Make a prediction before calculating. Then open the solution and find the detail that defeats your first intuition. Dice and coins in these models are fair, and repeated trials are independent unless stated otherwise.
P(A) denotes the probability of event A; P(A | B) means the probability of A given B. The complement is the opposite event: P(not A) = 1 − P(A). Expected value is a theoretical average over many repetitions, not a promise about one experiment.
Ten challenges
Monty Hall: should you switch doors?
Three doors hide a prize and two goats. The prize is equally likely to be behind each door. You pick one. The host knows where the prize is, always opens another door hiding a goat and always offers a switch; when two doors are available, the host chooses randomly. Should you switch?
Show solution and reasoning
Yes: switching wins with probability 2/3; staying wins with 1/3. Your initial choice is correct in one case out of three, and switching loses then. In the other two cases you initially choose a goat, so the host must leave the prize door closed: switching wins.
The two remaining doors are not equivalent: the eliminated one was chosen using information. A host who opens a door randomly without knowing where the prize is defines a different experiment.
Further reading: Wolfram MathWorld.
Birthdays: are 23 people enough?
In a group of 23 people, what is the probability that at least two share a birthday? Assume independent birthdays uniformly distributed over 365 days, with no leap day.
Show solution and reasoning
Calculate the complement: all birthdays different. The second person must avoid one day, the third two, and so on.
P(at least one match) = 1 − (365/365)(364/365)…(343/365) ≈ 0.5073 = 50.73%Any pair counts: there are
23 · 22 / 2 = 253pairs. We cannot simply add 253 probabilities, because the events overlap. Matching one specified birthday is a different problem.Further reading: Wolfram MathWorld.
Bertrand’s three boxes
Three boxes contain two gold coins, two silver coins, and one of each respectively. Choose a box uniformly, then one of its two coins uniformly. The coin drawn is gold. What is the probability that the other coin is also gold?
Show solution and reasoning
2/3, not 1/2. Before observing the coin, each of the six physical coins has probability 1/6 of being drawn. Of the three possible gold coins, two come from the all-gold box and one from the mixed box. These three cases remain equally likely after observing gold.
Counting the two possible boxes is not enough: the all-gold box is twice as likely to produce the observation.
Two children: at least one is a boy
In a simplified model, each child is a boy or a girl with probability 1/2, independently. Select a family uniformly from families with exactly two children and at least one boy. What is the probability that both are boys?
Show solution and reasoning
1/3. In birth order, BB, BG, GB and GG were equally likely. The selection rule excludes only GG; three cases remain, with just one favourable case.
If instead we know the older child is a boy, only BB and BG remain: the answer is 1/2. Meeting a randomly selected child is also a different protocol. “I have a son” alone does not specify how the information was obtained.
De Méré: two apparently equivalent bets
Which is more likely: at least one six in four rolls of one die, or at least one double six in twenty-four rolls of two dice? All dice have six sides.
Show solution and reasoning
Use the probability of never succeeding:
P₁ = 1 − (5/6)⁴ ≈ 51.77% P₂ = 1 − (35/36)²⁴ ≈ 49.14%The first exceeds 50%; the second does not. Yet their expected numbers of successes agree:
4/6 = 24/36 = 2/3. The average number of successes is not the probability of at least one: an experiment can contain several successes.Further reading: Wolfram MathWorld.
The cloakroom: nobody gets their own hat
Ten people hand in ten distinct hats. They are returned according to a permutation chosen uniformly from all 10! possibilities. What is the probability that nobody receives their own hat?
Show solution and reasoning
About 36.79%. It is not
(9/10)¹⁰: assignments are dependent because each hat is handed out only once.Inclusion-exclusion gives:
P(no correct hat) = 1 − 1/1! + 1/2! − 1/3! + … + 1/10! = 1334961/3628800 ≈ 0.367879Here n! is the product of integers from 1 to n, and 0! = 1. For n people the alternating sum ends at 1/n!; as n increases it tends to
1/e ≈ 0.367879, not zero.Further reading: Wolfram MathWorld.
Two heads or heads-tails: the same wait?
Toss a coin until HH (heads-heads) first appears. In a separate experiment wait for HT (heads-tails). Count every toss from the start, including the final toss. Are the average waiting times equal, since either pattern has probability 1/4 in a specified pair of tosses?
Show solution and reasoning
No: HH takes 6 tosses on average, HT only 4. For HH, let E be the initial expected wait and F the remaining wait after a head:
E = 1 + E/2 + F/2 F = 1 + E/2 ⇒ E = 6After a head, a tail resets progress. For HT, another head instead leaves a useful head:
E = 1 + E/2 + F/2 F = 1 + F/2 ⇒ F = 2, E = 4The 1 counts the next toss; completing the pattern leaves zero remaining wait. Successive two-toss windows overlap, so they are not independent trials.
Simpson: better in every group, worse overall
Two methods A and B tackle easy and hard tasks. These invented data show successes out of attempts:
Tasks A B Easy 81/90 = 90% 10/10 = 100% Hard 2/10 = 20% 27/90 = 30% B has the higher rate in both groups. Can A have the higher overall rate?
Show solution and reasoning
Yes: A totals 83/100 = 83%, B 37/100 = 37%. Do not take the unweighted average of percentages: add successes and attempts.
A mostly tackled easy tasks; B mostly tackled hard ones. Group weights differ, allowing the ranking to reverse. These descriptive rates alone do not establish which method causes better results: we need information about task assignment.
Further reading: Wolfram MathWorld.
Gambler’s ruin: is a fair game enough?
You start with 10 tokens; your opponent has 90. Each independent toss of a fair coin transfers one token from one player to the other. Stop when one player owns all 100 tokens. Do you have a 50% chance of winning everything?
Show solution and reasoning
No: your chance of winning everything is 10%, and of losing your tokens 90%. Let u(i) be the probability of reaching 100 before 0 starting with i:
u(0) = 0, u(100) = 1 u(i) = [u(i − 1) + u(i + 1)]/2 ⇒ u(i) = i/100 ⇒ u(10) = 0.10The recurrence makes successive differences equal, so the solution is a straight line. Each toss is fair, but the starting balances are asymmetric. Your expected final balance remains 10:
0.10 · 100 + 0.90 · 0 = 10. This is not gambling advice.Further reading: Wolfram MathWorld.
Buffon’s needle: finding π by dropping a needle
Parallel lines on a plane are spaced distance d apart. A needle of length ℓ ≤ d falls with uniform orientation and a centre uniformly distributed relative to the lines, independently of its angle. If ℓ = d, what is the probability of touching or crossing a line?
Show solution and reasoning
2/π ≈ 63.66%, not 50%. Let x be the distance from the centre to the closest line, uniform between 0 and d/2. Let θ be the acute angle with the lines, uniform between 0 and π/2. The needle meets a line when
x ≤ (ℓ/2) sin θ.P = (2/π) ∫₀^(π/2) (ℓ/d) sin θ dθ = 2ℓ/(πd)With N independent drops and K crossings, estimate
π ≈ 2ℓN/(dK), provided K > 0. This is a random estimate, not an exact identity: few drops or nonuniform orientations can give poor results. It connects probability to Monte Carlo methods.Further reading: Wolfram MathWorld.
The common thread is the model: how we select, what information we receive and when we stop. Simulations can check calculations, but must reproduce these exact rules.