In this article
The four moves behind most answers
Four moves solve most of these questions, in order: write down the equally likely outcomes, condition on what you are told, pick a shortcut such as the complement, symmetry or linearity of expectation, and check the answer with a small simulation. The first group below leans on the second and third moves.
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- Write down the equally likely outcomes, then condition on what you are told.
- Pick a shortcut such as the complement, symmetry or linearity of expectation.
- Check the answer with a small simulation.
Counting and basic chance
The probability and statistics hub lists the neighbouring topics for these basics.
What is the chance of at least one six in four rolls of a die?
About 51.8%, exactly 671 in 1,296.1 Use the complement, the probability that the event does not happen: no six on one roll is 5 in 6, so no six in four rolls is 5 in 6 to the fourth power, which is 625 in 1,296. Subtract from one. Adding four sixths gives 2 in 3, which is wrong because the events overlap.
Why it is asked: it tests whether you reach for the complement instead of adding overlapping cases. Read the full explanation: Grinstead and Snell, chapter 1.
What is the probability that one fair die shows a higher number than another?
Five in twelve, about 41.7%.1 Of the 36 equally likely pairs, 6 are ties, so 30 are decided, and by symmetry the two dice win equally often: 15 each. The chance that the first die is higher is 15 in 36, which reduces to 5 in 12, and a tie has probability 1 in 6.
Why it is asked: it checks that you use symmetry on the decided outcomes instead of listing all pairs. Read the full explanation: Grinstead and Snell, chapter 1.
What is the probability of an odd number of heads in 200 fair coin flips?
Exactly one half.1 Fix the first 199 flips. Whatever count they give, the last coin makes the total odd with probability one half, so the answer does not depend on 200 at all. Summing the binomial terms exactly gave one half for 1, 2, 3, 10 and 200 flips.
Why it is asked: it separates candidates who start summing 100 binomial terms from those who find the symmetry. Read the full explanation: NIST e-Handbook, binomial distribution.
How many people must be in a room for a shared birthday to be more likely than not?
Twenty-three.1 Compute the chance that all birthdays differ: the second person avoids the first with probability 364 in 365, the third avoids both with 363 in 365, and so on. At 23 people the chance all differ is 0.4927, so a shared birthday has probability 0.5073. At 22 it is 0.4757. This ignores leap years and assumes birthdays are equally likely. By 57 people the chance reaches 0.9901.
Why it is asked: intuition says about 183, and the gap shows how pairs, not people, drive the result. Read the full explanation: Grinstead and Snell, birthday problem.
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- The chance of a shared birthday is 0.1169 at 10 people and 0.5073 at 23 people.
- It reaches 0.9901 at 57 people.
You have a coin that lands heads 70% of the time and need a fair coin flip. How do you get one?
Flip the coin twice. Heads then tails counts as heads, tails then heads counts as tails, and any other result means flip twice again.1 The two mixed outcomes have the same probability, 0.7 times 0.3, whatever the bias, so the output is fair. Each pair succeeds with probability 0.42, so it takes about 4.76 flips per fair bit. A seeded run of 200,000 bits gave 50.1% heads and 4.77 flips.
Why it is asked: it tests whether you can build fairness from symmetry without knowing the bias. Read the full explanation: Grinstead and Snell, chapter 1.
Knowing it is half the job. Saying it is the other half.
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Start Free Trial →Conditional probability
A family has two children and at least one is a girl. What is the chance both are girls?
One in three.1 List the four equally likely pairs in birth order: girl-girl, girl-boy, boy-girl and boy-boy. Knowing at least one girl removes boy-boy and leaves three, of which one is girl-girl. If you are told the older child is a girl, only two pairs remain and the answer is one half. The wording of the information changes the answer.
Why it is asked: it checks that you condition on exactly what was said. Read the full explanation: Grinstead and Snell, conditional probability.
A bag holds one fair coin and one two-headed coin. You pick one at random and flip heads. What is the chance it is the two-headed coin?
Two in three.1 Conditional probability is the chance of an event given that another has happened. Heads from the two-headed coin has probability one half: you pick that coin with probability one half, and it always shows heads. Heads from either coin has probability three quarters, one half from the two-headed coin plus one quarter from the fair coin. Divide one half by three quarters and you get two thirds.
Why it is asked: it is the smallest Bayes-style update. Read the full explanation: Bayes theorem interview questions.
A jar holds 1,000 coins and one has two heads. You pick one at random and flip ten heads in a row. What is the chance it is the two-headed coin?
1,024 in 2,023, about 50.6%.1 Before flipping, the two-headed coin has 1 chance in 1,000. A fair coin shows ten heads with probability 1 in 1,024. Weigh the evidence: the trick coin counts 1 times 1, the 999 fair coins together count 999 times 1 over 1,024. Odds of 1 to 999 over 1,024 are 1,024 to 999. Ten heads turn a 1-in-1,000 long shot into a coin flip.
Why it is asked: it shows how fast evidence beats a small prior. Read the full explanation: Bayes theorem interview questions.
Three bowls hold two blue balls, two red balls, and one of each colour. You draw a blue ball from a random bowl. What is the chance the next ball from that bowl is blue?
Two in three, not one half.1 Label the three blue balls. Drawing a blue ball picks one of them with equal chance: two are in the all-blue bowl, one is in the mixed bowl. So the bowl is all blue with probability two thirds, and then the second ball is blue for certain. If it is the mixed bowl, the second ball is red.
Why it is asked: it tests counting balls, not bowls. Read the full explanation: Grinstead and Snell, conditional probability.
A person goes to the gym every Monday and Tuesday and on each other day with probability 30%. Spotted at the gym on a random day, what is the chance it is Monday?
Two in seven, about 28.6%.1 Weigh each day by how likely a gym visit is: Monday counts 1, Tuesday 1 and each of the five other days 0.3, so the total is 3.5. Monday's share is 1 divided by 3.5, which is 2 over 7. Without the information, Monday would be 1 in 7.
Why it is asked: it checks that you weight the days by their chance of the evidence. Read the full explanation: Bayes theorem interview questions.
Expectation and waiting times
What is the expected number of rolls until the first six?
Six.1 A fair die shows a six with probability one in six on each roll, and the wait for the first success has an expected length equal to the reciprocal of that probability. The expected value is the probability-weighted average of the possible outcomes. Note that the chance of a six within six rolls is only 66.5%, so the average is not a promise.
Why it is asked: it tests the geometric distribution and the gap between a mean and a likely outcome. Read the full explanation: Grinstead and Snell, chapter 6, expected value.
What is the expected number of people who get their own hat back when hats are shuffled at random?
One, whatever the number of people.1 Linearity of expectation says the expected value of a sum is the sum of expected values, even when the parts are dependent. Each person gets their own hat with probability one over n, and there are n people. Checking all 720 shuffles of six hats gave 720 matches in total, a mean of exactly one.
Why it is asked: it is the standard demonstration that linearity needs no independence. Read the full explanation: Grinstead and Snell, chapter 6, expected value.
Numbers 1 to 5 are placed at random around a circle. What is the expected number that are smaller than both neighbours?
Five thirds, about 1.67.1 Take one position and its two neighbours. Among these three numbers, each is equally likely to be the smallest, so the middle one is the smallest with probability one third. By linearity of expectation, add the five positions: five times one third. Checking all 120 arrangements gave 200 in total, a mean of 5 over 3.
Why it is asked: it is linearity of expectation with indicators on dependent positions. Read the full explanation: Grinstead and Snell, chapter 6, expected value.
How many rolls does it take on average to see all six faces of a die?
14.7 rolls, exactly 147 over 10.1 Split the wait into stages. Getting the first new face takes 1 roll, the second takes 6 over 5 rolls on average, the third 6 over 4, and so on up to 6 over 1 for the last. Add the six stage means: 6 times the sum of 1 over k for k up to 6.
Why it is asked: this coupon-collector problem tests breaking a wait into geometric stages. Read the full explanation: Grinstead and Snell, chapter 6, expected value.
You keep rolling a die. What is the expected number of rolls until six sixes in a row?
55,986.1 Each time a non-six breaks the streak you start over. Working from the end, the wait for a streak of one is 6 rolls, for two it is 6 plus 36, which is 42, for three 258, and in general it is the sum of 6 to the power j for j up to the streak length. For six sixes that sum is 55,986. A seeded run of 200,000 trials of the three-in-a-row case gave 258.6 against the exact 258.
Why it is asked: it tests recursion on the streak length instead of brute force. Read the full explanation: Grinstead and Snell, chapter 6, expected value.
In fair coin flips, which comes first more often, heads-heads-heads or heads-tails-heads?
Heads-tails-heads comes first 3 times in 5, so heads-heads-heads wins only 2 in 5.1 Track only the latest flips that still matter. After heads-heads, a tail leaves you at heads-tails with a fresh chance at the rival, but after heads-tails a head ends the game against heads-heads at once. A seeded run of 400,000 games gave heads-heads-heads first in 39.96%. Two patterns equally likely at any position still need not be equally likely to come first.
Why it is asked: it shows that overlap in a pattern changes who wins a race. Read the full explanation: Grinstead and Snell, chapter 6, expected value.
Random numbers between 0 and 1 are added one at a time until the total passes 1. What is the expected number of numbers added?
The number e, about 2.718.1 The chance that the first n numbers still total less than 1 is 1 over n factorial, the volume of a corner of the unit cube. The expected count is the sum of those chances for n from 0 up, which is the series for e. A seeded run of one million trials gave 2.7191. The textbook poses the same experiment in section 6.1, exercises 32 and 33.
Why it is asked: it is a classic where the answer is a famous constant and the proof is a volume argument. Read the full explanation: Grinstead and Snell, section 6.1.
A coin starts a prize at 1 and doubles it on every head until the first tail, when you collect. What would you pay to play?
Fair value in pure expected terms is infinite, but a sensible answer is a few units.1 The prize is 2 to the power k with probability 1 over 2 to the power k plus 1, so every k adds one half to the average and the sum never ends. This is the St. Petersburg paradox, Example 6.3 in the textbook. In a million seeded plays the median prize was 2 and 93.75% of games paid 8 or less, while the mean of 10.11 was lifted by rare large wins. If the bank can pay at most 2 to the 20th, about a million, the fair value is just 11.
Why it is asked: it tests whether you can say what an average hides and when to stop trusting it. Read the full explanation: Grinstead and Snell, Example 6.3.
You pick a person at random, then keep picking until someone is taller than the first. What is the expected number of additional picks?
Infinite.1 The first person is the tallest of the first n plus one people with probability 1 over n plus 1, which is the chance you are still waiting after n picks. These chances form the harmonic series, which has no finite sum. Half the time one pick is enough, and the median wait was 2, yet the average is unbounded. If you stop after at most 100 picks, the average is 5.19, and a seeded run gave 5.21.
Why it is asked: it tests whether you notice an expectation that does not exist instead of quoting a number. Read the full explanation: Grinstead and Snell, chapter 6, expected value.
Stopping and geometric puzzles
If a one-metre stick is broken at two random points, what is the probability the three pieces form a triangle?
One quarter.1 Three lengths form a triangle exactly when each is shorter than half the stick. In the square of possible break points that region is one quarter of the area. A seeded simulation of one million sticks gave 0.2505.
Why it is asked: it tests turning a physical story into an area. Read the full explanation: Grinstead and Snell, chapter 2, continuous probability.
What is the chance that a random chord of a circle is longer than the side of the inscribed equilateral triangle?
It depends on how the chord is chosen: one third with random endpoints, one half with a random distance from the centre, one quarter with a random midpoint.1 This is Bertrand's paradox, which the textbook notes is not truly a paradox but an ambiguous question. Simulations gave 0.333, 0.501 and 0.251. The phrase random chord is ambiguous until the selection method is stated.
Why it is asked: it tests whether you ask what random means before you answer. Read the full explanation: Grinstead and Snell, Bertrand's paradox.
Text description of the diagram
- Random endpoints give 0.333, a random radius distance gives 0.501 and a random midpoint in the disc gives 0.251.
- The phrase random chord is ambiguous until the selection method is stated.
You roll a die repeatedly, keeping a running total. What is the probability that the throw that takes the total past 100 is a 2?
About 9.5%, close to 2 in 21.1 A throw is more likely to be the one that crosses a high line if it is big, in proportion to its size, so face j has chance j over 21, where 21 is the sum of the faces. A dynamic programme over every total up to 100 gave 0.095238 for a 2, and six on top at 0.2857. A seeded run of 400,000 games gave 0.0944.
Why it is asked: it tests the idea that the crossing step is size-biased, not an ordinary roll. Read the full explanation: Grinstead and Snell, chapter 6, expected value.
One player wins each point with probability 30%. At 9-9 in a game won by two clear points, what is the chance this player wins?
9 in 58, about 15.5%.1 From a tie, the player wins by taking the next two points, with probability 0.09, or loses by dropping them, with probability 0.49, and otherwise returns to a tie. So the chance of winning is 0.09 over 0.09 plus 0.49, which is 9 over 58. A seeded run of 400,000 games gave 0.1554. A modest edge per point gets amplified in the long game.
Why it is asked: it checks that you cancel the looping back to a tie instead of summing a long series. Read the full explanation: Grinstead and Snell, chapter 12, random walks.
Three cards read n, n plus 1 and n plus 2. You flip one, see it, and may keep it or flip a card from the other two. What is the best strategy?
Keep the highest card, switch on the lowest, and either choice on the middle one: the average value is n plus 1.5.1 This assumes you know n. If you see n, the other two average n plus 1.5, so switch. If you see n plus 2, nothing beats it, so keep. If you see n plus 1, the other two average n plus 1, so you are indifferent. Never switching averages n plus 1, so knowing n adds half a unit. A seeded run of 400,000 games gave 1.4995 for n equal to 0.
Why it is asked: it tests defining the strategy as a rule on each possible observation. Read the full explanation: Grinstead and Snell, chapter 6, expected value.
Follow-up questions
Which probability interview questions should I practise first?
Start with the complement, conditioning and linearity of expectation, because each unlocks many problems: the six-in-four-rolls, two-children and hat questions above are the smallest examples of each. Add the waiting-time and stopping puzzles once those feel automatic.
How do I prepare for probability and statistics interview questions?
Work problems aloud, state the sample space before computing, and check each answer by a quick simulation or a limiting case. The probability and statistics hub and the how to answer interview questions page cover structure.
How do I prepare for the probability maths in an equity research interview?
Expect fair-price and expected-value games, simple Bayes updates and the difference between correlation and causation. Practise the prize-doubling question above, see where such questions sit in the investment banking guide and read the page on correlation versus causation for the statistics side.
Do I need to memorise formulas for probability interviews?
Few. Most answers rest on counting, the complement, conditioning and linearity. The numbers worth knowing cold are the die mean of 3.5, the geometric wait of one over the success chance, and the binomial count for coin flips. The mean versus median page covers another common distinction.
- Write the outcomes and the information given before you compute anything.
- The complement, symmetry and linearity of expectation solve most of these problems without long sums.
- Expectation gives an average, not a likely outcome: six rolls on average for a six, yet only 66.5% within six rolls, and some waits have no finite average at all.
- Ambiguous wording, such as a random chord or two children, changes the answer, so ask what random means.
- Check each answer with a small simulation or a limiting case.
Sources
- Grinstead and Snell's Introduction to Probability (The CHANCE Project, free version dated 4 July 2006 of the AMS second edition, 2003) — Dartmouth College / American Mathematical Society (GNU Free Documentation License version), 4 Jul 2006
- 1.3.6.6.18. Binomial Distribution, NIST/SEMATECH e-Handbook of Statistical Methods — National Institute of Standards and Technology (read 2026-10-24), 24 Oct 2026
- fractions: Rational numbers, Python 3 documentation — Python Software Foundation (page shows last updated Oct 24, 2026), 24 Oct 2026
