Quant Interview Cheat Sheet

Order matters, no repeats. Podium finishes, passwords, seat assignments.

Order does not matter, no repeats. Committees, poker hands, subsets.

Each of $k$ slots independently picks one of $n$ options.

Non-negative integer solutions of $x_1+\cdots+x_k=n$ — i.e. $n$ identical items into $k$ labelled boxes. Require every box non-empty and it becomes $\binom{n-1}{k-1}$.

Distinct arrangements of a word with repeated letters, or splitting $n$ items into labelled groups of sizes $n_i$.

Fix one item to kill the rotational symmetry. This anchoring trick reappears constantly in circle problems.

Choosing who is in equals choosing who is out.

Condition on whether a distinguished item is included. The template for most combinatorial recursions.

The number of subsets of an $n$-set.

Split a selection across two pools and sum over how many come from the first.

Condition on the largest element chosen.

For three: $|A\cup B\cup C| = |A|+|B|+|C| - |A\cap B| - |A\cap C| - |B\cap C| + |A\cap B\cap C|$.

Use it when the complement is not simpler. Truncating after the first term gives the union bound.

$[\,\cdot\,]$ is the nearest integer. $D_1..D_6 = 0,\,1,\,2,\,9,\,44,\,265$, and $D_n/n! \to e^{-1} \approx 0.368$ fast — the "nobody gets their own hat" probability is essentially $1/e$ from $n=5$ onwards.

Inclusion–exclusion over which targets are missed. This is the engine behind "every face appears" and occupancy questions.

Right/up steps from $(0,0)$ to $(m,n)$: choose which of the $m+n$ steps are up.

$1,\,1,\,2,\,5,\,14,\,42,\,132$. Counts paths that never cross the diagonal, balanced bracketings, and non-crossing pairings.

Within 1% by $n=10$. Use it to compare factorial-scale quantities without computing them.

The sequential view of any without-replacement draw.

Independent is not the same as mutually exclusive. Two disjoint events of positive probability are maximally dependent: one occurring rules the other out.

Partition on the piece of information you wish you had.

Posterior odds = prior odds × likelihood ratio. Far faster than the fraction form under time pressure, and it composes: independent pieces of evidence multiply their likelihood ratios.

Almost always faster than a union. The trigger words are at least, some, any, collision, max, min.

A free upper bound with no independence assumption. Good enough for most rare-event sanity checks.

Prevalence $\pi$, sensitivity $s$, specificity $t$. Rebuild it as counts out of 10,000 rather than algebra and the base-rate fallacy disappears.

Neither implies nor is implied by unconditional independence. Coin flips from an unknown-bias coin are conditionally independent given the bias and dependent without it.

Mean $p$, variance $p(1-p)$ — maximised at $p=\tfrac12$. One trial; the atom every indicator argument is built from.

Mean $np$, variance $np(1-p)$. Fixed number of independent trials, count the successes. Heads in $n$ flips; reds in $n$ draws with replacement.

Mean $1/p$, variance $(1-p)/p^{2}$. Memoryless: $\Prob(X > m+n \mid X > m) = \Prob(X > n)$.

Mean $(1-p)/p$, same variance $(1-p)/p^{2}$. Identical process, different convention — read the question to see which one it wants. The two means differ by exactly 1.

Mean $r/p$, variance $r(1-p)/p^{2}$. It is a sum of $r$ independent geometrics, which is why both moments are just $r$ times the geometric ones.

Population $N$ with $K$ successes, $n$ drawn without replacement. Mean $n\frac{K}{N}$ — same as the binomial. Variance $n\frac{K}{N}\!\left(1-\frac{K}{N}\right)\!\frac{N-n}{N-1}$; that last finite-population correction is what makes it tighter than the binomial. This is the card-problem distribution.

Mean $\lambda$, variance $\lambda$ — equal, which is the diagnostic. The limit of $\mathrm{Bin}(n,p)$ as $n \to \infty$, $p \to 0$ with $np \to \lambda$. Sums of independent Poissons are Poisson.

Mean $\frac{n+1}{2}$, variance $\frac{n^{2}-1}{12}$. A fair $n$-sided die.

Mean $\frac{a+b}{2}$, variance $\frac{(b-a)^{2}}{12}$.

Mean $1/\lambda$, variance $1/\lambda^{2}$. The only continuous memoryless law. Gaps between Poisson arrivals; the minimum of independent exponentials is exponential with the summed rate.

Roughly 68% / 95% / 99.7% of mass lies within 1 / 2 / 3 standard deviations.

Two-sided 90% / 95% / 99% intervals use $z = 1.645$ / $1.96$ / $2.576$.

Variances add; standard deviations do not. Scaling time by $t$ scales volatility by $\sqrt{t}$.

Mean $k/\lambda$, variance $k/\lambda^{2}$. Time until the $k$-th Poisson arrival.

The conjugate prior for a coin bias: a $\mathrm{Beta}(\alpha,\beta)$ prior plus $h$ heads and $t$ tails gives a $\mathrm{Beta}(\alpha+h,\ \beta+t)$ posterior. A uniform prior is $\mathrm{Beta}(1,1)$, which is why the posterior mean after $h$ heads in $n$ flips is $\frac{h+1}{n+2}$ (Laplace).

The mean sits above the median $e^{\mu}$. This gap is the volatility drag that makes compounded returns lag their arithmetic average.

No independence required. Write the quantity as a sum of indicators, price each indicator with a probability, and add.

The bridge that turns counting into expectation.

Shifts move the mean and leave the variance alone.

And in general $\E[g(X)] \ne g(\E[X])$ — see Jensen below.

Unlike expectation, variances add only when the terms are uncorrelated. Forgetting the covariance term is the single most common quant-interview slip.

Zero correlation is not independence: with $X \sim N(0,1)$, $X$ and $X^{2}$ are uncorrelated and completely dependent. Correlation only measures the linear part.

Condition on the first step, the hidden state, or the piece of information you wish you had — then average it away.

Read it as within-group variation plus between-group variation.

The fastest route to the expected maximum of several draws, because $\Prob(\max \ge k)$ is easy and the full pmf is not.

Requires $N$ to be a stopping time for the i.i.d. sequence, with $\E[N] < \infty$. It breaks if $N$ is allowed to look at the $X_i$ it is about to include — which is exactly how "guaranteed profit" betting systems smuggle in their error.

Convex payoffs love variance; concave utilities hate it. Reversed for concave $g$.

Distribution-free, therefore loose. If you know the shape, use it instead.

Equivalently $|\rho| \le 1$.

Variance $\frac{1-p}{p^{2}}$. Expected rolls to a first six: 6.

With $\gamma \approx 0.5772$. For a fair die, $6H_6 = 14.7$ rolls to see all six faces. Variance $\approx \frac{\pi^{2} n^{2}}{6}$ — the distribution has a long right tail, so the mean understates the typical bad case.

For a fair coin this is $2^{k+1} - 2$: 6 flips for $HH$, 14 for $HHH$, 30 for $HHHH$.

Self-overlapping patterns wait longer. After a failed $HH$ you keep nothing; after a failed $HT$ you still hold an $H$. Attack any of these with states indexed by the length of the currently matching prefix.

Crosses 50% at $k \approx 1.177\sqrt{n}$ — for $n = 365$ that is $k = 23$. Expected draws to the first repeat $\approx \sqrt{\pi n / 2} + \tfrac{2}{3} \approx 24.6$.

Solve backwards from the last decision. One reroll of a fair die: continuation value is 3.5, so keep 4, 5, 6 and reroll otherwise, giving $\E = 4.25$.

Success probability tends to $1/e \approx 0.368$. The lesson interviewers want is that "observe, then commit" beats both greed and patience.

The reason most "is this game fair?" answers are one line. It needs the boundedness or integrability condition — dropping it is precisely why the double-your-bet martingale system is not free money.

Linear in the starting point. Your chance of doubling before busting is just your fraction of the total stake.

Maximised in the middle, and surprisingly long: from 50 with barriers at 0 and 100 the game lasts 2,500 steps on average.

A small edge compounds hard: with $r < 1$ the probability approaches $1 - r^{i}$ for large $N$, so an infinitely rich opponent still cannot ruin you with certainty.

Valid for $p \ne q$.

Reflect the start across the barrier. This one bijection gives the ballot problem, Catalan paths, and first-passage counts.

For $a$ votes for A and $b < a$ for B, counted in random order.

Decays slowly, which is why symmetric walks keep coming back.

Distance grows like $\sqrt{n}$, not $n$.

A free sanity check: after an odd number of steps you cannot be at the origin.

A symmetric walk returns to its start with probability 1 on a line or a grid, but only about 34% of the time in three dimensions.

For an $m$-sided die: $\E = \frac{m+1}{2}$, $\Var = \frac{m^{2}-1}{12}$.

Variance scales with $n$ because the dice are independent; the standard deviation scales with $\sqrt{n}$.

Triangular. $\Prob(7) = \tfrac16$ is the mode.

Then use the tail sum. For two fair d6, $\E[\max] = \frac{161}{36} \approx 4.47$.

By symmetry $\Prob(A>B) = \Prob(B>A) = \frac{1 - 1/6}{2}$. Note that "beats" can be intransitive once the dice are non-standard — that is the Efron dice trap.

Linearity over one indicator per face — the indicators are dependent and it does not matter.

Grows like $\sqrt{n}$: 4 flips give 1.5, 100 flips give about 8.

$52! \approx 8.07 \times 10^{67}$ orderings.

Linearity, no hypergeometric needed.

The $k$ marked cards cut the deck into $k+1$ gaps of equal expected length. First ace in a 52-card deck: $\frac{53}{5} = 10.6$.

0.00015%

0.0015%. Ten starting ranks × 4 suits.

0.024%. $13 \times 48$.

0.144%. $13 \cdot \binom{4}{3} \cdot 12 \cdot \binom{4}{2}$.

0.197%. $4\binom{13}{5} = 5{,}148$, minus the 40 straight flushes.

0.393%. $10 \cdot 4^{5} = 10{,}240$, minus 40.

2.113%. $13\binom{4}{3}\binom{12}{2}4^{2}$.

4.754%. $\binom{13}{2}\binom{4}{2}^{2} \cdot 44$.

42.257%. $13\binom{4}{2}\binom{12}{3}4^{3}$.

50.118%. The nine counts above sum with this to exactly $\binom{52}{5}$ — that is your check.

Anchor on each point in turn and ask whether the other $n-1$ fall in the semicircle clockwise from it; the $n$ cases are disjoint up to a measure-zero tie. For an arc covering a fraction $a \le \tfrac12$ of the circle, the same argument gives $n\,a^{\,n-1}$.

The complement of "all three in a semicircle", which is $3/2^{2} = 3/4$.

Uniform in area, not in radius — the density of the radius is proportional to $r$. Half the area sits outside $r = R/\sqrt{2}$.

Random endpoint pair gives $\tfrac13$; uniform distance from the centre along a radius gives $\tfrac12$; uniform midpoint in the disk gives $\tfrac14$. The lesson is to state the sampling rule before computing — say this out loud in an interview.

So $\E[\max] = \frac{n}{n+1}$ and $\E[\min] = \frac{1}{n+1}$. The $n$ points cut the interval into $n+1$ gaps, each with mean length $\frac{1}{n+1}$.

And $\E[\max] = \tfrac23$, $\E[\min] = \tfrac13$.

Two uniform cut points; the condition is that no piece exceeds $\tfrac12$.

Needle of length $L$, lines spaced $d$ apart.

Multiply the two legs. For a forbidden point, subtract this from the total.

Dyck paths. Derived from the reflection principle: $\binom{2n}{n} - \binom{2n}{n+1}$.

Choose two of the $m+1$ vertical lines and two of the $n+1$ horizontal lines.

When a tiling or reachability question smells impossible, look for a two-colouring the move preserves. Counting the colour imbalance usually finishes the proof in one line.

Four times the data halves the error bar.

The $n-1$ (Bessel) corrects for having spent one degree of freedom estimating $\bar x$.

Needs finite variance. Convergence is fast for symmetric distributions and slow for heavy-tailed ones — which is the caveat interviewers are listening for.

Apply a $\pm 0.5$ continuity correction. Rule of thumb: $np$ and $n(1-p)$ both above roughly 10.

Large $n$, small $p$. Better than the normal in the rare-event tail.

Use $t$ rather than $z$ for small $n$.

The fitted line always passes through $(\bar x, \bar y)$, and $R^{2} = \rho^{2}$.

With $|\rho| < 1$, predictions are pulled toward the mean. The name of the technique comes from this effect, not the other way round.

As $n \to \infty$ the idiosyncratic term vanishes and volatility floors at $\sigma\sqrt{\rho}$. Diversification removes specific risk, never common risk.

$1/7 = 0.\overline{142857}$ cycles, and every multiple of $1/7$ is a rotation of the same six digits.

16% of 25 is the same as 25% of 16, which is 4.

$47 \times 53 = 50^{2} - 3^{2} = 2491$. The workhorse for any pair straddling a round number.

$85^{2} = 100 \cdot 8 \cdot 9 + 25 = 7225$.

$103 \times 108 = 100 \cdot 111 + 24 = 11{,}124$.

$47 \times 11 = 4\,|\,11\,|\,7 = 517$.

From $x=7$, $\sqrt{50} \approx 7 + \tfrac{1}{14} = 7.071$ — correct to four decimals in one step.

Good while $|nx|$ stays well below 1. Add $\binom{n}{2}x^{2}$ for a second digit.

The rule of 72: 6% a year doubles in about 12 years.

A 10% loss then a 10% gain is $-1\%$, not flat. The cross term is where money goes.

So $2^{20} \approx 10^{6}$ and $2^{30} \approx 10^{9}$ — the fastest way to size anything exponential.

Also: 1 mile ≈ 1.61 km, a year ≈ 2,000 working hours, and a typical rich-country household is 2–3 people. Estimate in orders of magnitude, state your assumptions, and quote a range.

Compute this first and say it out loud before quoting anything.

You must be happy to trade on either side. If you would not, your fair is wrong or your spread is too tight.

The conditioning is the whole point. If a counterparty only lifts you when they know something, then $\E[\text{value} \mid \text{lifted}] > \text{fair}$ and the naive $\text{ask} - \text{fair}$ overstates your edge.

Widen for high payoff variance, informed counterparties, large size, thin markets, and inventory you already do not want. Tighten to win flow when you are confident and flat.

Getting filled is itself information. The trades you win are disproportionately the ones you should have lost.

Move the midpoint against your position so you are more likely to trade out than in. Skewing is a different lever from widening; use the right one.

Raw implied probabilities sum above 1; the excess is the bookmaker margin. Normalise before comparing to your own estimate.

At net odds $b$ to 1. Even money needs 50%, 2-to-1 needs 33.3%.