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

After the revolution, each of the 66 citizens of a certain city, including the king, has a salary of 1. King cannot vote, but has the power to suggest changes - namely, redistribution of salaries. Each person's salary must be a whole number of dollars, and the salaries must sum to 66. He suggests a new salary plan for every person including himself in front of the city. Citizens are greedy, and vote yes if their salary is raised, no if decreased, and don't vote otherwise. The suggested plan will be implemented if the number of "yes" votes are more than "no" votes. The king is both, selfish and clever. He proposes a series of such plans. What is the maximum salary he can obtain for himself?

Accepted Answer

63. Continuing from the hint, The king begins by proposing that 33 citizens have their salaries doubled to $2, at the expense of the remaining 33 (himself included). Next, he increases the salaries of 17 of the 33 salaried voters (to $3 or $4) while reducing the remaining 16 to $0. In successive turns, the number of salaried voters falls to 9, 5, 3, and 2. Finally, the king bribes three paupers with $1 each to help him turn over the two big salaries to himself, thus finishing with a royal salary of $63. It is not difficult to see that the king can do no better at any stage than to reduce the number of salaried voters to just over half the previous number; in particular, he can never achieve a unique salaried voter. Thus, he can do no better than $63 for himself, and the six rounds above are optimal Hence the answer is $ 63 Note: It is possible to reach $65 if the King could vote (that would be a variant of this puzzle).

Question 3

Two bullets are loaded into a gun's round barrel consecutively. The barrel has a capacity of 6. The gun is fired once, but no bullet is shot. Does rolling the barrel (shuffling) before next shot increase the probability of firing a bullet?

Accepted Answer

Yes, shuffling increases the probability of firing a bullet from 25% to 33.3%. Since the two bullets are loaded consecutively in a 6-chamber barrel, after one empty shot, there are 4 remaining positions, of which 1 has a bullet next. So without shuffling, P = 1/4 = 25%. With shuffling, there are 2 bullets among 6 slots, so P = 2/6 = 33.3%.

Question 4

How do you place 50 good candies and 50 rotten candies in two boxes such that if you choose a box at random and take out a candy at random, it better be good! We need to maximize the probability of getting a good candy when selecting a random box and a random candy from it.

Accepted Answer

Place 1 good candy in one box and all the remaining (49 good and 50 rotten candies) in the second box. This gives a probability of (1/2) * 1 + (1/2) * (49/99) = 74.74%.

Question 5

In a world where everyone wants a girl child, each family continues having babies till they have a girl. What do you think will the boy-to-girl ratio be eventually? Assuming probability of having a boy or a girl is the same and there is no other gender at the time of birth.

Accepted Answer

1:1. Suppose there are N couples. First time, N/2 girls and N/2 boys are born. N/2 couples retire, and the rest try again. Next time, N/4 couples give birth to N/4 girls and N/4 boys. Thus, even in the second iteration, the ratio is 1:1. This ratio will always remain the same, no matter how many times people try to give birth to a favored gender.

Question 6

The probability of having accidents on a road in one hour is 3/4. What is the probability of accidents in half an hour?

Accepted Answer

1/2. Assuming disjoint time intervals are independent: P(no accident in 1 hour) = P(no accident in first half) * P(no accident in second half). So 1 - 3/4 = p^2, giving p = 1/2. Hence the probability of an accident in half an hour = 1 - 1/2 = 1/2.

Question 7

Spiderman has two close friends, Mary Jane & Gwen Stacy. After every mission, he rushes to the central subway. One line heads towards Mary's place, and another towards Stacy. Trains from each line leave every 10 minutes. Spiderman being impartial always boards the first train that leaves. However, he observes that he ends up visiting Mary Jane nine times more often than Gwen Stacy. Can you decipher why?

Accepted Answer

The gap from M train to G train is 1 minute, and from G to M is 9 minutes. Since both trains leave at the same frequency of 10 minutes, the solution lies in the gap between the two consecutive trains. Suppose train M (towards Mary) leaves at minutes: 0, 10, 20, ... And train G (towards Gwen) leaves at minutes: 1, 11, 21, ... Spiderman arrives at a random time. He catches train G only if he arrives in the 1-minute window between M and G departures. Otherwise he catches M. So P(visiting Mary) = 9/10 and P(visiting Gwen) = 1/10, a ratio of 9:1.

Question 8

I bought two fair six-sided dice. Keeping the first one untouched, I modified the second die, by erasing each face and writing a number of my own. Now, the sum of outcomes of the two dice is equally likely an integer between 1 and 12. Can you deduce the numbers written on the modified die? Note: For each dice, each face is equally likely to turn up after a toss.

Accepted Answer

0 0 0 6 6 6. The sum of 1 and 12 ensures that there is at least one 0 and one 6. Since the sum ranges from 1 to 12 only, the faces of the modified die can only have an integer from 0 to 6. P(R + M = 1) = P(R = 1) * P(M = 0) = 1/6 * P(M = 0). Since each sum from 1 to 12 must have probability 1/12, and P(R + M = 1) = 1/12, we get P(M = 0) = 1/2, meaning 3 faces show 0. By similar logic with the sum of 12, P(M = 6) = 1/2, meaning 3 faces show 6. So the modified die has faces: 0, 0, 0, 6, 6, 6.

Question 9

A father claims about snowfall last night. First daughter tells that the probability of snowfall on a particular night is 1/8. Second daughter tells that 5 out of 6 times the father is lying! What is the probability that there actually was a snowfall?

Accepted Answer

1/36. Let S = Snowfall occurred, and C = father is claiming snowfall. P(S|C) = P(C|S) * P(S) / P(C). P(S) = 1/8. P(C|S) = 1 - 5/6 = 1/6 (he tells truth when it snowed). P(C|not S) = 5/6 (he lies when it didn't snow). P(C) = P(C|S)*P(S) + P(C|not S)*P(not S) = (1/6)(1/8) + (5/6)(7/8) = 1/48 + 35/48 = 36/48 = 3/4. P(S|C) = (1/6 * 1/8) / (3/4) = (1/48) / (3/4) = 1/36.

Question 10

Two witches make a nightly visit to an all-night coffee shop. Each arrives at a random time between 0:00 and 1:00. Each one of them stays for exactly 30 minutes. On any one given night, what is the probability that the witches will meet at the coffee shop?

Accepted Answer

3/4. Plot the graph of X and Y, where X and Y denote the time each witch arrives. This will form a 60x60 square for total feasible region. The probability that they meet is |X-Y|<= 30. Draw this as favorable region, by joining line (30,0) with (60,30) and (0,30) with (30,60). Clearly the interior of this region has area 3/4 th of total. Hence the probability = 3/4 = 0.75

Question 11

Let X be a random variable that takes value 0 or 1 with 50% probability each. You need to define a new random variable Y as a function of X, such that Y has the value 1 with 25% probability and 0 otherwise.

Accepted Answer

Y = X * X' where X' is an independent copy of X. Given P(X=1) = P(X=0) = 0.5, then P(Y=1) = P(X=1) * P(X'=1) = 0.5 * 0.5 = 0.25.

Question 12

In this gambling game, a player can buy a ticket for Rs 1 on any number from 1 to 6. Three identical and unfair dice are rolled. If the booked number appears on 0, 1, 2 or 3 dice, player wins Rs 0, 1, 2 or 3 respectively, without returning the original Rs 1. What is expected money you can win after buying a ticket for Rs 1?

Accepted Answer

1/2. We bet Rs 1 on each number 1-6. In any case, we get Rs 3 back. That is 1/2 per ticket. Hence the expected amount of money we can win is 1/2. A tedious way to arrive at this answer is to calculate the probability of getting 1, 2 or 3 faces common to our booking.

Question 13

A and B are in a team called AB, playing against C. If AB team wins they win Rs 3, nothing otherwise. The Game: A and B are placed in 2 separate rooms far away. A will toss a coin and B will also toss a coin; A will have to guess the outcome of B's toss and B will guess A's. If both guesses are right, team AB wins Rs 3, nothing otherwise. Should they play the game, by paying Rs 1 at the start?

Accepted Answer

Yes, they should play. The probability of winning is 1/2. All possibilities: A gets H, B gets H; A gets H, B gets T; A gets T, B gets H; A gets T, B gets T. If both always guess their own coin's face, they win whenever the coins match (HH or TT), which is 2 out of 4 cases = 1/2. The expected pay-off is 1/2 * 3 - 1 = 0.5, which is positive.

Question 14

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 15

Snow-particles are falling on the ground one after another. A particular snowflake turns out to be of type "Stellar Dendrite" with probability 'p' if its previous particle was also Stellar Dendrite, and with probability 'q' if previous one was something else. If a snowflake is picked from ground, what is the probability that it is Stellar Dendrite?

Accepted Answer

q/(1-p+q). Let x be the probability that a snowflake picked from the ground is Stellar Dendrite. When a new snowflake falls, the last snowflake was Stellar Dendrite with probability x. So the probability the new snowflake is Stellar Dendrite = x*p + (1-x)*q. For the composition to remain constant: xp + (1-x)q = x. Solving: xp + q - xq = x, so q = x - xp + xq = x(1 - p + q). Therefore x = q/(1 - p + q).

Question 16

p and q are two points chosen at random between 0 & 1. What is the probability that the ratio p/q lies between 1 & 2?

Accepted Answer

0.25. Let the points be x and y respectively. We want x/y between 1 and 2. The limits are y = x and y = x/2. The desired region is the area between these lines within the unit square. Area = integral from 0 to 1 of min(x, 1) - x/2 dx = integral from 0 to 1 of x/2 dx = 1/4 = 0.25.

Question 17

Roll a die, and you get paid what the dice shows. But if you want, you can request a second chance & roll the die again; get paid what the second roll shows instead of the first. What is the expected value?

Accepted Answer

4.25. The expected value of a single roll is (1+2+3+4+5+6)/6 = 3.5. You should re-roll if your first roll is below 3.5 (i.e., 1, 2, or 3) and keep it if it's 4, 5, or 6. Expected value = P(keep) * E(keep) + P(reroll) * E(reroll) = (1/2)(5) + (1/2)(3.5) = 2.5 + 1.75 = 4.25.

Question 18

A very innocent monkey throws a fair die. The monkey will eat as many bananas as are shown on the die, from 1 to 5. But if the die shows '6', the monkey will eat 5 bananas and throw the die again. This may continue indefinitely. What is the expected number of bananas the monkey will eat?

Accepted Answer

4. Let E be the expected number of bananas. Either the die shows 1-5 (probability 5/6) and the monkey eats that many, or it shows 6 (probability 1/6) and the monkey eats 5 and rolls again. E = (5/6) * (1+2+3+4+5)/5 + (1/6) * (5 + E) = (5/6)(3) + (1/6)(5+E) = 5/2 + 5/6 + E/6. Solving: 5E/6 = 5/2 + 5/6 = 20/6, so E = 4.

Question 19

A chess tournament has K levels and N = 2^K players with skills P_1 > P_2 > ... > P_N. At each level, random pairs are formed and one person from each pair proceeds to the next level. When two opponents play, the one with better skills always wins. What is the probability that players P_1 and P_2 will meet in the final level?

Accepted Answer

2/N. Player P_1 always wins, so P_1 is guaranteed to reach the final. P_2 reaches the final if and only if P_2 is in the opposite half of the bracket from P_1 (otherwise P_2 would face P_1 before the final and lose). The probability that P_2 is in the opposite half is (N/2) / (N-1). Wait, more precisely: P_2 meets P_1 in the final iff P_2 never faces P_1 until the final, and P_2 beats everyone else (since P_2 is second best). P_2 is in the opposite half of the bracket from P_1 with probability (N/2)/(N-1). But since the bracket is random, P(P_2 in opposite half) = 1/2... Actually the answer is 2/N by symmetry of bracket placement and the deterministic outcomes.

Question 20

A coin is tossed 10 times and the output written as a string. What is the expected number of HH? Note that in HHH, number of HH = 2. (eg: expected number of HH in 2 tosses is 0.25, 3 tosses is 0.5)

Accepted Answer

Let the expected number of HH for n tossed is E(n). So, probability that an (n-1) toss experiments, ends in T is 1/2. So, E(n) = 1/2 * E(n-1) + 1/2 * ( 1/2 * (E(n-1)+1) + 1/2 * (E(n-1))) (The first case when it ends in T. & The second case when it ends in H. In the second case, if you get an H then, you get 1 more HH. ) So, E(n) = E(n-1) + 1/4, us E(2) = 1/4 So, E(n) = (n-1)/4 For n=10, E(10) = 2.25

Question 21

The stick drops and breaks at a random point distributed uniformly across the length. What is the expected length of the smaller part? Solution The probability that the break point belongs to first half is 1/2. In this case, the expected length of the small part is L/4. Its L/4 in the other case too. Hence E = 1/2 * L/4 + 1/2 * L/4 = L/4

Accepted Answer

The probability that the break point belongs to first half is 1/2. In this case, the expected length of the small part is L/4. Its L/4 in the other case too. Hence E = 1/2 * L/4 + 1/2 * L/4 = L/4

Question 22

There are 26 black(B) and 26 red(R) cards in a standard deck. A run is number of blocks of consecutive cards of the same color. For example, a sequence RRRRBBBRBRB of only 11 cards has 6 runs; namely, RRRR, BBB, R, B, R, B. Find the expected number of runs in a shuffled deck of cards.

Accepted Answer

27. We form the answer for general deck of 2n cards, with n red cards and n black cards. Consider any sequence X1,X2..X2n (of R & B). Now define Y1 = 1; Yi = 1 if Xi and X(i-1) are of different colors Yi = 0 otherwise. We want to find E[Y1 + ..+ Y2n], using linearity of expectation, = E(Y1) +...+E(Yn). Now note that, E[Y1] = 1 and E[Yi] = E[Yj] for the rest (by symmetry) Also E[Yi] = E(Y2) = P(X2 is diff from X1) = (number of ways first two are RB or BR) / (Total number of orientations) = 2 * [ (2n-2)choose(n-1) ] / [ 2n choose n ] = n/(2n-1) Ans = 1+(2n-1) * E[Y_i] = n+1 This is 27 for deck of 52 cards.

Question 23

An ant is standing on one corner of a cube & can only walk on the edges. The ant is drunk and from any corner, it moves randomly by choosing any edge! What is the expected number of edges the ant travels, to reach the opposite corner?

Accepted Answer

10. We define 4 equivalent states based on distance from start: State 0 (starting corner), State 1 (adjacent to start), State 2 (shares edge with opposite corner), State 3 (opposite corner, target). From State 0: 3 edges lead to State 1. E_0 = 1 + E_1. From State 1: 1 edge back to State 0, 2 edges to State 2. E_1 = 1 + (1/3)E_0 + (2/3)E_2. From State 2: 2 edges to State 1, 1 edge to State 3. E_2 = 1 + (2/3)E_1. Solving: E_2 = 1 + (2/3)E_1, E_1 = 1 + (1/3)E_0 + (2/3)(1 + (2/3)E_1) = 5/3 + (1/3)E_0 + (4/9)E_1. So (5/9)E_1 = 5/3 + (1/3)E_0, E_1 = 3 + (3/5)E_0. E_0 = 1 + 3 + (3/5)E_0 = 4 + (3/5)E_0. (2/5)E_0 = 4, E_0 = 10.

Question 24

What is the expected distance of any point on Earth and the north pole? Take Earth radius 1. Clarification: Shortest distance cuts through the sphere, instead of lying on surface. Further thinking: Is this question same as choosing two random points on unit sphere and asking their expected distance?

Accepted Answer

4/3. 2 * sin(x/2)is the distance of north pole from a point on the ring at angle x from the z axis. So, integrate from 0 to pi, (2 * pi * sin(x) * 2 * sin(x/2) dx) and divide by the total area which is 4 * pi. Answer:4/3 Another approach is to imagine a horizontal ring of dy thickness at distance y from N (north pole). Area of ring = 2 * pi * dy Probability of choosing point on this ring = dy/2 Distance of N & a point on ring = sqrt(2y) Exp length = integral (y=0 to 2) of sqrt(2y) dy/2 = 4/3 And yes, taking two random points on surface of sphere and asking their expected distance is same as this very question.

Question 25

A stick is broken into 3 pieces, by randomly choosing two points along its unit length, and cutting it. What is the expected length of the middle part?

Accepted Answer

1/3. Double integral of |x-y|dxdy gives 1/3 as answer. This is same as one would expect from a broken pencil. Palak's Solution: Integrate from 0 to 1, x * x/2 + (1-x) * (1-x)/2 = 1/3 logic: if one cut is at distance x from left, with probability x, the second cut comes before it, and expected length of middle piece is x/2.. Similarly with prob (1-x) it, middle piece is expected to have length (1-x)/2. Thus adding and integrating from 0 to 1.

Question 26

There are n letters and n envelopes. You put the letters randomly in the envelopes so that each letter is in one envelope. (Effectively a random permutation of n numbers chosen uniformly). Calculate the expected number of envelopes with the correct letter inside them.

Accepted Answer

1. Let X_i be the indicator random variable: 1 if the i-th letter ends up in the i-th envelope, 0 otherwise. E[X_i] = P(X_i = 1) = 1/n. By linearity of expectation, E[total correct] = sum of E[X_i] = n * (1/n) = 1.

Question 27

You have 100 noodles in your soup bowl. You are told to take two ends of some noodles (each end on any noodle has the same probability of being chosen) in your bowl and connect them. You continue until there are no free ends. What is the expected number of loops? What is the probability of making one large loop which includes every noodle?

Accepted Answer

First calculate expected number of loops: Denote X1 = identity variable, which takes value 1 every time a (long) noodle becomes a loop by connecting to its end, (if its own end is connected to some other noodle, then that other noodle's end will be considered). Thus, number of loops = X1 +...+ X100, thus, expected number of loops = E(X1) + ... + E(X100). This will give answer = 1/(2N-1) + 1/(2N-3) + ... + 1/3 + 1 where N = 100. This formula can also be proved using induction. Calculating Probability of single large loop: If you followed last result, this one is simple, this time all the identitiy variables are zero except last one. Thus probability is [1 - 1/(2N-1)] * [ 1 - 1/(2N-3)] * ... * [1-1/3] = Product ( i = 2 to N) [ (2i-2) / (2i-1)]

Question 28

Given the set of numbers from 1 to n: { 1, 2, 3 .. n } We draw n numbers randomly (with uniform distribution) from this set (with replacement). What is the expected number of distinct values that we would draw?

Accepted Answer

E(n) = n[1-(1-1/n)^n]. Notice that number of distinct points in the produced set is same as number of distinct points selected from the given set {1..n} . For that we denote indicator function Si, that is, Si = 1 if the interger i is taken into produced set. Let S denote the number of distinct points selected, S = S1+..+Sn E(S) = E(S1) +...+ E(Sn) and E(Si) = 1 * P(Si) = 1 - (Probability that 'i' is not chosen in any draw) = 1 - ( Prob that is i is not chosen in one 1st draw)^n = 1 - ( 1 - 1/n )^n Thus E(S) = n * ( 1 - ( (n-1) / n)^n )

Question 29

What is the expected number of cards that need to be turned over in a regular 52-card deck in order to see the first ace?

Accepted Answer

53/5. Define X1,X2…X48, such that Xi = 1 if ith card turns over before 4 aces, 0 otherwise. Thus, total number of cards turned to see first ace = 1 + sum (Xi) using linearity of expectation, E(X) = 1 + sum ( E(Xi)) Now consider the ith card and the four aces, all the orders are equally likely: X,A,A,A,A Or A,X,A,A,A, or 3 others. They are equally likely when we have no knowledge about position of these Aces with respect of the deck. Hence we have 5 equally likely slots for ith card: 1 A 2 A 3 A 4 A 5. We are interested in 1st slot. Hence E(Xi) = 1/5 for all i = 1 to 48 Thus E(X) = 1 + 48/5 = 53/5 = 10.6 This can be done by recursive equation f(n) = (4/n) * 1 + ((n-4)/n) * (1+f(n-1)), where f(n)=expected cards to flip in a deck of n cards, but I definitely can't solve this without a computer. A shorter explanation is to consider the 52 cards uniformly distributed over (0,1), so on average they're at k/53 for k=1,2,3,…,52. The four aces are on average at 1/5, 2/5, 3/5, 4/5. So 0.2=k/53 implies k=10.6, done!

Question 30

x and y are two random points selected uniformly between 0 and 1. Using them, create a point uniformly random in a circle of radius 1. (uniform means that the probability density is constant)

Accepted Answer

Set θ = 2πx and r = sqrt(y), then take the point as (r*cos(θ), r*sin(θ)). The sqrt is needed because the area element in polar coordinates is r*dr*dθ. If we used r = y directly, points would cluster near the center. Taking r = sqrt(y) ensures uniform distribution over the disk.

Question 31

N points are chosen at random on the circumference of a circle. A convex n-gon (n sided polygon) is drawn by joining these n points. What is the probability that the center of circle lies inside the region of the n-gon?

Accepted Answer

1 - n/2^(n-1). The polygon excludes the center if and only if all points lie within a semicircle. Fix one point; the probability that each of the remaining n-1 points lies in a specific semicircle is (1/2)^(n-1). Since any of the n points could define the semicircle, by inclusion-exclusion (no overlap for uniform points on a circle), P(all in some semicircle) = n * (1/2)^(n-1). So P(center inside) = 1 - n/2^(n-1).

Question 32

On pressing a button, a random number is generated uniformly between 0 & 1. You keep on generating these numbers until their sum exceeds 1. What is the probability that you need to press the button more than n times? What is the expected number of times you need to press the button?

Accepted Answer

Probability: 1/n! ; Expected times: e. Probability of more than n throws is equivalent to saying X1+..+Xn <=1, this is the volume of n dimensional symplex from origin, its volume is 1/n! and can be proved by induction. This can be used to calculate expected number of button presses, sum( 1/n! * n ) = sum( 1/(n-1)!) = e^1 Otherwise, set up a (lag)differential equation for f(x), the expected number of draws needed for the sum to exceed x. For x=0,f(x)=1. For x>0, suppose a draw gave number t, then f(x)= 1 + integral (t=0 to x)f(x-t)dt, which is same as f(x) = 1 + integral(t=0 to x) f(t)dt, differentiate wrt x, we get f'(x) = 0 + f'(x). This has the solution f(x)=e^x and x=1 gives e. To take derivative, we used leibniz integral rule: d/dx ( integral( a(x) to b(x) ) f(t) dt = f(b(x)) * b'(x) - f(a(x)) * a'(x) = f(x) * 1 - f(x) * 0 = f(x)

Question 33

You are initially located at the origin on the x-axis. You start a random walk with an equal probability of moving left or right, one step at a time. What is the probability that you will reach point a before reaching point -b? Assume a and b are natural numbers, a > 1, b > 1.

Accepted Answer

b/(a+b). Let P_x be the probability of reaching a starting from position x. We have P_a = 1 and P_{-b} = 0. By the random walk recurrence, P_x = (1/2)P_{x-1} + (1/2)P_{x+1}. The solution is linear: P_x = (x+b)/(a+b). At x=0: P_0 = b/(a+b).

Question 34

You are taking out candies one by one from a jar that has 10 red candies, 20 blue candies, and 30 green candies in it. What is the probability that there is at least 1 blue candy and 1 green candy left in the jar when you have taken out all the red candies? Assume that the candies of the same color are indistinguishable from one another.

Accepted Answer

7/12. Consider the last red candy drawn. The probability that the very last candy overall (position 60) is not red times... More directly: P(last red comes before last blue AND last green) = P(last among {last red, last blue, last green} is not red). The last red is at position 10 from the end among reds. Actually: P = (30/60)(20/30) + (20/60)(30/40) = (1/2)(2/3) + (1/3)(3/4) = 1/3 + 1/4 = 7/12.

Question 35

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 36

A & B are alternately picking balls from a bag without replacement. The bag has k black balls and 1 red ball. Winner is the one who picks the red ball. Who is more likely to win, the on who starts first, or second?

Accepted Answer

Begin!. A & B keep on picking the balls without looking at their color. After all balls have been picked, the one who starts the game will have more balls (if k=even, total balls=odd) and hence higher probability of winning. (prob of A winning in first chance = 1/(k+1)) + (prob of A winning in 3rd chance = (1-(1/k+1)) * (1-1/k) * 1/(k-1) = 1/(k+1)) + ... + (prob of A winning in 2r+1-th chance = 1/(k+1)) + ... . When k=2n+1, there are n+1 such terms, giving the prob as (n+1) * 1/(k+1)=1/2. When k=2n, there are n+1 such terms, giving prob as (n+1) * 1/(k+1)=(n+1)/(2n+1)>1/2. Hence, doesnt matter who starts first when k is odd. The first player has higher chance of winning when k is even

Question 37

A postman brought N letters to a house with two letter-boxes. Since the two boxes were empty, he puts 1 mail in each of the two mail boxes. Then he chooses one of boxes with probability proportional to number of letters present in that box, and puts the 3rd letter in it. He does this for all subsequent letters. What is the expected number of letters in the box with lower letters?

Accepted Answer

Suppose I have a stack of 2 black cards and one red card. Initially I put red between 2 black cards. Now I add black cards randomly between any two cards (so, initially it is either above or below red). Note that the probability that I add the card above the red card, when x-1 is the number of cards above red and y-1 is the number of cards below red is x/(x+y). Let the problem be: if red card is dividing the black cards into two sets, what is the expected number of black cards in the smaller section. So, we see that the two problems are equivalent. Now this way, we are getting all possible combinations in which one red and n black cards can be mixed, we see that the probability that the red card is at height h is independent of h. So, the probability that the smallest box contains n/2 letter or 1 letter (or any number of letters between 1 and n/2) are all same. So, expected number of letters in the smaller box is asymptotically n/4

Question 38

You have an opportunity to make one bid on an object, whose value to its owner is, as far as you know, uniformly random integer between $0 and $100. What you do know is that you are so much better at operating the widget than he is, that its value to you is 80% greater than its value to him. If you offer more than the widget is worth to the owner, he will sell it. But you get only one shot. How much should you bid? For example, if its actual value is $10, you bid & win at $11, and sell it for $18, making profit. But if you bid more than $18, you make lose! But since you don't know how much its actual price is, how do you bid in order to make some profit?

Accepted Answer

CSEBLOG: We should bet only 0! (dont bet!) Suppose I bet $x and get the widget. So, the value of it for the owner would be $y, uniformly distributed between 0 and x. So, its value for me is $1.8y. Expected value for me is 1.8* Expected value of y = 1.8*x/2= 0.9x So, if I get, expected value of the widget for me is 0.9x $ paying x $. If x is less, i.e I am not getting it, I did not gain/lose anything. So, overall I am losing. So, I should not bid.

Question 39

On a given highway, trucks arrive at the station according to a Poisson process with Lambda = 0.1/minute. This means that after a truck is just passed, the time for the next truck to arrive is an exponential random number with average arrival time of 10 minutes. Your car just broke on this highway, and you are waiting for the next truck for hitchhiking, what is your expected waiting time? On average how many minutes ago the last truck left?

Accepted Answer

Using memoryless property of exponential distribution, the expected waiting time is 10 minutes. This also holds backwards, hence the expected time last truck passed is also 10 minutes. But this does not violate the total inter-arrival time of 10 minutes, because if your car breaks at a random time, you are more likely to be in long interval than a short one. I don't have time today, remind me to post a more mathematical explanation.

Question 40

A very sharp, consistently skillful blind archer aimed for the center of a circular board and shot 2 arrows. He is expected to hit the aim, but doesn't hit it for sure. The archer is told that his first shot is better than second. He tried one more shot. What is the probability that this 3rd shot is the best shot among 3? (ie, Probability that 3rd arrow lands closer to center than his first two shots?)

Accepted Answer

1/3. Suppose x1, x2 and x3 are the distances of the arrows from center. As the archer is consistent, we can use symmetry, i.e., there are six equally likely cases: x1<x2<x3, (or 5 others, which are its permutations). Since archer is told that x1<x2, we are left with following equally likely cases: x1<x2<x3 , x2<x1<x3 , x1<x3<x2 Among these three, one is favorable. Hence The probability of last shot being best is 1/3. Notice that if there were (N-1) arrows, with first being better than rest, and then he shoots Nth arrow. The probability that Nth shot is best is 1/N.

Question 41

In a room stand N armed and angry people. At each chime of a clock, everyone simultaneously spins around and shoots a random other person. The persons who get shot fall dead and the survivors spin and shoot again at the next chime. Eventually, either everyone is dead or there is a single survivor. As N grows, what is the limiting probability that there will be a survivor? Warning: This is an open problem.

Accepted Answer

The probability appears to converge to approximately 1/2 as N grows, based on simulation evidence, but no closed-form proof exists. Simulations show an oscillating pattern with expanding wavelength that converges toward 0.5.

Question 42

The people in a country are partitioned into clans. In order to estimate the average size of a clan, a survey is conducted where 1000 randomly selected people are asked to state the size of the clan to which they belong. How does one compute an estimate average clan size from the data collected? Solution This is more of a puzzle-to-ponder rather than a puzzle to learn. In my opinion, best estimator for average is sum( n )/ sum( #n/n), where #n is the number of people with clan size as 'n', and this sum is over all the values of 'n' we receive.

Accepted Answer

This is more of a puzzle-to-ponder rather than a puzzle to learn. In my opinion, best estimator for average is sum( n )/ sum( #n/n), where #n is the number of people with clan size as 'n', and this sum is over all the values of 'n' we receive.

Question 43

You are given an urn with 100 balls (50 black and 50 white). You pick balls from urn one by one without replacements until all the balls are out. A black followed by a white or a white followed by a black is "a colour change". Calculate the expected number of colour-changes if the balls are being picked randomly from the urn.

Accepted Answer

50. There are 99 possible color-change positions (between consecutive draws). Let X_i be an indicator variable, 1 if the i-th and (i+1)-th balls differ in color, 0 otherwise. E[X_i] = P(different colors at positions i and i+1). By symmetry, this equals 1 - P(same color) = 1 - (50*49 + 50*49)/(100*99) = 1 - 4900/9900 = 1 - 49/99 = 50/99. By linearity of expectation, E[S] = 99 * (50/99) = 50.

Question 44

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 45

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 46

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 47

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 48

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 49

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 50

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 51

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 52

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 53

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 54

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 55

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 56

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 57

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 58

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 59

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 60

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 61

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 62

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 63

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 64

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.

Quantitative Finance

Quantitative Finance

Quantitative Finance

Quantitative Finance