is not longer than the sum of the other two (one side $> \frac{1}{2}$ ) and the difference between two pieces is not longer than the third. Consider the midpoint $M$ of the original stick: if both cuts are made on the same side of $M$ (probability $\frac{1}{2})$ , then one side will have length greater than $\frac{1}{2}$ and a triangle is not possible. Let’s say we break the stick at two points $x$ and $y$ with $x < y$ and $x$ and $y$ are on different side of $M$ . There is probability $\frac{1}{2}$ that $x$ is farther left on it half of the stick than $y$ is right on its half or the vice versa. In this case, we can’t form a triangle either because the difference between two sides will be longer than the third. Then only $(1 - \frac{1}{2}) \cdot ( 1 - \frac{1}{2}) = \frac{1}{4}$ of breaks can form a triangle (my justification of this seems a bit shady and I’ll work on a more analytical solution another time).

Problem 5. Consider a game where there are two blue balls and two red balls in a bag. Each ball is removed from the bag without replacement and if you guess the color of the ball correctly you earn a dollar. How much would you pay to play this game? Solution. On your first draw you have a $\frac{2}{4}$ chance of guessing correctly. On the second draw, you have a $\frac{2}{3}$ chance of guessing correctly so long as you guess the color opposite of the first ball. By the third ball, either you have selected two balls of the same color with probability $\frac{1}{3}$ or you selected two balls of different color with probability $\frac{2}{3}$ . In the first case you are guaranteed to know the colors of the next ball and in the second you have a $50\%$ chance of guessing the next ball correctly. Finally, the probability of guessing correctly on the last ball is $1$ . Then calculating the expected value of the game from our individual draws:

\mathbb{E}[x] = (\frac{2}{4}\cdot 1) + (\frac{2}{3}\cdot 1) + (\frac{1}{3} \cdot 1 + \frac{2}{3}\cdot\frac{1}{2} \cdot 1) + 1 = \frac{17}{6}

So I would pay $ $\frac{17}{6}$ to play this game.

Day Two: June 22

Today’s problems focus on stochastic processes and Markov chains. A couple of these problems come from David Aldous’s STAT 150 page. I also came across a pretty sick compilation of different resources this morning.

Problem 1. A calculating machine uses the digits $0$ and $1$ and transmits one of these digits through several stages. At every stage, there is a probability $p$ that the digit will be changed when it leaves and probability $q = 1 - p$ that it will not. Suppose the machine starts at state 0. What is the probability that the machine will display the correct digit after 2 stages? Solution. We can identify the Markov chain for this problem as below:

\begin{matrix} & 0 & 1\\ 0 & q & p\\ 1 & p & q \end{matrix}

which produces the transition matrix

P= \begin{pmatrix} q & p\\ p & q \end{pmatrix}

We are interested in finding the probability that the state remains in state 0 after two time periods, which is indexed by the $(1,1)$ -entry of $P^2$ . Then,

P^2 = \begin{pmatrix} q & p\\ p & q \end{pmatrix}\begin{pmatrix} q & p\\ p & q \end{pmatrix} = \begin{pmatrix} q^2 + p^2 & 2qp\\ 2pq & p^2 + q^2 \end{pmatrix}

and our desired probability is $q^2 + p^2$ .

Problem 2. A transmitter is sending binary code (+ and - signals) that must pass through three relay signals before being sent on to the receiver. At each relay station, there is a $25\%$ chance that the signal will be reversed. Suppose $+$ signals make up $60\%$ of the messages sent. If a $+$ signal is received, what is the probability that a $+$ was sent? Solution. This was a problem in the notes of my statistics class last semester and was solved using a tree diagram. I decided to take a stochastic approach:

Let $A$ be the event that a $+$ was transmitted and $B$ be the event that $+$ is received after it passes through the third relay. We’re interested in finding $P(A \, | \, B)$ , which by Bayes’ Theorem can be written as

P(A \, | \, B) = \frac{P(B\, | \, A)P(A)}{P(B)} = \frac{P(B\, | \, A)P(A)}{P(B \, | \, A)P(A) + P(B \, | \, A^c)P(A^c)}

Note that $A^c$ is the event that a $-$ was transmitted. Let’s consider the Markov chain for this problem:

\begin{matrix} & + & -\\ + & 0.75 & 0.25\\ - & 0.25 & 0.75 \end{matrix}

giving us the transition matrix

P= \begin{pmatrix} 0.75 & 0.25\\ 0.25 & 0.75 \end{pmatrix} \to P^3 = \begin{pmatrix} 0.5625 & 0.4365\\ 0.4365 & 0.5625 \end{pmatrix}

We can see that $P(B\, | \, A) = P^3_{1,1} = 0.5625$ and $P(B \, | \, A^c) = P^3_{2,1} = 0.4365$ . Then, we have

P(A \, | \, B) = \frac{0.5625 \cdot 0.6}{0.5625 \cdot 0.6 + 0.4365 \cdot 0.4} = 0.6585

Problem 3. Consider a Markov chain with two states: 1 and 2. If $p_{1,2} = a$ and $p_{2,1} = b$ . For which values of $a$ and $b$ do we have an absorbing Markov chain? Solution. An absorbing Markov chain is one in which every state can reach a state where, once entered, cannot be left. Our transition matrix looks something like this:

P= \begin{pmatrix} ? & a\\ b & ?\\ \end{pmatrix}

$a$ and $b$ are the probabilities of state $1$ changing to $2$ and vice versa. In order for a state to remain constant between time periods, we must have at least one of $a$ or $b$ be 0 so that upon reaching one of the states, we will leave it.

Problem 4. A Markov chain with the state space $\{1, 2, 3\}$ has the transition matrix

P= \begin{pmatrix} 1/3 & 1/3 & 1/3\\ 0 & 1/2 & 1/3\\ 0 & 0 & 1 \end{pmatrix}

Starting from each state, find the expected time until absorption occurs. Solution. Clearly, absorption happens at state 3. Let $\mathbb{E}[i]$ denote the expected time to reach state $3$ from $i = 1, 2, 3$ . Then, we have

\begin{aligned} \mathbb{E}[1] &= 1 + \frac{1}{3}\mathbb{E}[1] + \frac{1}{3}\mathbb{E}[2] + \frac{1}{3}\mathbb{E}[3]\\ \mathbb{E}[2] &= 1 + \frac{1}{2}\mathbb{E}[2] + \frac{1}{2}\mathbb{E}[3]\\ \mathbb{E}[3] &= 0 \end{aligned}

Walking through the case of $\mathbb{E}[1]$ , we get the recursion in the following manner: if we go forward one time period, then there is a $\frac{1}{3}$ chance that we are still at state $1$ and our expected time is now $1 + \mathbb{E}[1]$ . Similarly, we have a $\frac{1}{3}$ chance of being at state $2$ or $3$ respectively, which brings us to $\mathbb{E}[1] = 1 + \frac{1}{3}\mathbb{E}[1] + \frac{1}{3}\mathbb{E}[2] + \frac{1}{3}\mathbb{E}[3]$ .

Solving these equations gives us $\mathbb{E}[i] = \begin{cases} 5/2 & \text{if } i = 1\\ 2 & \text{if } i = 2\\ 0 & \text{if } i =3 \end{cases}$

Problem 5. A group of quants wants to determine their average salary but want to do so such that no person has access to anyone else’s salary. Can they accomplish this, and, if so, how? Solution. The quants can do this by having the first quant taking a random number and adding his salary to it and passing that number onto the second quant. Then each subsequent quant will add their salary to the sum and tell the number to the next quant. Finally, the first quant removes the random number they chose and divides by the number of total quants for the average salary.

Day Three: June 23

I threw in some statistics questions for today’s problems in addition to the usual bunch.

Problem 1. There are N lions and 1 sheep in a field. All the lions really want to eat the sheep, but the problem is that if a lion eats a sheep, it becomes a sheep. A lion would rather stay a lion than be eaten by another lion. There is no other way for a lion to die than to become a sheep and then be eaten. When is it safe for any lion to eat? Solution. Let’s consider this similarly to Problem 2 of Day 1. If there is one lion, then it can happily eat the sheep. If there are two lions, no lion would eat the sheep for fear of getting eaten by the other lion. If there are three lions, then a lion eating the sheep would reduce us to the two lion scenario, in which no lion would eat the sheep. We can continue on this train of thought for any number $N$ and conclude that the sheep will be eaten when $N$ is odd. I actually think this problem is actually quite similar to the game of Nim or other impartial combinatorial games in the sense that we’re thinking in terms of a recursion.

Problem 2. You draw cards from a deck without replacement. What is the expected number of cards you draw before you get an ace? Solution. I had to refer to Stack Exchange for this question. You can find the solution here. I was initiallly tempted to use $\mu_x$ for a negative binomial distribution but I later realized that cards are drawn without replacement. Let $D$ be the r.v. that represents the number of draws before we draw an ace and assign each of the non-ace cards in the deck a number between $1$ and $48$ . Then we can define the random indicator variable $X_i = 1$ if the card with label $i$ was drawn before any ace and $0$ otherwise. Then, we have that

D = X_1 + X_2 + \cdots + X_{48}

From linearity of expectation, we have

\mathbb{E}[D] = \mathbb{E}[X_1 + X_2 + \cdots X_{48}] = \mathbb{E}[X_1] + \mathbb{E}[X_2] + \cdots + \mathbb{E}[X_{48}]

Let’s consider $P(X_1 =1)$ . If we look card $1$ along with the $4$ aces, it’s pretty simple to see that $P(X_1 = 1) = \frac{1}{5}$ . It follows that $\mathbb{E}[X_1] = \frac{1}{5}$ . Symmetry guarantees that all the $X_i$ have the same distribution so $P(X_i = 1) = P(X_j = 1)$ for all $i,j\in[1,\dots, 48]$ . Then each $\mathbb{E}[X_i] = \frac{1}{5}$ and $\mathbb{E}[D] = \frac{48}{5}$ .

Problem 3. You roll a die up to three times. You can decide to stop and choose the number on the die (where the number is your payoff) during each roll. What’s your strategy? Solution. Trading firms often ask these types of questions, where you have to devise an optimal strategy to a given game with uncertainty. Work backwards from the end of the game, we know that the expected value of the last roll is $\frac{7}{2}$ . Then, for our second to last roll, it makes sense to reroll if we rolled a $1,2,3$ , and just keep the money and quit if we rolled a $4, 5, 6$ . Then, the expected value of our last two rolls is $\frac{1}{6}(4 + 5 + 6) + \frac{1}{2}\cdot\frac{7}{2} = 4.25$ (it should make sense why we multiply each EV by $\frac{1}{2})$ . Using similar logic to before, it makes sense to reroll if we rolled a $1,2,3,4$ and keep the money if we rolled a $5$ or $6$ . Then, our strategy is to roll once if we roll a $5$ or a $6$ , reroll otherwise. When rerolling, we’ll keep anything $4$ and beyond and reroll our final time otherwise.

Problem 4. A hat contains $100$ coins, $99$ of them are guaranteed to be fair and $1$ that has a $\frac{1}{2}$ chance to be double-headed. A coin is randomly chosen at random and flipped $7$ times. If it landed heads each time what is the probability that one of the coins is double-headed? Solution. Let’s use Bayes’ Theorem for this question. Let $A$ be the event that one of the $100$ coins is double-headed and $B$ be the event that a randomly chose coin lands heads on all $7$ flips. We’re interested in finding $P(A \, | \, B)$ . Then,

$P(A) = \frac{1}{2}$
$P(A^c) = \frac{1}{2}$
$P(B \, | \, A) = P( \text{ all head flips given unfair coin}) = \frac{1}{100} + \frac{1}{2^7}\cdot \frac{99}{100}$
$P(B \, | \, A^c) = P( \text{ all head flips given no unfair coin}) = \frac{1}{2^7}$

The third bullet point can be solved for by breaking down the components of the probability of rolling all $7$ heads in the case of having a guaranteed rigged coin, which is $\frac{1}{100}\cdot 1 + \frac{1}{2^7}\cdot \frac{99}{100}$ . Then we have

P(A \, | \, B) = \frac{(\frac{1}{100} + \frac{1}{2^7}\cdot \frac{99}{100})\cdot \frac{1}{2}}{(\frac{1}{100} + \frac{1}{2^7}\cdot \frac{99}{100})\frac{1}{2} + \frac{1}{2^7}\cdot\frac{1}{2}} = 0.3194

So if we roll 7 heads in a row, there is a $32\%$ chance that there is an unfair coin in the bunch.

Problem 5. Let $X, Y \sim N(0,1)$ be i.i.d. random variables. What is the probability that $X > 3Y$ ? Solution. This is a pretty simple problem that I’m using as statistics review. Lets define $Z := X - 3Y$ . Then we simply need to find $P(Z > 0)$ . We know that $Z$ is normally distributed as it’s a linear combination of standard normal variables so we only need to find the mean and standard deviation in order to find $P(Z > 0)$ . Note that

\mathbb{E}[Z] = \mathbb{E}[X] - 3\mathbb{E}[Y] = 0

Also,

\sigma(Z) = \sigma(X-3Y) = \sigma(X) + (-3)^2\sigma(Z) = 10

Note that since $Z$ is normally distributed about $0$ , we technically don’t even need $\sigma(Z)$ since $P(Z >0) = \frac{1}{2}$ for $Z \sim N(0, \sigma)$ . Then, $P(X > 3Y) = \frac{1}{2}$ .

Day Four: June 24

I didn’t get to this until pretty late this evening so I did some relatively simple questions for today.

Problem 1. In unprofitable times corporations sometimes suspend dividend payments. Suppose that after a dividend has been paid the next one will be paid with probability $0.9$ , while after a dividend is suspended the next one will be suspended with probability 0.6. In the long run what is the fraction of dividends that will be paid? Solution. This is another Markov chain problem so we’ll tackle it like we have before. We can call the two states of this chain State 1: “dividend paid” and State 2: “dividend suspended”. We’re provided that $p_{1,1} = 0.9$ and $p_{2,2} = 0.6$ , which implies that $p_{1,2} = 0.1$ and $p_{2,1} = 0.4$ . This gives us the transition matrix

P = \begin{pmatrix} 0.9 & 0.1 \\ 0.6 & 0.4 \end{pmatrix}

Let $\pi$ be the stationary distribution for our Markov chain. In the steady state,

\begin{bmatrix} \pi_1 & \pi_2 \end{bmatrix} = \begin{bmatrix} \pi_1 & \pi_2 \end{bmatrix}\begin{pmatrix} 0.9 & 0.1 \\ 0.6 & 0.4 \end{pmatrix}

Then, $0.9 \pi_1 + 0.6\pi_2 = \pi_1$ and $0.1\pi_1 + 0.4\pi_2 =\pi_2$ . Note also the $\pi_1 + \pi_2 = 1$ be definition. Then, solving this system of linear equations gives us that $\pi_1 = 0.8$ and $\pi_2 = 0.2$ and

\pi = \begin{bmatrix} \pi_1 & \pi_2 \end{bmatrix} = \begin{bmatrix} 0.8 & 0.2 \end{bmatrix}

so $80\%$ of dividends will be paid.

Problem 2. If you repeatedly flip a fair coin. What is the expected number of flips you need to do in order to get a head immediately followed by a tail? Solution. We know that we’ll end with $HT$ so let’s think about the tosses we make before this happens. If we roll a heads on our first try, then we have to roll only heads until we roll our final $HT$ . For this reason, if there are $m$ tosses before $HT$ , then it has to be of the form $T^kH^{m-k}$ where $k = 0,1, \dots, m$ . Then let $X$ be a random variable that describes the number of tosses until we get $T$ . We can flip $T$ on our first try ( $p = \frac{1}{2}$ ), or we could flip heads first and we’d effectively start the process over again ( $p = \frac{1}{2}$ ). Then,

\mathbb{E}[X] = \frac{1}{2}\cdot 1 + \frac{1}{2}(1 + \mathbb{E}[X])

Solving this gives us $\mathbb{E}[X] = 2$ . Now define $Y$ is the random variable that describes the number of tosses until we get our first $HT$ . We can either roll a tails on our first try ( $p=\frac{1}{2}$ ) and making it so that we have to start over again, or we can roll heads on our first try ( $p = \frac{1}{2}$ ), and all we need to do now is roll at $T$ , whose EV we just modeled with $X$ . Then,

\mathbb{E}[Y] = \frac{1}{2}(1 + \mathbb{E}[Y]) + \frac{1}{2}(1 + \mathbb{E}[X])

We can plug in $\mathbb{E}[X] = 2$ and find that $\mathbb{E}[Y] = 4$ ! Other solutions to this can be found here.

Problem 3. Let $X$ be an r.v. such that $\mathbb{E}[X^2] < \infty$ . What constant $c$ minimizes $\mathbb{E}[(X-c)^2]$ ? Solution. I thought that this would be a difficult problem but then I realized that it was extremely simple. Taking the derivative with respect to $c$ gives us

\begin{aligned} \frac{d}{dc}\mathbb{E}[(X-c)^2] &= \frac{d}{dc}\mathbb{E}[X^2 - 2cX + c^2] = \mathbb{E}[-2X + 2c] \\ &= -2\mathbb{E}[X] + 2c \end{aligned}

Setting this equal to $0$ makes it pretty clear that $c = \mathbb{E}[X]$ minimizes the expression. You could also solve for this by finding the $X$ that minimizes the the expression by differenting with respect to $X$ and get the same conclusion but the question specified finding a $c$ value. Also, it never hurts to plug this back in and make sure that it is actually a minimum.

Problem 4. What is the expected number of rolls of a fair die needed to get all six numbers? Solution. This was a nice brush-up on geometric distributions. It takes one roll in order to roll out first number. After our first roll, the number of rolls until a second number is rolled is geometrically distributed with success probability $p = \frac{5}{6}$ . Since $\mu_X = \frac{1}{p}$ for a geometric distribution, the expected number of rolls until a second number is rolled is $\frac{5}{6}$ . Continuing in this fashion, the number if rolls before each additional unique number geometrically distributed with probability $p = \frac{n}{6}$ where $n$ is the number of remaining numbers to be rolled. Then,

\mathbb{E}[6] = \sum_{i = 1}^6 \frac{6}{i} = 14.7

We can extend this further and say that for an $n$ -sided die,

\mathbb{E}[n] = n\sum_{i=1}\frac{1}{i}

Problem 5. You are on a game show, and there are 3 doors. Two of the doors conceal something worthless, and one door conceals a valuable prize. The game show host, Monty Hall, knows where the prize is. He lets you pick a door, then he opens one of the remaining two doors to reveal something worthless. He then offers you the chance to switch doors. Should you? Solution. This is the Monty Hall Problem. Lets call the doors $A, B, C$ . If you chose $A$ , you have $p =\frac{1}{3}$ of guessing the prize, and $p = \frac{2}{3}$ of guessing incorrectly. This probability of $A$ being correct does not change after the host offers to let you switch after he uncovers one of the other doors (lets say $B$ ). In allowing you to switch doors, the host is offering you the option to bet on or against whether your first guess was right with $p = \frac{1}{3}$ or that it was wrong $p= \frac{2}{3}$ . In essence, when you choose door $A$ there a $\frac{2}{3}$ chance that the prize door is either between doors $B$ or $C$ . This doesn’t change when you open up door $B$ to see nothing in it—instead it now means that there is a $\frac{2}{3}$ chance that door $C$ holds the prize so its better for you to switch.

Day Five: June 25

Today’s problems focus on betting games and review of probability distributions.

Problem 1. You play a game where you toss two fair coins in the air and win $1 each time you toss. However, if you have tossed 2 heads at least once, and 2 tails at least once, you lose all your winnings and the game ends. You can also choose to stop playing at anytime. What’s your strategy? Solution. I pulled this problem from stack exchange so you can find some solutions here. Solution 1 on this page derives the solution in a cleaner manner than the solution my friend and I came up with for this.

It takes at least 2 tosses in order to get $HH$ and $TT$ . Then, after we roll our first bad event we have probability $\frac{1}{4}$ to lose everything and probability $\frac{3}{4}$ to win an additional dollar. Coin tosses are independent so each $k^{th}$ toss we make after our first bad toss, we have probability $(\frac{3}{4})^k$ of gaining an additional $k$ dollars and $1 -(\frac{3}{4})^k$ of losing the whole thing. So if we get our first bad toss on toss $n$ (by which we have acquired $n$ dollars), our expected value for $n+k$ is $\mathbb{E}[n+k] = (\frac{3}{4})^k(n+k)$ . To find the optimal time to stop tossing, we need to find the boundary of $\mathbb{E}[n+k] \leq \mathbb{E}[n+k+1]$ . That is,

\begin{aligned} (\tfrac{3}{4})^k(n+k) &\leq (\tfrac{3}{4})^{k+1}(n+k+1)\\ 4(n+k) &\leq 3(n+k+1)\\ 4n + 4k &\leq 3 n + 3k + 3\\ n+ k &\leq 3 \end{aligned}

Note that it’s not always optimal to stop at a total number of rolls $\leq 3$ . The whole premise of this question revolves around the first bad toss. So if $n < 3$ (our first bad toss is on toss 1 or 2), we should keep tossing until we’ve tossed 3 times. Otherwise, we should just take our earnings and end the game.

Problem 2. A room of 100 people put their business cards in a hat, then each person randomly draws a business card. What’s the expected number of people who draw their own business card? Solution. Each person has a $\frac{1}{100}$ chance of drawing their particular name, assuming that everyone draws at the same time. Let $X_i$ be a Bernoulli r.v. with success being person $i$ drawing their own card. Then, $P(X_i = 1) = \frac{1}{n}$ from our first observation. It follows that

\mathbb{E}[X_i] = \sum_{x \in \mathbb{R}}xp(x) = \sum_{x=0}^1 xp(x) = 0 \cdot \frac{99}{100} + 1 \cdot \frac{1}{100} = \frac{1}{100}

Then,

\mathbb{E}[X_1 +\cdots + X_{100}] = \mathbb{E}[X_1] + \cdots + \mathbb{E}[X_{100}] = n(\tfrac{1}{n}) = 1

so we expect $1$ person to draw their own name. Note that $100$ is not a special number or anything—if we replace every instance of $100$ with $n$ in our argument, we observe that our EV is still $1$ person.

Problem 3. There are $100$ noodles in a soup bowl you are eating. You are blindfolded and told to take the ends of some noodles and connect them (each end is chose with equal probability). You do this until there are no more free ends. What is the expected number of circles created in this fashion? Solution. This is actually be solved in a stochastic manner. Let $f(n)$ be a random variable that describes the number of circles that are made from connecting noodles. Starting from $n=1$ noodle, you must connect the two ends together, forming $\mathbb{E}[f(1)] = 1$ circle. For $n=2$ noodles, there are 4 ends and you must make 2 connections, forming $\binom{4}{2} = 6$ possible connections. Of these, $2$ connections will connect the same ends of the same noodle together, giving us $1$ noodle and $1$ circle. The other four combinations give us a single noodle. Notice that we’re only doing so much as to connect two of the ends, which brings us to a $n=1$ case. Then the expected number of circles for $n=2$ is

\mathbb{E}[f(2)] = \tfrac{2}{3}(1 + \mathbb{E}[f(1)]) + \tfrac{4}{6}\mathbb{E}[f(1)]

Solving this gives us that $\mathbb{E}[f(2)] = \frac{4}{3} = 1 + \frac{1}{3}$ . Moving on to $n=3$ , we have $\binom{6}{2} = 15$ choices for connections. Of these, $3$ choices will yield $1$ circle and $2$ noodles and the other $12$ will simply yield $2$ noodles. Then, we have

\mathbb{E}[f(3)] = \tfrac{3}{15}(\mathbb{E}[1 + f(2)]) + \tfrac{12}{15}\mathbb{E}[f(2)]

Solving this yields $\mathbb{E}[f(3)] = \frac{23}{15} = 1 + \frac{1}{3} + \frac{1}{5}$ . There’s a pretty clear pattern that’s starting to appear. If we extend it to $n$ noodles in general, we have that

\mathbb{E}[f(n)] = 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} = \sum_{k = 1}^n \frac{1}{2k -1}

Plugging $n=100$ gives us $\mathbb{E}[100] \approx 3.28$ (I cheated a bit and used a calculator to find this sum).

Problem 4. Two archers shoot at a target. The distance of each shot from the center of the target is uniformly distributed from 0 to 1, independently of the other shot. What is the PDF (probability density function) of the distance from the center for the losing shot? Solution. Let $X$ be the distance from the center of the shot by Archer 1 and $Y$ be that of Archer 2. Let $Z = \max\{X,Y\}$ be the distance from the center of the losing shot. As $X$ and $Y$ are uniformly distributed over $[0,1]$ , we have that for all $z\in[0,1]$ ,

P(X\leq z) = P(Y\leq z) = \frac{z - 0}{1 - 0} = z

Then, as $X$ and $Y$ are independent,

\begin{aligned} F_Z(z) &= P(\max\{X,Y\}\leq z) = P(X \leq z \, , \, Y \leq Z)\\ &= P(X \leq z)P(Y \leq z) = z^2 \end{aligned}

When we differentiate we then get the pdf of $Z$

f_Z(z) = \begin{cases} 2z& \text{ if } z\in[0,1]\\ 0& \text{ otherwise} \end{cases}

This is an interesting problem that I haven’t seen before. I’ve seen another form of the question that asks for the pdf of the winning shot, which is a bit more tricky as $Z$ would be defined as $\min\{X, Y\}$ and it wouldn’t be clear that we could use the $P(X,Y \leq z)$ since the minimum requires only one of them to satisfy the condition.

Problem 5. Let $X_1, \dots X_n$ be independent random variables uniformly distributed on $[0,1]$ . Find $P(X_1 + \cdots + X_n \leq 1)$ ? Solution. Like Problem $3$ , I’m going to try to start from the $n=1$ case and try to build up a solution. Clearly, $P(X \leq 1) = 1$ by definition. $P(X_1 + X_2 \leq 1)$ is the probability that $X_1$ and $X_2$ , distributed over the unit square, lie on or below the line that separates the lower left and upper right triangles so $P(X_1 + X_2 \leq 1) = \frac{1}{2} = \frac{1}{2!}$ . Moving on to $n =3$ , $P(X_1 + X_2 + X_3)$ describes the probability that the set of points in the unit cube all lie in the bottom-left-front triangular pyramid (probability $\frac{1}{6}) = \frac{1}{3!}$ . Again, a bit of a pattern is showing up: it seems like $P(X_1 + \cdots + X_n) = \frac{1}{n!}$ . We can confirm this with induction of a more general case, that for $0 \leq t \leq 1$

P(X_1 + \cdots + X_n \leq t) = \frac{t^n}{n!}

For simplicity sake, I’ll call $S_n := X_1 + X_2 + \cdots + X_n$ . We’ve already proved the base case earlier in our solution. Then suppose our assumption holds for $n$ , then for $n+1$ :

\begin{aligned} P(S_{n+1}\leq t) &= \int_0^1P(S_n + x \leq t)f(x)\,dx = \int_0^t\frac{(t-x)^n}{n!}\, dx \\ &= \frac{t^{n+1}}{(n+1)!} \end{aligned}

and thus our claim is proved by induction. This last part does require a bit of massaging but should be clear after applying a bit of calculus/analysis. Then, it follows that $P(S_n \leq 1) = \frac{1}{n!}$

Day Six: June 26

Most of the questions today come from A Practical Guide to Quantitative Finance Interviews and past interview questions.

Problem 1. Bond $A$ will default next year with $p = 0.5$ and bond $B$ will default next year with $p= 0.3$ . What is the range of probability that at least one bond defaults and what is the range of their correlation? Solution. The upper bound on the the probability of at least one bond defaulting takes place when $A$ defaulting and $B$ defaulting are mutually exclusive, in which case,

P(\text{A or B defaults}) = P(A) + P(B) = 0.5 + 0.3 = 0.8

For the lower bound, we consider the case when $A$ and $B$ are completely dependent events. Then,

P(\text{A or B defaults}) = \max\{A, B\} = 0.5

Then $P(\text{A or B defaults}) \in [0.5, 0.8]$ . To calculate the correlation between $A$ and $B$ , let $I_A$ and $I_B$ be indicator variables for the events that $A$ and $B$ default and let $\rho$ be their correlation. We have that $\mathbb{E}[I_A] = 0.5$ and $\mathbb{E}[I_B] = 0.3$ . Since $A$ and $B$ are Bernouli random variables, we have that $\sigma_{I_A} = p_A(1 - p_A) = 0.5^2 = 0.25$ . Similarly, $\sigma_{I_B} = 0.21$ . We have that

\begin{aligned} P(\text{A or B defaults}) &= \mathbb{E}[I_A] + \mathbb{E}[I_B] - \mathbb{E}[I_AI_B]\\ &= \mathbb{E}[I_A] + \mathbb{E}[I_B] - (\mathbb{E}[I_A]\mathbb{E}[I_B] - Cov(I_A, I_B))\\ &= 0.5 + 0.3 - (0.5 \cdot 0.3 - \rho\sigma_{I_A}\sigma{I_B})\\ &= 0.65 - \frac{\sqrt{0.21}}{2 \rho} \end{aligned}

Plugging in the bounds of $P(\text{A or B defaults})$ , we have $\rho \in \left[-\sqrt{\tfrac{3}{7}}, \sqrt{\tfrac{3}{7}}\right]$ .

Problem 2. What is the expected number of dice rolls until we get two $6$ ’s in a row? Solution. This is similar to our expected value of $HT$ problem. Let $X$ be a random variable that describes the number of throws to get two sixes in a row. There are three scenarios that we should consider:

Case 1: throwing two consecutive sixes ( $p = \frac{1}{36}$ )
Case 2: throwing a six on the first toss and a different number on the second ( $p = \frac{5}{36}$ )
Case 3: nothing throwing a six on the first toss ( $p = \frac{5}{6}$ )

In Case 1, were are done and our ev is $2$ . In Case 2, rolling a different number means we’re exactly back where we started and our ev is $2 + \mathbb{E}[X]$ since we “wasted” our first two throws. In Case 3, we’re back where we started and we’ve only used up the first throw, giving expecation of $1 + \mathbb{E}[X]$ . Then,

\begin{aligned} \mathbb{E}[X] &= \tfrac{1}{36}\cdot 2 + \tfrac{5}{36}(2 + \mathbb{E}[X]) + \tfrac{5}{6}(1 + \mathbb{E}[X])\\ &= \tfrac{42}{36} + \tfrac{35}{36}\mathbb{E}[X] \end{aligned}

Solving this gives us $\mathbb{E}[X] = 42$ so we expect to roll $42$ times total in order to get our first string of consecutive $6$ ‘s.

Problem 3. You flip a weighted coin with $P(\text{heads}) = 0.6$ four times. Every time that a heads is rolled $A$ wins $\$1$ and every time a tails is flipped, $B$ wins $\$1$ . What is the expected value of a bet where if A wins overall, you get his winnings, if B wins overall you get 0. Solution. This questions is from DRW’s Glassdoor page. It’s a bit unclear what the question means by expected value of a bet so I’ll be solving for the expected value of the game. Additionally, since it doesn’t state what happens if $A$ and $B$ tie, I’ll be increasing the number of rolls to five. Note that $A$ wins if we roll $3,4,5$ heads and $B$ wins if we roll $0, 1, 2$ heads. Let $X$ be a Bernoulli random variable with success being flipping heads ( $p= 0.6$ ). Then, $X$ follows a binomial distribution and we can caculate the expected value of the game as follows:

\begin{aligned} \mathbb{E}[X] &= \sum_{x\in \mathbb{R}}xp(x) = \sum_{x= 3}^5xb(x; 5, 0.6) = \sum_{x= 3}^5x\binom{n}{x}p^x(1-p)^{5-x}\\ &= 3 \binom{5}{3}(0.6)^3(0.4)^2 + 4 \binom{5}{4}(0.6)^4(0.4)^1 + 5 \binom{5}{5}(0.5)^5\\ &\approx 2.23 \end{aligned}

Then, the expected value of the game is $2.23$ , so it would make sense of us to play is so long as our cost of playing is less than $\$2.23$ .

Problem 4. You’re presented with two empty jars and $100$ marbles on a table, $50$ of which are white and the other $50$ are black. You are to put all $100$ of the marbles into the two jars in any way you choose. I then blindfold you and shakethe jars up to ensure good mixing and rearrange the placing of the jars randomly. You can then request the left-hand jar or the right-hand jar. You get to choose exactly one jar and and can withdraw at most one marble from the jar and you do not get a second chance if you are unhappy with your choice. How many of each color marble should you put in each jar to maximize the probability your blindfolded random draw obtains a white marble? Solution. At first I thought that both jars needed to contain $50$ marbles, in which case it didn’t matter how we distributed the black and white marbles. Due to the mixing and the randomization of which of the two jars you’re picking, you have $p=0.5$ of picking either jar. Say you have $x$ white marbles in one jar and $50-x$ in another jar. Then, your probability of picking a white marble is $\frac{1}{2} \cdot \frac{x}{50} + \frac{1}{2} \cdot \frac{50 - x}{50}$ which gives a $50\%$ chance of picking a white marble no matter how you distribute them.

I then realized that we could distribute the marbles however we’d like, meaning that we could keep at little as $0$ marbles in one jar and all $100$ in the other jar. Instead of divising a function to describe how to optimize for this problem I decided to approach it with several cases. We know we have $p = \frac{1}{2}$ of choosing either jar. We want to maximize the chance of picking a white marble regardless of which jar you choose. This can be done by placing $1$ white marble in the first jar, and the other $99$ marbles in the second jar. This ensures that we have $p = \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{49}{99} \approx 0.75$ of picking a white marble. Just testing out some other placements of marbles it’s clear that this is the optimal configuration. Approaching the problem analytically, we simply need to find the maximum of the function

f(x,y) = \tfrac{1}{2} \cdot \tfrac{x}{y} + \tfrac{1}{2} \cdot \tfrac{50 -x}{100 -y}

where $x$ is the number of white marbles we place in the first jar and $y$ is the total number of marbles we place in the first jar.

Problem 5. I will spin a fair roulette wheel with five sections. Four of the five sections pay $1$ and the other pays $5$ . If the cost is $\$1.5$ per spin should you play the game? What if you can only spin once and the payoffs to enter and cost per spint were multiplied by $1000$ ? Solution. This is a really simple EV problem. Let $X$ be the payoff of playing the game. Then,

\mathbb{E}[X] = \tfrac{4}{5} \times 1 + \tfrac{1}{5} \times 5 = 1.8 > 1.5

When you have the option of playing as much as you like, it makes sense to play the game as the actual payoff converges to the expected payoff in the long run so we expect to make $\$0.30$ profit every $5$ times we play.

If we can only spin once, our EV for the game doesn’t change. However, this scenario guages how risk-averse we might be: $80\%$ of the time we expect to lose $50$ cents, and $20\%$ of the time we expect to win $\$3.50$ beyond what we spent to play the game. Since the cost of the game is relatively low, I would probably still take the bet. However, when the numbers are scaled up by $1000$ , we either have a net loss of $\$500$ or net gain of $\$3500$ . While the problems hasn’t changed in nature, the risk of playing the game has increased dramatically, and depending on my initial amount of money, I would be much more cautious when playing.

Day Seven: June 27

Todays problems focus on betting games and expected value.

Problem 1. Suppose there are $11$ pirates who are placing their treasure in a safe. As a democratic group of pirates, they have decided that any majority of the pirates should be able to open the safe. They decide to implement this by placing several locks on the the safe, which all need to be openned to open the safe. Each lock can have multiple keys but each key can only open one lock. What is the smallest number of locks needed to secure the safe in this manner? In addition, how many keys should each pirate carry if each pirate can carry multiple keys? Solution. Note that any combination of $5$ pirates should not be able to open the safe. That is, there must be a unique lock for every combination of $5$ pirates such that none of these pirates have the key to this lock. Then, there should be a total of $\binom{11}{5} = 462$ locks. Each of these locks should have $6$ copies in order maintain the majority criterion and these keys should be evenly distributed between all $11$ pirates, giving us $\frac{462 \cdot 6}{11} = 252$ keys per pirate.

Problem 2. There is a population of ants that for each time period has $p = 0.2$ of dying out (E), $p = 0.5$ of staying the same (S), and $p= 0.3$ of doubling in size. Is the ant population likely to die long term? Solution. This is a spinoff of a question regarding amoeba populations from A Practical Guide. Lets call $P(E)$ the probability that the ant population goes extinct. We can break this down into the different conditional probabilities: $P(E\, | S)$ , $P(E\, | E)$ , and $P(E\, | D)$ . $P(E \, |\, S) = P(E)$ since the state is the same as the beginning state, $P(E\, |\, D) = P(E)^2$ (assuming each additional population of ants is independent to one another) as there are now two populations of ants, and $P(E \, | \, E) = 1$ . Then,

P(E) = 0.5 \cdot P(E) + 0.3 \cdot P(E)^2 + 0.2 \cdot 1

Solving for this gives us that $P(E) = \frac{2}{3}$ so it is quite likely the ant population dies out. The question doesn’t clarify what it means to double in size long term and $E$ , $D$ , and $S$ are not collectively exhaustive events so it’s quite tricky to calculate the other two outcomes. One could potentially use a Markov chain to simulate what happens in the future

Problem 3. A gambler starts with an initial fortune of $i$ dollars. On each successive game, the gambler wins $\$1$ white probability $p$ , $0 < p < 1$ , or loses $\$1$ with probability $q = 1-p$ . He will stop if he ever accumulates $N$ dollars or loses all his money. What is the probability that he will end up with $N$ dollars? Solution. This is the gambler’s ruin problem. Let $P_i$ denote the probability that the gambler’s fortune will reach $N$ instead of $0$ for each initial state $0 \leq i \leq N$ . Note that $P)0 = 0$ and $P_N = 1$ . The next after $i$ is going to be $i+$ with probability $p$ or $i-1$ with probability $q$ . Then, $P_i = pP_{i+1} + qP_{i-1}$ . As $p +q =1$ , we have that

P_{i+1} - P_i = \frac{q}{p}(P_i - P_{i-1}) = \left(\frac{q}{p}\right)^2(P_{i-1} - P_{i-2}) = \left(\frac{q}{p}\right)^i (P_1 - P_0)

We can then build an expression for $P_i$ from the ground up starting from $P_2$ :
$P_1 = pP_2 + qP_0 \to P_2 = \frac{1}{p}P_1 = \left[1 + \frac{q}{p}\right]P_1$
$P_2 = pP_3 + qP_1 \to P_3 = \frac{1}{q} P_2 - \frac{q}{p}P_1 = \left[1 + \frac{q}{p} + \left(\frac{q}{p}\right)^2\right] P_1$
$\cdots$
$P_i = \left[1 + \frac{q}{p} + \cdots + (\frac{q}{p})^{i-1}\right]P_1$ Note that the terms in the brackets is just a finite geometric sum. Applying this to $P_N$ ,

P_N = 1 = \left[1 + \frac{q}{p} + \cdots + \left(\frac{q}{p}\right)^{N-1}\right]P_1 = \begin{cases} \frac{1- (\frac{q}{p})^N}{1 - \frac{q}{p}}P_1& \text{ if } \frac{q}{p} \neq 1 \\ NP_1& \text{ if } \frac{q}{p} = 1 \end{cases}

We can manipulate this to solve for $P_i$ through $P_1$ :

P_1 = \begin{cases} \frac{1- \frac{q}{p}}{1 - \left(\frac{q}{p}\right)^N}P_1& \text{ if } \frac{q}{p} \neq 1 \\ \frac{1}{N}& \text{ if } \frac{q}{p} = 1 \end{cases} \text{ and } P_i = \begin{cases} \frac{1- \left(\frac{q}{p}\right)^i}{1 - \left(\frac{q}{p}\right)^N}P_1& \text{ if } \frac{q}{p} \neq \frac{1}{2} \\ \frac{i}{N}& \text{ if } \frac{q}{p} = \frac{1}{2} \end{cases}

Given even odds of winning and losing each round, our probability of coming out with $N$ dollars is proportional to our beginning state. What’s interesting is that even if we decrease the odds of success slightly, say changing it from $\frac{1}{2}$ to $\frac{38}{80}$ , there is a drastic decrease in the ability to come out with $N$ dollars versus losing everything. I hope to include a Markov chain version of this problem with specific constraints and work through that sometime soon.

Problem 4. Suppose you roll a dice and are paid the value that you roll. If you roll a $5,6$ you can roll again and if you roll $1,2,3,4$ the game ends. How much would you pay to play this game? Solution. Let $X$ be the payoff of playing this game. There is $p = \frac{2}{3}$ that you roll a losing number and your payoff is what you rolled the first round. In this case, our expected value is the average of $\{1,2,3,4\} = \frac{5}{2}$ . Also, there is a $\frac{1}{3}$ chance where we roll a $5$ or a $6$ and have the chance to roll again, bringing us back to our initial state. Then, we have

\mathbb{E}[X] = \frac{2}{3}\cdot \frac{5}{2} + \frac{1}{3}(\frac{11}{2} + \mathbb{E}[X])

Solving this gives us $\mathbb{E}[X] = \frac{21}{4}$ . Then I would pay up to $\$5.25$ to play this game.

Problem 5. Your recruiter loads two bullets separated by one chamber into a 6-chamber revolver. He then spins it,points it at his head and shoots but survives since he fired an empty chamber. You’re given the gun and must fire at your head but you may choose to spin the barrel first or take the shot directly. Unfortunately, another bullet will be loaded into the gun if you decide to spin. Do you spin the barrel? Solution. If we spin, we have a $\frac{3}{6}$ chance of dying as we have randomized the location of the next shot relative to the chambers. If we don’t there are five chambers left, two of which have bullets. However, these bullets are staggered by one chamber so the number of configurations of the two bullets is limited. Let’s number the chambers in order from $1$ through $6$ . Let’s say that the bullets are in chambers $1$ and $3$ . Given that your recruiter survived, he must have fired from chambers $2, 4, 5$ or $6$ , which occur equally likely. If he fired chamber $2$ or $6$ , then the next shot will have a bullet and if he fired $4$ or $5$ , then the next shot will be safe. Then you have $50\%$ chance of dying if we shoot immediately, which is equal to the chance of dying if we decided to spin. Then it does not matter whether we spin or not since we have an equal chance of surviving the next round regardles of how we act.

Bonus. You have $x$ red cards and $y$ blue cards. I flip them over one at a time. The probability of flipping a particular color is proportional to the amount of those colored cards left. You start with $1$ and every flip you can bet some proportion of your money on red or blue. If you win the bet, you gain twice your bet, but if you lose the bet, you gain nothing. What is the strategy that maximizes expectancy and minimizes variance? Solution. I’ll be typing up the solution later!

Week One Review

This week, I covered mostly probability games, Markov chains, and some combinatorial brain teasers. I realized that I still have a lot intution to build on how analytically to approach each problem, which I’ll keep in mind in future weeks. I hope to be posting some more challenging problems as well, since the ones I’m currently working on are not overly difficult. Next week I hope to include some topics on options and market making and reviewing some statistics and algorithms to add some variety to the problems I’ve been doing. I’m also doing readings on game theory so hopefully I’ll be able to include some more nuanced questions on winning strategies and other related topics.

Thanks for tuning in and below are a couple of resources that I used this week that might be helpful:

A Practical Guide to Quantitative Finance Interviews
Quant/FM/Data Science Compilation
Linear Algebra, Probability, and Financial Math
General Problems & Brain Teasers
Heard on the Street: Quantitative Questions from Wall Street Job Interviews