Probability interview questions

The complement is three tails, probability $\tfrac18$. So $1-\tfrac18=\tfrac78$.

Six of the 36 outcomes sum to 7: $(1,6),(2,5),\dots,(6,1)$. So $6/36=\tfrac16$.

Sums of 10, 11, 12 occur in $3+2+1=6$ ways out of 36.

$\tfrac16(1+2+\dots+6)=\tfrac{21}{6}=3.5$.

The count is geometric with success probability $p=\tfrac16$, so the mean is $1/p=6$.

Let $E_0$ be the expected remaining rolls when the last roll was not a six, $E_1$ when it was. Then $E_0=1+\tfrac56E_0+\tfrac16E_1$ and $E_1=1+\tfrac56E_0$. Substituting gives $\tfrac1{36}E_0=\tfrac76$, so $E_0=42$.

With states "no progress" ($E_0$) and "last toss H" ($E_1$): $E_0=1+\tfrac12E_1+\tfrac12E_0$, $E_1=1+\tfrac12E_0$. Solving gives $E_0=6$.

Wait for the first H (expected 2 tosses), then wait for the first T after it (another 2). Once you have an H, a later T always completes HT, so $2+2=4$. It is less than HH because HT never loses progress.

Equally likely families: BB, BG, GB, GG. "At least one boy" removes GG and leaves three, of which one is BB.

Only BB and BG remain possible, so the answer is $\tfrac12$. The information is about a specific child, which changes the conditioning.

Your first pick is right with probability $\tfrac13$. Switching wins exactly when it was wrong, which happens with probability $\tfrac23$.

$P(\text{all distinct})=\prod_{k=0}^{n-1}\tfrac{365-k}{365}$ first drops below $\tfrac12$ at $n=23$, where it is about $0.493$.

A triangle forms iff every piece is shorter than $\tfrac12$. In the unit square of break points this region has area $\tfrac14$.

$P(\max\le t)=t^2$, so the density is $2t$ and $\mathbb E=\int_0^1 2t^2\,dt=\tfrac23$.

$\min+\max=X+Y$, so $\mathbb E[\min]=1-\tfrac23=\tfrac13$.

For $n$ uniforms $\mathbb E[\max]=\tfrac{n}{n+1}$. The $n$ points split $[0,1]$ into $n+1$ gaps of equal expected length.

$|X-Y|=\max-\min$, so $\mathbb E=\tfrac23-\tfrac13=\tfrac13$.

For $n$ points the answer is $n/2^{n-1}$: each point can be the "first" of a clockwise semicircle containing the rest, with probability $2^{-(n-1)}$, and these events are disjoint. For $n=3$, $\tfrac34$.

Using $n/2^{n-1}$ with $n=4$ gives $4/8=\tfrac12$.

Coupon collector: $6\left(1+\tfrac12+\tfrac13+\tfrac14+\tfrac15+\tfrac16\right)=6\cdot2.45=14.7$.

Your wealth is a martingale; by optional stopping $10p=3$, so $p=0.3$.

$W_n^2-n$ is also a martingale, which gives $\mathbb E[T]=k(N-k)=3\cdot7=21$.

With $r=q/p=\tfrac23$, $P=\dfrac{1-r^k}{1-r^N}=\dfrac{1-\tfrac23}{1-\tfrac{8}{27}}=\dfrac{1/3}{19/27}=\tfrac{9}{19}\approx0.474$.

$\tfrac4{52}\cdot\tfrac3{51}=\tfrac1{221}\approx0.0045$.

Choose a suit and five of its 13 cards: $4\binom{13}{5}=5148$ hands out of $\binom{52}{5}=2{,}598{,}960$.

$1-\left(\tfrac56\right)^4=1-\tfrac{625}{1296}=\tfrac{671}{1296}$. This is the de Méré bet, slightly favourable.

$1-\left(\tfrac{35}{36}\right)^{24}\approx0.491$. De Méré's second bet, slightly unfavourable.

Bayes: $\dfrac{0.01\cdot0.99}{0.01\cdot0.99+0.99\cdot0.05}=\dfrac{0.0099}{0.0594}=\tfrac16$.

$\tfrac35\cdot\tfrac24+\tfrac25\cdot\tfrac14=\tfrac{6}{20}+\tfrac{2}{20}=\tfrac25$.

Each letter is correct with probability $\tfrac1{10}$; by linearity of expectation the total is $10\cdot\tfrac1{10}=1$, for any $n$.

Derangements: $P=\sum_{k=0}^{52}\tfrac{(-1)^k}{k!}$, which is $1/e$ to many decimal places.

Each face is missed with probability $\left(\tfrac56\right)^6$, so $\mathbb E=6\left(1-\left(\tfrac56\right)^6\right)\approx3.99$.

For a geometric variable $\operatorname{Var}=\dfrac{1-p}{p^2}=\dfrac{3/4}{1/16}=12$.

Reroll iff the first roll is below the value of rerolling, $3.5$: keep 4, 5, 6. $\mathbb E=\tfrac{4+5+6}{6}+\tfrac12\cdot3.5=4.25$.

With two rolls left the value is $4.25$, so on the first roll keep only 5 or 6: $\tfrac{5+6}{6}+\tfrac46\cdot4.25=\tfrac{28}{6}=\tfrac{14}{3}$.

$\tfrac16(1+4+9+16+25+36)=\tfrac{91}{6}$. Note this is more than $3.5^2$, by Jensen.

$P(\max=k)=\tfrac{2k-1}{36}$, so $\mathbb E=\tfrac1{36}\sum_{k=1}^6k(2k-1)=\tfrac{161}{36}$.

Independence gives $\mathbb E[XY]=\mathbb E[X]\,\mathbb E[Y]=3.5^2$.

Differences 0 to 5 occur 6, 10, 8, 6, 4, 2 times. $\tfrac{0+10+16+18+16+10}{36}=\tfrac{70}{36}$.

Buffon: with centre distance $x\sim U[0,\tfrac12]$ and angle $\theta\sim U[0,\tfrac\pi2]$, it crosses iff $x\le\tfrac12\sin\theta$. The probability is $\tfrac{4}{\pi}\int_0^{\pi/2}\tfrac12\sin\theta\,d\theta=\tfrac2\pi$.

$\sum_{n\ge1}2^n\cdot2^{-n}=\sum 1=\infty$. The St Petersburg paradox: nobody would pay very much, which is why utility and finite bankrolls matter.

HHH wins only if the first three tosses are HHH. Any earlier T means that the first HH to follow completes THH before HHH can.

For $k$ heads in a row the answer is $2^{k+1}-2$; for $k=3$, 14.

$P(N=0)=e^{-\lambda t}=e^{-2}$.

The exponential distribution is memoryless, so the remaining life is again exponential with mean 1000.

$\min$ is $\operatorname{Exp}(2)$ with mean $\tfrac12$; after it, by memorylessness, the other needs another $1$ on average. $\tfrac12+1=\tfrac32$.

$1.96$ is the two-sided 95% point, so each tail holds 2.5%.

$\operatorname{cov}=\operatorname{var}X=1$ and $\operatorname{sd}(X+Y)=\sqrt2$, so $1/\sqrt2$.

The longer piece $L$ is uniform on $[\tfrac12,1]$. Breaking it at $a$, all three pieces are below $\tfrac12$ iff $a\in(L-\tfrac12,\tfrac12)$, a window of length $1-L$, so the conditional probability is $\tfrac{1-L}{L}$. Then $\int_{1/2}^{1}2\cdot\tfrac{1-L}{L}\,dL=2\ln2-1$.

Runs $=1+$ number of changes between consecutive tosses. Each of the 9 adjacent pairs differs with probability $\tfrac12$: $1+\tfrac92=5.5$.

It returns iff the two steps differ: UD or DU, 2 of 4 paths.

Whatever the first card, 39 of the remaining 51 are another suit: $\tfrac{39}{51}=\tfrac{13}{17}$.

The count is geometric with $p=\tfrac12$, so the mean is $1/p=2$.

$\tfrac{6\cdot5\cdot4}{6^3}=\tfrac{120}{216}=\tfrac59$.

$\tfrac{13}{52}\cdot\tfrac{12}{51}=\tfrac14\cdot\tfrac4{17}$.

Conditioned on $X>\tfrac12$, $X$ is uniform on $\left(\tfrac12,1\right)$, whose mean is the midpoint.

The longer piece is $\max(U,1-U)$, which is uniform on $\left[\tfrac12,1\right]$: mean $\tfrac34$.

Derangements of 4 objects: $D_4=9$ out of $4!=24$.

Let $p$ be the first player's chance. The first roll wins at once with probability $\tfrac16$; otherwise both players miss (probability $\tfrac{25}{36}$) and the game restarts. So $p=\tfrac16+\tfrac{25}{36}p$, giving $p=\tfrac6{11}$. Going first is worth about nine percentage points over a coin toss, which is why the order of play gets negotiated in real games.

A given box stays empty with probability $0.9^{10}\approx0.349$. By linearity, the expected number of empty boxes is $10\times0.349\approx3.49$. No box needs to be independent of any other for this to work, and they are not. Roughly a third of the boxes stay empty, close to the $1/e$ you get as the number of boxes grows.

The $k$-th arrival is a record exactly when it is the largest of the first $k$, and by symmetry that happens with probability $\tfrac1k$. The indicators are not independent, but expectation is additive regardless, so the mean is $H_{10}=1+\tfrac12+\dots+\tfrac1{10}\approx2.93$. Records are logarithmically rare: a thousand numbers produce only about $7.5$ of them, which is why "best so far" events feel so much scarcer than people expect.

Conditioning on the position $k$ of the best candidate, the strategy succeeds when $k>r$ and the best of the first $k-1$ lies in the rejected block, which has probability $\tfrac{r}{k-1}$. Summing, $P(r)=\tfrac{r}{100}\sum_{k=r+1}^{100}\tfrac1{k-1}\approx x\ln(1/x)$ with $x=r/100$. That is maximised at $x=1/e$, giving $r=37$ and a success probability of about $37\%$ — strikingly high for a problem where you see each candidate once.

Imagine a gambler arriving before every toss, betting £1 on H, then on T, then on H, doubling through and stopping on any loss. This is a fair game, so the casino's expected profit is zero: after $\mathbb E[T]$ tosses it has taken $\mathbb E[T]$ pounds in stakes and must pay out what the winners hold. When HTH completes, one gambler has run the full three-step parlay and holds $£8$, and one has just matched the trailing H and holds $£2$. So $\mathbb E[T]=8+2=10$. The overlap is what costs you: HTH can begin again on its own last letter.

Same accounting as for HTH. When HHT completes, only the gambler who backed the whole pattern is still solvent, holding $£8$; nobody is mid-parlay, because a T cannot start HHT. So $\mathbb E[T]=8$. Two patterns of the same length need not have the same waiting time — HHT has no self-overlap and so arrives sooner than HTH, which is the fact that makes the next problem possible.

Choose THH. For HHT to come first the sequence must open with HH, because any T before the first HH puts a T in front of a later HH and hands you THH. The opening HH has probability $\tfrac14$. So you win with probability $\tfrac34$. The game is non-transitive: every pattern has a beater, so the second player always has the advantage, and the first player's "strongest-looking" choice is irrelevant.

The sum of $n$ independent uniforms is at most 1 exactly when the point $(x_1,\dots,x_n)$ lies in the simplex $x_i>0,\ \sum x_i\le1$, whose volume is $\tfrac1{n!}$. So $P(N>n)=\tfrac1{n!}$, and for a non-negative integer variable $\mathbb E[N]=\sum_{n\ge0}P(N>n)=\sum_{n\ge0}\tfrac1{n!}=e$. One of the few places $e$ appears without a limit or a logarithm in sight.

By inclusion–exclusion the number of permutations with no fixed point is $n!\sum_{k=0}^{n}\tfrac{(-1)^k}{k!}$, so the probability is $\sum_{k=0}^{10}\tfrac{(-1)^k}{k!}\approx0.36788$. That is $1/e$ to five decimal places, and it is essentially constant from $n=6$ onwards — the answer does not care whether there are ten people or ten thousand.

Follow the cycle through 1: $1\to\pi(1)\to\pi(\pi(1))\to\cdots$. At each step the next element is uniform over those not yet visited plus the option of closing the cycle back at 1. By symmetry, 2 is equally likely to appear before the cycle closes as after it, so the answer is $\tfrac12$ for every $n\ge2$. The same argument drives the 100-prisoners puzzle, where everything depends on cycle lengths.

Each prisoner opens the box with his own number, then the box named by what he finds, and follows that chain. The arrangement is a permutation, and this chain is the cycle containing him, so he succeeds precisely when that cycle has length at most 50. Everyone succeeds together exactly when the permutation has no cycle longer than 50. A permutation of $2n$ has at most one cycle longer than $n$, and the probability of a cycle of length $\ell>n$ is $\tfrac1\ell$, so the failure probability is $\sum_{\ell=51}^{100}\tfrac1\ell=H_{100}-H_{50}\approx0.6882$. Success is $\approx0.3118$, against $2^{-100}$ for independent guessing. Correlating the failures, rather than reducing them, is the whole trick.

The three pieces are exchangeable, so think of the two break points as splitting a circle of circumference 1 at three points into three arcs — the standard trick is to note the shortest of $n$ such spacings has mean $\tfrac1{n^2}$. Directly: $P(\min>t)=(1-3t)^2$ for $0\le t\le\tfrac13$, so $\mathbb E[\min]=\int_0^{1/3}(1-3t)^2dt=\tfrac19$. The middle piece averages $\tfrac{5}{18}$ and the longest $\tfrac{11}{18}$, which sum to 1 as they must.

An inscribed angle is obtuse exactly when its opposite arc exceeds a semicircle, and at most one of the three arcs can. So the triangle is right or obtuse precisely when the three points lie in some semicircle, which happens with probability $3/2^{2}=\tfrac34$. The triangle is acute with probability $\tfrac14$. Most random inscribed triangles are obtuse, which surprises people who picture the equilateral case.

Count faces, not cards. Three faces are red, and two of them belong to the double-red card, so the answer is $\tfrac23$. The intuitive "$\tfrac12$" treats the two cards as equally likely once red is seen, but the double-red card had twice the chance of showing red in the first place.

Label each child by sex and weekday: 14 equally likely types, 196 ordered pairs. Pairs with no Tuesday boy number $13^2=169$, leaving 27. Of those, the both-boys pairs are the $14^2-13^2=27$ minus... count directly: both boys with at least one Tuesday boy is $7\cdot7\cdot2-1\cdot1=$ $2\cdot7-1=13$ out of the 7 weekday choices each way, i.e. 13. So $\tfrac{13}{27}\approx0.481$. The weekday looks irrelevant yet moves the answer from $\tfrac13$ towards $\tfrac12$, because the more specific the condition, the less it double-counts the two-boy case.

Face $j$ fails to appear with probability $(\tfrac56)^{10}$, so it appears with probability $1-(\tfrac56)^{10}\approx0.8385$. Summing the six indicators gives $6\left(1-(\tfrac56)^{10}\right)\approx5.03$. Indicators plus linearity again sidestep the dependence entirely; trying to track the joint distribution of which faces appear is a great deal of work for the same number.

The exponential is memoryless: however long it has been since the last bus, the wait for the next is still exponential with mean 10. The gap you land in is not a typical gap — you are more likely to fall inside a long one — and in fact the gap containing you has mean 20, split into 10 behind and 10 ahead. That is the inspection paradox, and it is why the average passenger's wait exceeds half the average headway.

Memorylessness: $X-5$ conditioned on $X>5$ is again exponential with mean 1, so $\mathbb E[X\mid X>5]=6$. The exponential is the only continuous distribution with this property, which is exactly why it models waiting times with no ageing.

Real roots require $b^2\ge4c$, i.e. $c\le b^2/4$, which is at most $\tfrac14$ and so never clipped by $c\le1$. The area is $\int_0^1\tfrac{b^2}4db=\tfrac1{12}$. A random quadratic with small positive coefficients almost always has complex roots — the discriminant is a demanding condition.

The $n$ points cut $[0,1]$ into $n+1$ gaps which are exchangeable, so each has mean $\tfrac1{n+1}$. The $k$-th smallest point sits at the end of $k$ of them, giving $\mathbb E=\tfrac{k}{n+1}$. With $n=9$ and $k=3$ that is $0.3$. No integration required once you see the gaps are symmetric.

Fix the first endpoint and rotate the triangle so it sits there. The chord beats the triangle's side exactly when the second endpoint falls on the opposite third of the circumference, so the probability is $\tfrac13$. Choosing the chord by its midpoint instead gives $\tfrac14$, and by its distance from the centre gives $\tfrac12$ — Bertrand's paradox. "At random" is not a specification until you say which quantity is uniform.

Immediate wins: $\tfrac{6}{36}+\tfrac{2}{36}=\tfrac{8}{36}$. With point $p$, the duel between $p$ and 7 is won with probability $\tfrac{n_p}{n_p+6}$, where $n_p$ is the number of ways to roll $p$. Points 4 and 10 give $\tfrac39$, 5 and 9 give $\tfrac4{10}$, 6 and 8 give $\tfrac5{11}$. Weighting by $\tfrac{n_p}{36}$ and summing gives $\tfrac{244}{495}\approx0.4929$ — a house edge of $1.41\%$, among the slimmest on the floor, which is precisely why the table is loud.

Only rolls showing 7 or 8 matter; everything else is a repeat of the same question. There are 5 ways to make 8 and 6 to make 7, so the answer is $\tfrac5{11}$. Conditioning on the relevant event and discarding the rest turns an infinite sum into a ratio.

Group the corners by distance from the start: $A$ (the start), $B$ (three neighbours), $C$ (three at distance 2), $D$ (the far corner). From $A$ you always go to $B$; from $B$, one third back to $A$ and two thirds to $C$; from $C$, two thirds to $B$ and one third to $D$. With $a,b,c$ the expected times from each class, $a=1+b$, $b=1+\tfrac13a+\tfrac23c$, $c=1+\tfrac23b$. Solving gives $a=10$. Lumping states by symmetry turns an eight-state chain into three equations.

$P(\text{all heads})=2^{-10}$ and $P(\text{at least one head})=1-2^{-10}=\tfrac{1023}{1024}$, so the ratio is $\tfrac1{1023}$. Conditioning removes exactly one of the 1024 sequences, and the answer is the reciprocal of what remains.

A named suit is void in your hand when all 13 cards come from the other 39, which has probability $\binom{39}{13}/\binom{52}{13}\approx0.01279$. So each suit is present with probability $\approx0.98721$, and the four indicators sum to $4\times0.98721\approx3.949$. Roughly one hand in 78 is missing a named suit, and about one in 20 is void in something — which is why bridge bidding spends so much effort on distribution.

Plot arrival times in the unit square. They meet when $|x-y|\le\tfrac14$, a band about the diagonal. The two corner triangles outside it each have legs $\tfrac34$, so the miss probability is $2\cdot\tfrac12(\tfrac34)^2=\tfrac9{16}$ and they meet with probability $\tfrac7{16}$. A fifteen-minute grace period in a one-hour window buys less than even odds, which is worse than most people guess.

This is Pólya's urn. The probability of any particular sequence with $k$ blacks and $n-k$ whites is $\dfrac{k!\,(n-k)!}{(n+1)!}$ — it depends only on the counts, not the order, so the sequences are exchangeable. Multiplying by the $\binom nk$ orders gives $\tfrac1{n+1}$ for every $k$. The number of blacks is uniform on $\{0,1,\dots,n\}$: reinforcement makes the long-run fraction converge, but to a limit that is itself uniform on $[0,1]$.

Let $p$ be the probability. Conditioning on the first step, $p=\tfrac12+\tfrac12p^2$, so $(p-1)^2=0$ and $p=1$. The walk is recurrent: it reaches every level with certainty. The expected time to do so, however, is infinite — a distinction that catches people out, and the reason the "double until you win" system is not free money.

There are $\binom{30}{2}=435$ pairs, each matching with probability $\tfrac1{365}$, so the expectation is $\tfrac{435}{365}\approx1.19$. The expected count passing 1 near $n=28$ is the honest reason the birthday paradox works: it is pairs that grow quadratically, not people.

$\tfrac{26}{52}\cdot\tfrac{25}{51}\cdot\tfrac{24}{50}=\tfrac2{17}\approx0.118$. With replacement it would be $\tfrac18=0.125$; removing each red card makes the next one slightly less likely.

Six of the 36 outcomes are ties. The other 30 split evenly between "first larger" and "second larger" by symmetry, so the answer is $\tfrac{15}{36}=\tfrac5{12}$. Counting the ties first and using symmetry on the rest is faster than listing cases.

The first toss shows one face. From then on you wait for the other face, which takes a geometric number of tosses with mean 2. Total $1+2=3$.

Let $a_n$ count length-$n$ sequences with no HH. A valid sequence ends in T (any valid sequence of length $n-1$ before it) or in TH (any valid sequence of length $n-2$ before that), so $a_n=a_{n-1}+a_{n-2}$ with $a_1=2$, $a_2=3$. Then $a_3=5$, $a_4=8$, and $8/16=\tfrac12$. The counts are Fibonacci numbers, so the probability falls roughly like $0.81^n$.

For fixed $X=x$, the chance that $Y>x^2$ is $1-x^2$. Averaging over $x$: $\int_0^1(1-x^2)\,dx=1-\tfrac13=\tfrac23$. Geometrically it is the area of the unit square above the parabola.

$\max(X,Y)=\tfrac12(X+Y)+\tfrac12|X-Y|$. The first term has mean 0. $X-Y$ is normal with variance 2, and for $W\sim N(0,\sigma^2)$, $\mathbb E|W|=\sigma\sqrt{2/\pi}$. So $\mathbb E[\max]=\tfrac12\sqrt2\sqrt{2/\pi}=\tfrac1{\sqrt\pi}$. Writing a maximum as mean plus half the absolute difference is a trick worth keeping.

$\mathbb E[Z\,\mathbf 1_{Z>0}]=\int_0^\infty z\,\varphi(z)\,dz=\varphi(0)=\tfrac1{\sqrt{2\pi}}$, because $\varphi'(z)=-z\varphi(z)$. Dividing by $P(Z>0)=\tfrac12$ gives $\sqrt{2/\pi}$. This number reappears as the mean absolute deviation of a normal, and in the expected payoff of an at-the-money option.

By symmetry, compute $P(Y>2X)$ and double it. The region $y>2x$ in the unit square is a triangle with vertices $(0,0)$, $(0,1)$, $(\tfrac12,1)$, area $\tfrac14$. Doubling gives $\tfrac12$.

The tempting answer is 3, from treating the die as a three-sided die $\{2,4,6\}$. That is wrong: the condition throws away every long sequence, because long sequences are much more likely to contain an odd number. Think of each roll as ending the experiment when it shows 1, 3, 5 or 6 (probability $\tfrac23$), and ask which of those endings happened. The number of rolls is geometric with success probability $\tfrac23$, independent of which ending occurred, so the conditional mean is $\tfrac1{2/3}=\tfrac32$. Conditioning on a rare event changes the distribution of everything it touches.

All three match with probability $\tfrac28=\tfrac14$, so each round settles it with probability $\tfrac34$. The number of rounds is geometric with mean $\tfrac43$.

Bayes: $\dfrac{\tfrac12\cdot1}{\tfrac12\cdot1+\tfrac12\cdot\tfrac18}=\tfrac89$. Each further head doubles the odds in favour of the double-headed coin: $1{:}1$ before tossing, $8{:}1$ after three heads.