Q: You have a shuffled deck of 52 cards. You turn over cards one at a time. I'll pay you $1 for each red card. You can stop at any time and keep your earnings, but if you turn over all 52 cards you get $26. What is the optimal strategy and expected value?

$26. No matter when you stop, the expected payout is always $26. At any point, the conditional expected value of the remaining cards is always proportional — the fraction of red cards remaining always has the same expectation. So every stopping strategy yields the same expected value of $26.

Q: A drunk man starts at the origin on a number line. Each step he goes +1 or -1 with equal probability. What is the expected number of steps to reach +1 or -1 from the origin?

1. From the origin, the very first step takes him to either +1 or -1 with certainty. So the expected number of steps is exactly 1.

Q: You have 10 pairs of socks in a drawer (20 socks total), each pair a different color. You grab socks one at a time in the dark. What is the expected number of socks you need to grab to get a matching pair?

By the pigeonhole principle, you're guaranteed a match after at most 11 socks. For the expected value: the probability of no match after k draws is (20/20)(18/19)(16/18)..., using a birthday-problem-style calculation. The expected number is approximately 5.7 socks.

Q: 100 passengers board a plane with 100 assigned seats. The first passenger sits randomly. Each subsequent passenger sits in their seat if available, otherwise picks randomly. What is the probability the 100th passenger sits in their assigned seat?

1/2. By symmetry, when the last passenger boards, only seat 1 or seat 100 can be empty. Every random choice treats these two seats with equal probability, so there is a 50/50 chance.

Q: What is the expected number of times you need to roll a fair die to see all 6 faces?

14.7. This is the coupon collector's problem. After seeing k distinct faces, the probability of seeing a new face is (6-k)/6, so the expected rolls for the next new face is 6/(6-k). Sum: 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 1 + 1.2 + 1.5 + 2 + 3 + 6 = 14.7.

Q: If you roll two fair six-sided dice, what is the probability the sum is exactly 7?

1/6. There are 36 total outcomes. The combinations summing to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — that's 6 outcomes. 6/36 = 1/6.

Q: When rolling two fair six-sided dice, what is the probability that at least one shows a 5?

11/36. The complement (no 5s) has probability (5/6)^2 = 25/36. So P(at least one 5) = 1 - 25/36 = 11/36.

Q: A family has two children. At least one is a girl. What is the probability both are girls?

1/3. Possible combinations: BB, BG, GB, GG. Eliminating BB leaves BG, GB, GG. Only GG has both girls, so 1/3.

Question 1

You flip a fair coin until you get two heads in a row. What is the expected number of flips?

Accepted Answer

6. Let E be the expected number of flips. From the start, you flip once (cost 1). If tails (prob 1/2), you restart. If heads (prob 1/2), you flip again. If the second is heads, you're done (total cost so far: 2 from this branch). If tails, you restart. Setting up the equation: E = 1 + (1/2)E + (1/2)(1 + (1/2)(0) + (1/2)E) simplifies to E = 6.

Question 2

100 prisoners are each assigned a unique number 1-100. Boxes labeled 1-100 each contain a random number. Each prisoner may open 50 boxes. They all go free if every prisoner finds their own number. They can strategize beforehand but not communicate after. What strategy maximizes survival probability?

Accepted Answer

Each prisoner starts by opening the box matching their number, then follows the chain (opens the box with the number they found). They succeed if their chain length is <= 50. The probability that the longest cycle in a random permutation exceeds 50 is about ln(2) = 0.69, so the survival probability is about 31%. This is vastly better than the (1/2)^100 probability from random guessing.

Question 3

You have a shuffled deck of 52 cards. You turn over cards one at a time. I'll pay you $1 for each red card. You can stop at any time and keep your earnings, but if you turn over all 52 cards you get $26. What is the optimal strategy and expected value?

Accepted Answer

$26. No matter when you stop, the expected payout is always $26. At any point, the conditional expected value of the remaining cards is always proportional — the fraction of red cards remaining always has the same expectation. So every stopping strategy yields the same expected value of $26.

Question 4

A drunk man starts at the origin on a number line. Each step he goes +1 or -1 with equal probability. What is the expected number of steps to reach +1 or -1 from the origin?

Accepted Answer

1. From the origin, the very first step takes him to either +1 or -1 with certainty. So the expected number of steps is exactly 1.

Question 5

You have 10 pairs of socks in a drawer (20 socks total), each pair a different color. You grab socks one at a time in the dark. What is the expected number of socks you need to grab to get a matching pair?

Accepted Answer

By the pigeonhole principle, you're guaranteed a match after at most 11 socks. For the expected value: the probability of no match after k draws is (20/20)(18/19)(16/18)..., using a birthday-problem-style calculation. The expected number is approximately 5.7 socks.

Question 6

100 passengers board a plane with 100 assigned seats. The first passenger sits randomly. Each subsequent passenger sits in their seat if available, otherwise picks randomly. What is the probability the 100th passenger sits in their assigned seat?

Accepted Answer

1/2. By symmetry, when the last passenger boards, only seat 1 or seat 100 can be empty. Every random choice treats these two seats with equal probability, so there is a 50/50 chance.

Question 7

What is the expected number of times you need to roll a fair die to see all 6 faces?

Accepted Answer

14.7. This is the coupon collector's problem. After seeing k distinct faces, the probability of seeing a new face is (6-k)/6, so the expected rolls for the next new face is 6/(6-k). Sum: 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 1 + 1.2 + 1.5 + 2 + 3 + 6 = 14.7.

Question 8

If you roll two fair six-sided dice, what is the probability the sum is exactly 7?

Accepted Answer

1/6. There are 36 total outcomes. The combinations summing to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — that's 6 outcomes. 6/36 = 1/6.

Question 9

When rolling two fair six-sided dice, what is the probability that at least one shows a 5?

Accepted Answer

11/36. The complement (no 5s) has probability (5/6)^2 = 25/36. So P(at least one 5) = 1 - 25/36 = 11/36.

Question 10

A family has two children. At least one is a girl. What is the probability both are girls?

Accepted Answer

1/3. Possible combinations: BB, BG, GB, GG. Eliminating BB leaves BG, GB, GG. Only GG has both girls, so 1/3.

Question 11

What is the expected value of a single roll of a fair six-sided die?

Accepted Answer

3.5. E = (1+2+3+4+5+6)/6 = 21/6 = 3.5.

Question 12

Drawing two cards from a 52-card deck without replacement, what is the probability both are Aces?

Accepted Answer

1/221. P = (4/52) x (3/51) = 12/2652 = 1/221.

Question 13

In a room of 23 people, what is the approximate probability that at least two share the same birthday?

Accepted Answer

Approximately 50.7%. Calculate P(no match) = (365/365)(364/365)(363/365)...(343/365) ≈ 0.493. So P(match) ≈ 1 - 0.493 = 0.507.

Question 14

If you roll 10 fair six-sided dice, what is the probability that the sum is divisible by 6?

Accepted Answer

1/6. No matter what the first 9 dice show, exactly one value on the 10th die (from 1-6) makes the total divisible by 6. Each face is equally likely, so P = 1/6.

Question 15

You are on a game show with three doors. Behind one is a car; behind the others, goats. You pick Door 1. The host opens Door 3 (a goat). Should you switch to Door 2?

Accepted Answer

Yes, you should always switch. The key insight is that the host knows where the car is and will always reveal a goat. When you first pick, you have a 1/3 chance of being right. The host then deliberately opens a door he knows has a goat, which concentrates the remaining 2/3 probability onto the other unopened door. Switching wins whenever your initial pick was wrong (2/3 of the time). If the host had opened a door randomly and happened to show a goat, switching would not help — it is the host's knowledge that creates the advantage.

Question 16

Three ants sit at the corners of an equilateral triangle. Each randomly picks clockwise or counter-clockwise and walks along the edge. What is the probability none collide?

Accepted Answer

1/4. There are 2^3 = 8 direction combinations. No collision only if all go the same way (all CW or all CCW). That's 2/8 = 1/4.

Question 17

You flip a fair coin 17 times, your friend flips 16 times. What is the probability you flip more heads?

Accepted Answer

1/2. After 16 flips each, by symmetry P(you lead) = P(friend leads). The 17th flip breaks ties in your favor or maintains your lead, resulting in exactly 50%.

Question 18

If you roll three fair six-sided dice, what is the probability the numbers are in strictly increasing order?

Accepted Answer

5/54. There are 216 total outcomes. Choose 3 distinct numbers from 6: C(6,3) = 20 ways, each with exactly one increasing arrangement. P = 20/216 = 5/54.

Question 19

What is the probability of drawing two cards of the same suit from a 52-card deck without replacement?

Accepted Answer

4/17. The first card can be anything. Then 12 of the remaining 51 cards match its suit. P = 12/51 = 4/17.

Question 20

You call three friends in Seattle, each tells the truth with probability 2/3. All three say it's raining. If the base probability of rain is 10%, what is the actual probability it's raining?

Accepted Answer

About 47%. Using Bayes: P(3 Yes|Rain) = (2/3)^3 = 8/27. P(3 Yes|No Rain) = (1/3)^3 = 1/27. P(Rain|3 Yes) = (8/27 × 0.1) / (8/27 × 0.1 + 1/27 × 0.9) = 8/17 ≈ 47%.

Question 21

On average, how many times must you roll a fair die before seeing a 6?

Accepted Answer

6. This is a geometric distribution with p = 1/6. Expected value = 1/p = 6.

Question 22

In tennis, you win each point with probability 60%. What is the probability of winning from Deuce?

Accepted Answer

About 69%. Let P = probability of winning from deuce. P = W^2 / (1 - 2W(1-W)) where W = 0.6. P = 0.36/0.52 ≈ 0.692.

Question 23

What is the probability of rolling a sum of 11 with two fair six-sided dice?

Accepted Answer

1/18. Only (5,6) and (6,5) sum to 11. That's 2/36 = 1/18.

Question 24

What is the probability of drawing a red card from a standard 52-card deck?

Accepted Answer

1/2. Hearts and Diamonds are red (26 cards). 26/52 = 1/2.

Question 25

In a room of 10 people, if everyone shakes hands with everyone else exactly once, how many handshakes occur?

Accepted Answer

45. C(10,2) = 10 × 9 / 2 = 45.

Quantitative Finance

Quantitative Finance

Quantitative Finance

Quantitative Finance