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Interview Preparation

"拳不离手， 曲不离口"

Books

  1. A Practical Guide to Quantitative Finance Interviews - Xinfeng Zhou
  2. Heard on the Street: Quantitative Questions from Wall Street Job Interviews - Timothy Falcon Crack
  3. Fifty challenging problems in probability with solutions - Frederick Mosteller

Tips

  1. Broad knowledge base
  2. Practice interview skills
  3. Listen carefully
  4. Speak your mind
  5. Make reasonable assumptions

2 Brainteasers

2.2 Logic and Reasoning

River Crossing

  • ABCD need to cross a river
  • bridge can hold at most 2 people
  • cannot cross the bridge without a torch but only have 1 torch
  • A takes 10min to cross, B takes 5min, C takes 2min, D takes 1min
  • Minimum time to cross the bridge?

Give the torch to the fastest person: 10 + 1 + 5 + 1 + 2 = 19

10min should go with 5min: 2 (C+D) + 1 (D) + 10 (A+B) + 2 (C) + 2 (C + D) = 17min

Birthday Problem:

  • Boss A birthday is one of the following 10 dates.
  • A told only month
  • C told only day
  • You: Dont know A's birthday, C doesnt know it either
  • C: didnt know A's birthday before, but now i know it
  • You: Now i know it too

Sep 1, Dec 1 Dec 2 Mar 4, Jun 4 Mar 5, Sep 5 Jun 7 Mar 8, Dec 8

I didnt know -> its not dec 2 or jun 7 (single month) C doesnt know it -> march has 3 dates that has another date, so even if i know the month i cant know which is it. so it should be either sep 1 or dec 1. You -> if i know the month, dec 1 cannot be the option, so its likely to be sep 1.

Answer: Let D be day of the month of A's birthday, where \(D \in \{1,2,3,4,5,6,7,8\}\). If birthday is on a unique day, then C will know A's birthday immediately (cancel Dec 2 , Jun 7). So it cant be December or June.

C knows it must be either March or September. Day must be unqiue in March, so cannot be March 5 or Sep 5. So the birthday must be Sep 1, March 4, or March 8. March 4 and 8 have the same month, so if its march, you cannot figure out A's birthday. Since you can still figure out A's birthday, it must be sep 1.

Card game

A casino offers a card game using a normal deck of 52 cards. The rule is that you turn over two cards each time. For each pair, if both are black, they go to the dealer's pile; if both are red, they go to your pile; if one black and one red, they are discarded. The process is repeated until you two go through all 52 cards. If you have more cards in your pile, you win $100; otherwise (including ties) you get nothing. The casino allows you to negotiate the price you want to pay for the game. How much would you be willing to pay to play this game?

  • EV question -> 0.25 to gain, 0.25 to lose, 0.5 to discard. 0.5 probability.
  • at the end of the day, probability to win is the same as dealer
  • ties -> lose, so you dont want to play. no edge.

Burning ropes

You have two ropes, each of which takes 1 hour to burn. But either rope has different densities at different points, so there's no guarantee of consistency in the time it takes different sections within the rope to burn. How do you use these two ropes to measure 45 minutes?

  • light 1 rope on both ends, will take 30mins to burn through. light rope 2 at the same time, which will become a 30min rope.
  • light other end of rope to become a 15min rope to get 45min.,

Defective Ball

You have 12 identical balls. One of the balls is heavier OR lighter than the rest (you don't know which). Using just a balance that can only show you which side of the tray is heavier, how can you determine which ball is the defective one with 3 measurements? Hint: first do for 9 identical balls and use only 2 measurements, knowing that one is heavier than the rest.

  • each weighing gives us 3 outcomes, so 3^3 = 27 means our splitting is feasible.

  • separate into 3 groups: 1234(1), 5678(5), 9101112(9)

  • 1: Weigh (1)-(5)
  • If 1 = 5: the splitting is straightforward because there are only 4 balls that you have to find the defect from. split and weigh 2: (9, 10) - (11, 8)
    • (9, 10) = (11,8): 12 defective. Weigh 3: (8) - (12) to get 12L or 12H
    • (9, 10) > (11,8): 11L or 9/10H. Weigh 3: (9) - (10) to get 11L/10H/10L
    • (9, 10) < (11,8): 11H or 9/10L. Weigh 3: (9) - (10) to get 11H/9L/10L
  • If 1 != 5: the problem is more complicated. Assume that 1 < 5, then we have that 1234L or 5678H, which gives us 8 options (3^2 > 8, so splitting is feasible). Since one weighing gives us 3 outcomes, each group can only contain at least 3 states. We want to make it such that we can only have 3 outcomes.
  • remove 3 states (4L, 78H) - each weigh can split 3 states.
  • (123L) - (56H), but you cannot resolve it at the end.
  • if we cross 3L and 5H, we then have (12L5H) - (3L6H) which gives us 4 combinations of possilities which can be resolved in the last weighing. (12L5H) - (3L6H9N) allows us to weigh them equally. Weigh 2: (12L5H) - (3L6H9N)
  • (12L5H) = (3L6H9N): defective is in 4L or 78H. Weigh 3: (7)-(8)
  • (12L5H) < (3L6H9N): defective is in 6H or 12L
  • (12L5H) > (3L6H9N): defective is in 5H or 3L
  • By symmetry, we just have to swap the values if 5 > 1.

Trailing Zeros: How many trailing zeros are there in 100! (factorial of 100)?

  • 1 x 2 x 3 x 4 x 5 = 120 -> pairs of 2 and 5 will give trailing 0
  • 2s will outnumber 5s, so we count the 5s
  • 20 numbers divisble by 5 (5, 10, 15,...,100)
  • 4 numbers divisible by 5^2 (25, 50, 75, 100)
  • 24 trailing 0s

25 horses

There are 25 horses, each of which runs at a constant speed that is different from the other horses'. Since the track only has 5 lanes, each race can have at most 5 horses. If you need to find the 3 fastest horses, what is the minimum number of races needed to identify them?

  • 1: race 1-5, 6-10, 11-15, 16-20, 21-25
  • top 3 from each group is accounted for now we need to compare between different groups. elimate bottom 2.
  • say their rank corresponds to their speed. (i.e. 1 is fastest in group, 5th is slowest in group)
  • If horse is ranked 1 in the group, then 2,3 might be in top 3. If horse is ranked 2 in the group, then only 3 might be in top 3. If ranked 3 in the group, then none will be in the group.
  • 2: race 1, 6, 11, 16, 21
  • If top 3 is 1, 6, 11, then 16 and 21 are eliminated. 8, 12, 13 are eliminated by ranking.
  • Left with 1,2,3, 6,7, 11. need to race 2,3,6,7,11

Infinite Sequences

If xxxxx ... = 2, where x^y = x, what is x?

  • \(\lim_{n \to \infty} x^{x^{x^{\cdots}}} = 2 \iff \lim_{n-1 \to \infty} x^{x^{x^{\cdots}}} = 2\)
  • as \(n \to \infty\), it should still yield the same results
  • \(x^{x^{x^{\cdots}}} = x^{(x^{x^{\cdots}})} = x^2 = 2\), so \(x = \sqrt{2}\)

HEARD ON THE STERET LOGIC QN1 You are given two glass jugs. Each contains the same volume, V , of liquid. One jug contains pure alcohol, and the other jug contains pure water. A modest quantity, Q, of water is poured from the water jug into the alcohol jug, which is then thoroughly mixed. The same modest quantity, Q, of (now diluted) alcohol is then poured back into the water jug to equalize the volumes of the jugs at their initial levels. The initial concentration of alcohol in the alcohol jug equals the initial concentration of water in the water jug (at 100%). What is the relationship between the final concentrations of alcohol in the alcohol jug and water in the water jug?

  • concentration of water in the water jug is the same as concentration of alcohol in the alcohol jug
  • the concentration will be \(V / (V+Q)\) for both jugs

HEARD ON THE STERET LOGIC QN2 You are a bug sitting in one corner of a cubic room. You wish to walk (no flying) to the extreme opposite corner (the one farthest from you). Describe the shortest path that you can walk. Be sure to mention direction, length, and so on

  • flatten the cube. answer is a triangle with length 2 height 1, so sqrt(5)
  • shortest path between two points is a "geosedic". in a spherical world, its the arc of a circle.

HEARD ON THE STERET LOGIC QN3 Fischer and Myron just stepped side-by-side onto a moving escalator. They are climbing up the stairs, and counting steps as they climb. Myron is climbing more quickly than Fischer. Myron climbs three steps in the time it takes Fischer to climb only two steps. Neither of them skips any steps. Myron steps off at the top, having counted 25 steps. He waits at the top for the slower Fischer, who steps off having counted only 20 steps. How many steps are showing on the escalator at any instant?

  • myron \(v_m = 25\), fischer \(v_f = 20\)
  • \(t_f / t_m = \frac{20}{v_f} / \frac{25}{v_m} = \frac{20}{25} / \frac{2}{3} = 1.2 = k\)
  • \(t_f = 1.2 t_m\)
  • equations
  • \(25 + x = D\)
  • \(20 + k \cdot x = D\)
  • \(k = 1.2\)
  • solving the equations yield \(D = 50\)

2.3 Thinking Outside the Box

Coin Piles: Same Heads Suppose that you are blind-folded in a room and are told that there are 1000 coins on the floor. 980 of the coins have tails up and the other 20 coins have heads up. Can you separate the coins into two piles so to guarantee both piles have equal number of heads? Assume that you cannot tell a coin's side by touching it, but you are allowed to turn over any number of coins.

  • separate into two piles. pile 1 = n, pile 2 = 1000 - n. let pile 1 = m heads, pile 2 = 20 - m heads
  • pile 1 has n - m tails
  • want to make n-m = 20 -m
  • flip n coins in all piles

Coin Piles: Same Configuration You have 12 coins, 6 heads and 6 tails. You are blindfolded and you need to split the coins into two piles such that each pile has the same combination. How do you do it?

  • Split the coins into two piles of 6 coins each. Then flip all the coins in one pile. This will ensure that both piles have the same number of heads and tails.
  • The idea is that if you have 2 coins, then you split them into 2 piles. To make them the same combination, you flip one of the coins. For 4 coins, you have HH/TT, TT/HH, HT/TH combination. Likewise, if you flip one side of the pile, you will have the same combination in both piles.

GREEN BOOK: Mislabelled Bags You are given three bags of fruits. One has apples in it; one has oranges in it; and one has a mix of apples and oranges in it. Each bag has a label on it (apple, orange or mix). Unfortunately, your manager tells you that ALL bags are mislabeled. Develop a strategy to identify the bags by taking out minimum number of fruits? You can take any number of fruits from any bags

  • picking fruit from apple label -> mix or orange. hard to determine.
  • pick fruit from mix label -> if orange, then it must be the orange bag. by elimination, then picking the apple bag will give you orange or mixed, but it must surely be mixed. then the orange bag will be apple. same goes for the apple bag.

Wise men A sultan has captured 50 wise men. He has a glass currently standing bottom down. Every minute he calls one of the wise men who can choose either to turn it over (set itupside down or bottom down) or to do nothing. The wise men will be called randomly,possibly for an infinite number of times. When someone called to the sultan correctly states that all wise men have already been called to the sultan at least once, everyone goes free. But if his statement is wrong, the sultan puts everyone to death. The wise men are allowed to communicate only once before they get imprisoned into separate rooms (one per room). Design a strategy that lets the wise men go free.

  • 49 men should flip the glass to face down first time he sees that its faced up. do nothing if glass is already faced down or has flipped it once.
  • 1 men should always flip it back up and do nothing if its already facing up.
  • after doing 49 flip ups, it means that wise men have all been called.

2.6 Pigeonhole Principle

Basically it says that if you have n holes and more than n+1 pigeons, at least 2 pigeons have to share one of the holes. The generalized version is that if you have n holes and at least mn+1 pigeons, at least m+1 pigeons have to share one of the holes.

Matching socks Your drawer contains 2 red socks, 20 yellow socks and 31 blue socks. Being a busy and absent-minded MIT student, you just randomly grab a number of socks out of the draw and try to find a matching pair. Assume each sock has equal probability of being selected, what is the minimum number of socks you need to grab in order to guarantee a pair of socks of the same color?

  • need only 4. 3 pigeon holes, need only 3 + 1 to guarantee a pair of the same colour.

Handshakes

You are invited to a welcome party with 25 fellow team members. Each of the fellow members shakes hands with you to welcome you. Since a number of people in the room haven't met each other, there's a lot of random handshaking among others as well. If you don't know the total number of handshakes, can you say with certainty that there are at least two people present who shook hands with exactly the same number of people?

at least two people present who shook hands with exactly same number of people

  • 26 people in the room
  • at minimum 25 handshakes - me with everybody
  • at least 2 people must have shaken with 1 person

Ants on a square

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?¹

  • circle with radius 1/7 can cover a square with 1/5
  • split square into 5x5 squares
  • we will have at least 1 square with 3 ants

2.X Others

You have a 5L jug and a 3L jug. You need to measure exactly 4L of water. How do you do it?

  • Iteration 1: Fill the 3L jug and pour it twice into the 5L jug. 1L in the 3L jug. Pour out 5L jug and pour 1L then 3L from the 3L jug.
  • Iteration 2: Fill 5L jug and pour into 3L jug. 2L in the 5L jug. Pour 2L into the 3L jug, then fill 5L jug and pour 1L into the 3L jug. You will have 4L in the 5L jug.

4 Probability

4.1 Basic

Coin Toss Game

Two gamblers are playing a coin toss game. Gambler A has (n+1) fair coins; B has n fair coins. What is the probability that A will have more heads than B if both flip all their coins?

  • if n=1, P(A > B) = P(A=2, B=0) + P(A=2, B=1) + P(A=1, B=0) = 0.25(0.5) + 0.25(0.5) + 0.5 (0.5) = 0.5

  • if n=2, P(A > B) = P(A=3, B=2) + P(A=3, B=1) + P(A=3, B=0) + P(A=2, B=1) + P(A=2, B=0) + P(A=1, B=0) + P(A=0) = 0.125 * 1 + 0.375 (0.75) + 0.25(0.125) + 0.125(0) = 0.5

  • 3 outcomes: X is A's n coins more than B's n coins. Y is A equal to B. Z is A less than B.

  • by symmetry, P(X) = P(Z) because same number of coins. so 2x + y = 1.
  • Only Y matters for the calculation. so we want to find x + (0.5)y = x + (0.5)(1 - 2x) = 0.5

Card Game

A casino offers a simple card game. There are 52 cards in a deck with 4 cards for each value (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A). Each time the cards are thoroughly shuffled (so each card has equal probability of being selected). You pick up a card from the deck and the dealer picks another one without replacement. If you have a larger number, you win; if the numbers are equal or yours is smaller, the house wins-as in all other casinos,the house always has better odds of winning. What is your probability of winning?

  • pick one, dealer picks another. win if larger, lose if numbers smaller of equal to.
  • you start with 52. dealer has 51 picks.

  • Brute force: If 2, 0/51 probability of winning. If 3, 4/51 probability of winning. summing 0, 4, 8, ..., 48. \(1/13 * (0/51 + 4/51 + ... + 48/51) = \frac{4}{13 * 51} * (1,2,...,12) = \frac{4}{13 * 51} * \frac{(12 * 13)}{2} = 8/17\)

  • By symmetry. same as coin problem. X = your number larger, Y = your number equal to dealer, Z = dealer number larger.

  • X = Z. We only need to figure out Y. after picking a card, only 3/51 cards will be the same. therefore \(P(X) = (1-P(Y)) / 2 = (1-3/51) / 2 = 8/17\)

Drunk passenger

A line of 100 airline passengers are waiting to board a plane. They each hold a ticket to one of the 100 seats on that flight. For convenience, let's say that the n-th passenger in line has a ticket for the seat number n. Being drunk, the first person in line picks a random seat (equally likely for each seat). All of the other passengers are sober, and will go to their proper seats unless it is already occupied; In that case, they will randomly choose a free seat. You're person number 100. What is the probability that you end up in your seat (i.e., seat #100)?

  • consider #1 and #100. Y = Seat #1 taken before #100. Z = Seat #100 taken before #1.
  • If any passenger takes #100 before taking #1, then you will not end up in your seat. if any passenger takes #1 before #100 taken, you will definitely end up in your seat. by symmetry, either outcome has probability of 0.5.
  • at every jump point, there is equal probability of "drunk guy" choosing between #1 or #100.

N points on a circle

Given N points drawn randomly on the circumference of a circle, what is the probability that they are all within a semicircle?

  • set point 1 to be at 12:00. For any points \(i \in \{2, ..., N\}\), the probability that they are on the semicircle, is \(1 / 2^{N-1}\).
  • \(P(\bigcup_{i=1}^{N} E_i) = \sum_{i=1}^{N} P(E_i) = N \cdot \frac{1}{2^{N-1}} = \frac{N}{2^{N-1}}\)
  • In general, the probability is then \(N x^{N-1}\) where \(x \leq 1/2\)

Quarterfinal problem

8 players are set to play in a 1v1 draw (quarter final). What is the probability that a unique player A plays another player B in the semi-final.

  • 2/7 spots to fill
  • 1/2 player A wins, 1/2 player B wins
  • 2/7 * 1/2 * 1/2 = 1/14

Consecutive Winning

A plays a game with B and A wins each game with 2/3 probability. Series ends when individual wins 2 games. What is the probability that A wins?

  • let \(P_{-2} = 0\) and \(P_{2} = 1\)
  • \(P_0 = 2/3 P_1 + 1/3 P_{-1}\)
  • \(P_{-1} = 1/3 P_{-2} + 2/3 P_0\)
  • \(P_1 = 2/3 P_2 + 1/3 P_0 = 2/3 + 1/3 P_0\)
  • \(P_0 = 2/3 (2/3 + 1/3 P_0) + 1/3 (2/3 P_0) = 0.8\)

HOTS Q.4 SAMPLE 8 You are dealt exactly two playing cards from a well-shuffled standard 52-card deck. The standard deck contains exactly four Kings. What is the probability that both of your cards are Kings?

  • \(\frac{4}{52} \times \frac{3}{51} = \frac{1}{221}\)

You are given a dice. Your payoff is the number that comes up on the dice. You can re-roll the dice once after each first roll, but you have to take the value of the second roll if you choose to re-roll. What is your expected payoff?

  • First roll EV: (4 + 5 + 6) / 3 = 5
  • Second roll EV: (1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5
  • Since you have probability of rolling 4,5,6 on the first roll, your expected payoff is 0.5 * 5 + 0.5 * 3.5 = 4.25

4.2 Combinatorial Analysis

Permutation

\[\frac{n!}{n_1! n_2! ... n_r!}\]

where \(n\) objects and \(n_1\), \(n_2\) are alike.

Combination $\(\binom{n}{r} = \frac{n!}{(n-r)!\,r!}\)$

Poker Hands

Poker is a card game in which each player gets a hand of 5 cards. There are 52 cards in a deck. Each card has a value and belongs to one of four suits. There are 13 values (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and four suits (spade, club, heart, diamond). What are the probabilities of getting hands with four-of-a-kind (four of the five cards with the same value)? Hands with a full house (three cards of one value and two cards of another value)? Hands with two pairs?

  • 5 card hand : 52 C 5 = 2598960 combinations
  • 4 of a kind - 13 choices * 48 (any 5th card) = 624
  • full house - 4C3 for triple with 13 choices, 4C2 for pair with 12 choices. 13 * 4C3 * 12 * 4C2 = 13 * 4 * 12 * 6 = 3744
  • hands with two pairs: choose pairs 13C2, 4C2 for first choices, 4C2 for second choices and remaining card 44 (52 - 4 * 2, which cannot have the same suit). 13C2 * 4C2 * 4C2 = 123552

Hopping Rabbit

A rabbit sits at the bottom of a staircase with n stairs. The rabbit can hop up only one or two stairs at a time. How many different ways are there for the rabbit to ascend to the top of the stairs?

  • \(n=1, 1\), \(n=2, 2\)
  • in general, \(f(n) = f(n-1) + f(n-2)\) for \(n \geq 3\)

Screwy pirates

Having peacefully divided the loot (in chapter 2), the pirate team goes on for more looting and expands the group to 11 pirates. To protect their hard-won treasure, they gather together to put all the loot in a safe. Still being a democratic bunch, they decide that only a majority - any majority - of them (26) together can open the safe. So they ask a locksmith to put a certain number of locks on the safe. To access the treasure, every lock needs to be opened. Each lock can have multiple keys; but each key only opens one lock. The locksmith can give more than one key to each pirate. What is the smallest number of locks needed? And how many keys must each pirate carry?

  • 11 pirates, more than 6 can open the safe together.
  • find a special lock where 5-pirate groups cannot have a key.
  • minimum number of locks is 11C5 = 462 locks. each lock has 6 keys, so each pirate must have \(\frac{462 * 6}{11} = 252\)

Combinations 5 men and 3 women sit around a table. How many ways can they be seated such that two women do not sit together.

  • arrange men around table which is (5-1)! = 24
  • 5 gaps between men - 5C3 = 10
  • arrange 3 women = 3! = 6
  • total = 24 * 10 * 6 = 1440

4.3 Conditional Probability and Bayes' formula

Boys and girls

Part A. A company is holding a dinner for working mothers with at least one son. Ms. Jackson, a mother with two children, is invited. What is the probability that both children are boys?

  • ms jackson has at least one son.
  • A = of 2 boys, B = event of at least 1 boy. 1 / 3 probability of 2 boys. (bb, bg, gb)
  • given that there is at least one boy in two children, what is the probability of both being boys

Part B. Your new colleague, Ms. Parker is known to have two children. If you see her walking with one of her children and that child is a boy, what is the probability that both children are boys?

  • parker has at least one boy. probability of having both boys is 0.5. (independent of the current boy)
  • given one child is a boy, what is the conditional probability the other is a boy

All-girl world?

In a primitive society, every couple prefers to have a baby girl. There is a 50% chance that each child they have is a girl, and the genders of their children are mutuallyindependent. If each couple insists on having more children until they get a girl and once they have a girl they will stop having more children, what will eventually happen to the fraction of girls in this society?

  • fraction always 50% and will stay stable at 50%.
  • \(E[boys] = E[girls] = \sum_{k=1}^\infty \frac{k}{2^{k+1}} = 1\), so they will be equal

Unfair coin You are given 1000 coins. Among them, 1 coin has heads on both sides. The other 999 coins are fair coins. You randomly choose a coin and toss it 10 times. Each time, the coin turns up heads. What is the probability that the coin you choose is the unfair one?

  • The unfair coin always lands heads. The fair coin lands heads with probability \(1/2^{10}\).
  • Let \(A\) = chose unfair coin, \(B\) = tossed 10 heads.
  • \(P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B \mid A)P(A) + P(B \mid A')P(A')} = \frac{1 \cdot \frac{1}{1000}}{1 \cdot \frac{1}{1000} + \frac{1}{2^{10}} \cdot \frac{999}{1000}} = \frac{1024}{2023}\)

Two players, A and B, alternatively toss a fair coin (A tosses the coin first, then B tosses the coin, then A, then B...). The sequence of heads and tails is recorded. If there is a head followed by a tail (HT subsequence), the game ends and the person who tosses the tail wins. What is the probability that A wins the game?

  • \(P(A)\) is probability that A wins. \(P(B) = 1 - P(A)\)
  • \(P(A) = 0.5 P(A \mid H) + 0.5 P(A \mid T)\)
  • \(P(A \mid T) = P(B) = 1 - P(A)\) because if A's first toss is T then counter rests and B becomes the first to toss.
  • \(P(A \mid H) = 0.5 \cdot 0 + 0.5 (1 - P(A \mid H))\) where \(P(B \mid H) + P(A \mid H) = 1\), \(P(A \mid H) = 1/3\)
  • \(P(A) = 0.5 (1/3) + (1-P(A)) = 4/9\)

Russian roulette series

Let's play a traditional version of Russian roulette. A single bullet is put into a 6-chamber revolver. The barrel is randomly spun so that each chamber is equally likely to be under the hammer. Two players take turns to pull the trigger-with the gun unfortunately pointing at one's own head-without further spinning until the gun goes off and the person who gets killed loses. If you, one of the players, can choose to go first or second, how will you choose? And what is your probability of loss?

  • probability of loss is half. so it doesnt matter.

Now, let's change the rule slightly. We will spin the barrel again after every trigger pull. Will you choose to be the first or the second player? And what is your probability of loss?

  • second.
  • let \(p\) = probability of first person losing. \(1-p\) is the probability of second person losing.
  • long run, \(p = 1 * 1/6 + (1-p) * 5/6 = 6/11\)

If instead of one bullet, two bullets are randomly put in the chamber. Your opponent played the first and he was alive after the first trigger pull. You are given the option whether to spin the barrel. Should you spin the barrel?

  • spin
  • 1/3 probability to lose if spin
  • 2/5 probability if dont spin

What if the two bullets are randomly put in two consecutive positions? If your opponent survived his first round, should you spin the barrel?

  • 1/4 chance of losing, 3/4 winning
  • spin means 2/3 chance of winning.
  • therefore should not spin.

Aces Fifty-two cards are randomly distributed to 4 players with each player getting 13 cards. What is the probability that each of them will have an ace?

  • solved by counting, \(\frac{52!}{13!13!13!13!}\) permutations
  • distribute ace first, so 4! ways. disribute rest of 48 cards, which is \(\frac{48!}{12!12!12!12!}\)
  • \(P(all has ace) = \frac{4! \cdot \frac{48!}{12!12!12!12!}}{\frac{52!}{13!13!13!13!}}\)
  • conditional probabiility method is that \(1 \times 39/51 \times 26/50 \times 13/49\)

Statistics

Suppose that X and Y are independent random variables each distributed standard normal: X \(\sim\) N(0, 1), and Y \(\sim\) N(0, 1). What are the variance and the standard deviation of X − Y .

  • mean = 0
  • variance is additive so 2, and std = \(\sqrt{2}\) assuming independence