Top 20 Probability Puzzles for Data Science Interviews

Explanation: This is the classic problem the Chevalier de Mere lost money on. The exact probability is the complement of never rolling a six: $1 - (5/6)^4 \approx 0.5177$. Our estimate of 0.5175 sits within Monte Carlo noise of that value. The key modelling move is to compute the complement of "no sixes" rather than trying to add up the messy overlapping cases of one, two, three, or four sixes.

Explanation: The number of rolls until the first six is a geometric random variable with success probability $p = 1/6$, and its expectation is $1/p = 6$. The simulated 5.986 confirms it. A common interview trap is to guess "about 3" by confusing the median with the mean; the geometric distribution is right-skewed, so its mean sits above its median. The vectorised shortcut mean(rgeom(1e5, 1/6) + 1) gives the same answer faster.

Explanation: Of the 36 equally likely ordered pairs, six of them sum to seven: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). So the exact probability is $6/36 = 1/6 \approx 0.1667$, matching the simulated 0.1662. Seven is special because it is the only total reachable from every face of the first die, which is why it is the most common roll and the anchor of the game of craps.

Explanation: The result surprises almost everyone: just 23 people give better-than-even odds of a shared birthday. The exact probability is $1 - \frac{365 \times 364 \times \cdots \times 343}{365^{23}} \approx 0.5073$, which our 0.5075 confirms. Intuition fails because people count comparisons to themselves, but the relevant quantity is the number of pairs, $\binom{23}{2} = 253$, and each pair is a fresh chance to collide.

Explanation: The simulation lands on 23, the textbook answer, because the probability first crosses one half at exactly that group size: the exact analytic value is about 0.507, and our 20,000-room estimate reads 0.5109 while 22 people fall just short. Sweeping a parameter and locating a threshold crossing is a reusable pattern: define a function of the parameter, vectorise it with sapply(), and use which() to find the first index that satisfies your condition.

Explanation: Seeing the whole curve dissolves the paradox: the probability rises steeply, passes 0.5 near 23 people, and is already above 0.97 by 50. The steepness comes from the pair count growing quadratically with group size, so each extra person adds many new comparisons. Plotting a simulated quantity across a parameter range, rather than reporting a single number, is often what turns a surprising result into an obvious one for a stakeholder.

Explanation: Switching wins twice as often, close to the exact $2/3$ versus $1/3$. The reason is that your first pick is right only $1/3$ of the time, so $2/3$ of the time the car is behind one of the other two doors, and the host obligingly removes the empty one, concentrating that whole $2/3$ onto the single remaining door. The host's knowledge is the hinge: if he opened a door at random and happened to miss the car, the advantage would vanish.

Explanation: The answer is $1/3$, not the tempting $1/2$. Of the four equally likely families BB, BG, GB, GG, the condition "at least one boy" removes only GG, leaving three cases, and just one of them (BB) has two boys. The trap is treating the second child as if it were independent of the information given; conditioning on "at least one boy" is a statement about the pair, not about one specific child.

Explanation: Now the answer is $1/2$, confirmed by the simulated 0.4997. Naming a specific child ("the older one") makes the other child's sex genuinely independent, so it is a boy half the time. The lesson that trips up candidates is that "at least one is a boy" and "the older one is a boy" are different pieces of evidence: the first leaves three families in play, the second leaves two, and that difference is the whole gap between $1/3$ and $1/2$.

Explanation: The exact answer is the famously counterintuitive $13/27 \approx 0.4815$, which our 0.4787 estimates. Counting ordered (sex, weekday) pairs where at least one is a Tuesday boy gives 27 configurations, and 13 of them have two boys. Adding a specific-sounding detail pushes the probability from $1/3$ toward $1/2$ because it makes the two-boy families easier to satisfy in two independent ways; the more specific the condition, the closer you drift to the "named child" answer.

Explanation: Most people, including many clinicians, guess about 95%, but the true posterior is only about 17%. Bayes gives $\frac{0.01 \times 0.99}{0.01 \times 0.99 + 0.99 \times 0.05} \approx 0.167$, matching the simulated 0.1645. With such a rare disease the healthy group is so large that its 5% false positives (about 49,500 people) swamp the 990 genuine positives. The base rate dominates, which is why a single positive on a rare-disease screen is rarely a diagnosis on its own.

Explanation: A second independent positive transforms a weak 17% into a strong 79%. The exact posterior is $\frac{0.01 \times 0.99^2}{0.01 \times 0.99^2 + 0.99 \times 0.05^2} \approx 0.798$, and 0.7926 confirms it. Independence is doing the heavy lifting: a healthy person clears both 5% hurdles only 0.25% of the time, so a double positive is far harder to fake by chance. This is the quantitative reason confirmatory testing works, and why interviewers love it as a follow-up to the single-test trap.

Explanation: The intuitive answer of $1/2$ is wrong; the correct probability is $2/3$, confirmed by 0.6674. The mistake is conditioning on the box when you should condition on the coin. Drawing a gold coin is twice as likely from the all-gold box as from the mixed box, so among all gold draws, two-thirds come from the box whose partner is also gold. Counting individual coins rather than boxes is the same subtle shift that makes Monty Hall work.

Explanation: The exact expectation is $6 \times (1 + \tfrac12 + \tfrac13 + \tfrac14 + \tfrac15 + \tfrac16) = 6 \times 2.45 = 14.7$, which the simulated 14.681 confirms. The intuition is that collecting the first new coupon is instant, but the last one takes an average of six packs on its own because you rarely hit the single missing type. That final stretch is why complete-the-set promotions are so much more expensive than they look.

Explanation: The histogram explains why a single average can mislead: the distribution is strongly right-skewed, with most shoppers finishing around 12 to 13 packs but a heavy tail of unlucky collectors needing 30, 40, or more. The mean of 14.7 sits to the right of the mode, dragged up by that tail. Whenever a waiting-time process has a rare but costly worst case, plotting the full distribution rather than quoting the mean is what keeps a plan realistic.

Explanation: The exact expected wait for HH is 6 flips, confirmed by 6.024. The surprise is that different patterns take different times even though each is equally likely on any given pair: HT has expected wait 4, while HH has 6. The gap exists because a failed HH attempt (getting HT) throws you all the way back, whereas overlapping patterns can partially reuse progress. This waiting-time asymmetry underlies Penney's game, where the second player can always pick a pattern that beats the first.

Explanation: The expected number of matches is exactly 1 no matter how many guests there are, and 1.005 confirms it. The clean way to see this is linearity of expectation: each guest recovers their own hat with probability $1/10$, and summing those ten probabilities gives 1, without ever untangling how the matches depend on each other. This decomposition trick, turning a hard joint question into a sum of easy marginal ones, is one of the most reused tools in interview probability.

Explanation: For a fair game the probability of hitting the target before ruin is simply the starting fraction of the way there, $k/N = 3/10 = 0.3$, matching the simulated 0.298. The elegant part is that the answer ignores the path entirely and depends only on the endpoints. It also explains why a small bankroll chasing a large target usually busts: starting at 3 out of 10, you reach the goal less than a third of the time even with perfectly fair odds.

Explanation: The exact expected duration of a fair gambler's ruin is $k(N - k) = 3 \times 7 = 21$ bets, confirmed by 20.99. The formula is pleasingly symmetric: games starting near the middle of the range last longest, while games starting near a boundary end quickly. Interviewers pair this with the win-probability question because it shows two very different quantities, the chance of winning and the time it takes, can both come out of the same simple random walk.

Explanation: The theoretical expected payout is infinite, because a payout of $2^k$ occurs with probability $(1/2)^{k+1}$, so every term contributes $1/2$ to the sum forever. The simulation shows why that is not a bug: the average keeps climbing as you add plays (7.45 at ten thousand plays, 10.11 at a million), dragged up by rare enormous jackpots rather than settling down. It is the standard demonstration that a distribution with no finite mean has a sample average that never converges, which is why nobody would pay a large fixed fee to play despite the "infinite" expectation.