Probability Classics
These questions appear constantly. Know the setups cold — not just the answers but the reasoning chain.
The Monty Hall Problem — You pick door 1. Host opens door 3 (goat). Switch or stay?
Switch. Staying wins 1/3 of the time. Switching wins 2/3 of the time. The host's action is not random — he always opens a goat door, which concentrates the remaining 2/3 probability onto the other door. Intuition fails here; condition on the host's strategy.
Birthday Problem — How many people needed for >50% chance two share a birthday?
23 people. P(all different) = 365/365 × 364/365 × 363/365 × ... After 23 terms, this product drops below 0.5. The counterintuitive answer comes from the quadratic number of pairs: 23 people = 253 pairs, each with a ~0.27% collision chance.
Dice: Roll two fair dice. Given the sum is at least 7, what's the probability the sum is exactly 7?
6/21. P(sum ≥ 7) = 21/36. P(sum = 7) = 6/36. Conditional probability: (6/36)/(21/36) = 6/21 ≈ 28.6%.
Bayes: A test is 99% accurate. Disease prevalence is 0.1%. You test positive. What's P(disease)?
≈ 9%. P(disease|positive) = P(positive|disease)·P(disease) / P(positive). Numerator: 0.99 × 0.001 = 0.00099. Denominator: 0.99×0.001 + 0.01×0.999 ≈ 0.01098. Result: 0.00099/0.01098 ≈ 9%. Base rate matters enormously — this is the core lesson for risk management too.
Expected Value: A fair coin. Win $2 on heads, lose $1 on tails. What's the EV per flip?
$0.50. EV = 0.5 × $2 + 0.5 × (−$1) = $1 − $0.50 = $0.50. Simple, but interview pressure causes mistakes. Always write out the full formula.
Conditional: You draw 2 cards without replacement. Given the first is an Ace, P(second is Ace)?
3/51 = 1/17. After drawing one Ace, 3 Aces remain among 51 cards. Conditional probability on the observed event (first is Ace).
Random Walk: A drunk man starts at 0. Each step ±1 with equal probability. E[steps to reach ±N]?
N². The expected first passage time from 0 to ±N is N². This is the optional stopping theorem result for symmetric random walks. Key insight for understanding how long drawdowns can persist in mean-zero strategies.
St. Petersburg Paradox: A game pays $2ⁿ where n = flips until heads. Fair price to play?
EV is infinite, but no rational person pays more than ~$20. EV = Σ (1/2ⁿ)·2ⁿ = Σ 1 = ∞. The resolution is utility — logarithmic utility makes the game worth a finite price. This underpins Kelly criterion and why we use log utility in position sizing.
Brain Teasers
12 balls, one is heavier or lighter. Find the odd ball in 3 weighings.
Divide into groups of 4. Weigh 4 vs 4. If balanced, odd ball is in the remaining 4. If unbalanced, it's in one of the weighed groups (and you know whether it's heavier or lighter). Use remaining weighings to binary search within the suspect group. Key insight: each weighing gives 3 outcomes, so 3 weighings = 3³ = 27 distinguishable outcomes — enough for 24 possibilities (12 balls × 2 directions).
How many piano tuners are in Chicago?
~100–200. Fermi estimation: Chicago pop ~2.7M → ~1M households → ~150k have pianos. Pianos need tuning ~1x/year. A tuner does ~4 pianos/day × 250 days = 1,000/year. 150k ÷ 1,000 ≈ 150 tuners. The method matters more than the number — show your decomposition chain clearly.
100 socks: 50 red, 50 blue. Minimum picks to guarantee a matching pair?
3. With 3 picks, by pigeonhole principle, at least two must be the same color. (Worst case: first two are one of each color; third must match one.)
A lily pad doubles daily. It takes 100 days to cover the pond. When was it half-covered?
Day 99. If it doubles daily and covers the full pond on day 100, it covered half the pond the day before. Exponential growth — the answer is always one step back from the end.
3-gallon and 5-gallon jug. How do you measure exactly 4 gallons?
Fill 5-gal. Pour into 3-gal until full. 5-gal now has 2 gal. Empty 3-gal. Pour 2 gal from 5-gal into 3-gal. Fill 5-gal again. Pour from 5-gal into 3-gal (needs 1 more). 5-gal now has exactly 4 gal. Classic water-pouring puzzle. Tests systematic thinking.