Quant puzzles, solved step by step

Puzzles: 100
Traced to a firm: 71
Topics: 12
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002You are flying to a city where it rains on 25% of days. You phone three friends who live there. Each tells the truth with probability 2/3, independently of the others, and all three say it is raining. What is the probability that it is actually raining?Conditional probability and BayesCoreJane StreetNew York · 2025
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Pick your answer before working it.
Show the worked solution
8/11, about 72.7%. If it is raining, all three say yes with probability (2/3)^3 = 8/27. If it is dry, all three must be lying, (1/3)^3 = 1/27. Weight each by how often it happens: 1/4 x 8/27 against 3/4 x 1/27, which is 8 parts to 3. Three agreeing witnesses move a 25% prior a long way, but not to certainty.
Why is the answer not simply 8/9?
Picture a clinic where a test is quite reliable but the illness is uncommon. A positive result makes the illness more likely, but how much more depends on how rare it was to begin with. The friends' agreement tells you how much more likely rain makes their answer than dry does, eight times, but it does not erase the fact that dry days are three times as common. 8/9 is the answer you get if rain and dry start level. Here they do not.
Rain covers a quarter of days and all three friends say yes on 8/27 of those, while dry days cover three quarters and all three lie on only 1/27 of them, so the shaded areas stand 8 to 3 and the chance of rain given three yeses is 8/11, about 72.7%.
How do you set it up so the arithmetic stays small?
Use odds rather than probabilities. Posterior odds are prior odds times the likelihood ratioHow many times more likely the evidence is if the hypothesis is true than if it is false., and both are easy numbers here. Prior odds of rain are 1 to 3. The likelihood ratio of three yeses is (2/3)^3 over (1/3)^3, which is 2 cubed, 8. So the posterior odds are 8 to 3, and the probability is 8 over 8 plus 3, 8/11. Each additional agreeing friend would double the odds again.
The relationship
$\frac{P(R\mid YYY)}{P(D\mid YYY)} = \frac{P(R)}{P(D)}\cdot\frac{(2/3)^3}{(1/3)^3} = \frac{1}{3}\cdot 8 = \frac{8}{3} \;\Rightarrow\; P(R\mid YYY) = \frac{8}{11}$
R, D rain and dry
YYY all three friends say yes
(2/3)^3 and (1/3)^3 the chance of three yeses when it rains, and when it is dry
What it says in wordsMultiply the prior odds by how much more likely the evidence is under rain, then turn the odds back into a probability.
What assumption is doing the work, and should you say it?
The calculation needs the friends to lie independently. If they could be coordinating a joke, three yeses are really one piece of evidence, and the answer falls back towards the one-friend figure of 2/5. Say the independence assumption out loud, then give 8/11. Interviewers often follow up by making one friend unreliable or by letting them talk to each other.
Where candidates lose it
The most common wrong answer is 8/9: the candidate compares the chance of three truths with the chance of three lies and forgets the weather's own odds. The rain prior is a quarter, and leaving it out quietly assumes it is a coin flip.
The second loss is writing out a full Bayes formula with 27ths and 108ths and losing the thread under time pressure. Odds times likelihood ratio gets 8 to 3 in two lines and is easier to check out loud.
What the interviewer asks next
What if only two of the three friends say yes?
How many agreeing friends would you need before you were 95% sure it is raining?
What changes if the friends can talk to each other before answering?
Asked at Jane Street, Generalist, New York, 2025 (Wall Street Oasis): There was a question about the probability of rain the next day that relied on a very in depth understanding of bayes theorem
014Five per cent of fund managers are skilled and beat the market in any given year with probability 60%; the rest are unskilled and beat it with probability 50%. Years are independent. A manager has beaten the market in exactly 8 of the last 10 years. What is the probability the manager is skilled?Conditional probability and BayesCoreCitadel SecuritiesMiami · 2022
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Roughly how likely is it that this manager is skilled?
Show the worked solution
About 12.7%. A skilled manager wins exactly 8 of 10 with probability 0.1209; an unskilled one with 0.0439, a likelihood ratio of about 2.75. Prior odds of skill are 5 to 95, 1 to 19. Posterior odds are 2.75 to 19, so the probability is 0.05 x 0.1209 / (0.05 x 0.1209 + 0.95 x 0.0439) = 0.127. The record helps, but luck has far more players.
Why does an impressive record move the needle so little?
Imagine a thousand people each tossing a coin ten times. About 55 of them will get eight heads or better with a fair coin. If a handful of the thousand had slightly biased coins, you still could not pick them out from the lucky crowd by one run of ten. Evidence moves a belief in proportion to how much more likely it is under one explanation than the other, and 8 wins in 10 is not much more likely from a 60% manager than from a 50% one. The ratio is about 2.75.
Eight wins in ten years has probability 12.1% for a 60% manager and 4.4% for a 50% manager, but after weighting by how common each type is, 5% against 95%, the chance that an eight-win manager is skilled is only 12.7%.
How do you set it up quickly?
Use odds. Prior odds of skill are 1 to 19; the likelihood ratioHow many times more likely the evidence is under one hypothesis than under the other. of the record is (0.6/0.5) to the 8 times (0.4/0.5) squared, which is 1.2 to the 8 times 0.64, about 2.75; multiply to get posterior odds of about 0.145. Converting, 2.75 over 2.75 + 19 is 12.7%. The binomial coefficient, 45, is the same in both likelihoods and cancels, so you never need it.
The relationship
$\frac{P(S\mid 8)}{P(U\mid 8)} = \frac{0.05}{0.95}\cdot\frac{0.6^8\,0.4^2}{0.5^{10}} \approx \frac{1}{19}\times 2.75 \;\Rightarrow\; P(S\mid 8) \approx 0.127$
S, U skilled and unskilled
0.6^8 0.4^2 the chance of one particular sequence of 8 wins and 2 losses for a skilled manager
0.5^{10} the same for an unskilled manager
What it says in wordsPrior odds of 1 to 19, times a likelihood ratio of 2.75, give a posterior of about 12.7%.
What does this say about picking managers?
When skill is rare and its edge is small, even a long, strong track record leaves luck as the likelier explanation. Using 8 or more wins instead of exactly 8 barely changes things: the answer becomes 13.9%. The honest limitation is that the model is stylised: real skill is not a fixed 60%, and survivorship means the managers you hear about were already filtered for good records, which pushes the true figure lower still.
Where candidates lose it
The common answer is around 80%, reading the record's win rate as the chance of skill. That skips the prior entirely, and with only 5% of managers skilled, the prior dominates.
The quieter trap is computing the full binomial probabilities, 45 x 0.6 to the 8 x 0.4 squared and so on, and getting lost in decimals. The coefficient cancels. Say odds and likelihood ratio and the arithmetic stays on one line.
What the interviewer asks next
How many years of 80% wins would you need before the manager is more likely skilled than not?
What if 20% of managers were skilled?
How does survivorship bias change the answer if you only ever see managers with good records?
Asked at Citadel Securities, Sales and Trading, Miami, 2022 (Wall Street Oasis): I got a question about Bayes' theorem applied to a practical scenario, which I handled decently
027A bag holds three dice: one fair, one that shows six half the time with its other faces equally likely, and one that never shows six. You draw one at random and roll it twice, getting two sixes. What is the probability it is the loaded die?Conditional probability and BayesCoreBelvedere TradingChicago · 2022
Try it first
Before you calculate: how likely is it now that you hold the loaded die?
Show the worked solution
90%. Each die starts at one in three. The chance of two sixes is 1/36 for the fair die, 1/4 for the loaded die and zero for the die with no six. Weight each by its prior: 1/108 for the fair die, 1/12 for the loaded die, nothing for the third. The loaded die's weight is nine times the fair die's, so its probability is 9/10.
What does the die that never shows six do to the answer?
A neighbour tells you a red car blocked the gate this morning. If one of your suspects owns only a blue scooter, that suspect is out, however likely they looked before. A hypothesis that cannot produce the evidence gets zero weight afterwards, no matter what its prior was. The no-six die could never give two sixes, so it drops out, and the question becomes a contest between the fair die and the loaded die, which started level at one third each.
Each die starts at one third; multiplying by the chance of two sixes gives weights of 1/108 for the fair die, 1/12 for the loaded die and zero for the no-six die, so the loaded die ends at 90% and the fair die at 10%.
How much does each surviving die's likelihood count?
Now compare how easily each remaining die produces what you saw. The fair die gives two sixes 1 time in 36. The loaded die gives a six half the time, so two in a row 1 time in 4. With equal priors, the posterior odds are just the ratio of the likelihoodsThe probability of the observed evidence under each hypothesis, before any prior is applied.: 1/4 against 1/36, which is 9 to 1. Nine parts in ten is 90%.
The relationship
$P(L \mid 66) = \frac{\tfrac13\cdot\tfrac14}{\tfrac13\cdot\tfrac1{36} + \tfrac13\cdot\tfrac14 + \tfrac13\cdot 0} = \frac{9}{10}$
L the loaded die was drawn
66 the evidence: two sixes in two rolls
1/3 the prior for each die
1/36, 1/4, 0 the chance of two sixes from the fair, loaded and no-six dice
What it says in wordsThe loaded die's share of all the ways two sixes can happen is nine tenths.
Check it by counting, the safer habit under pressure. Imagine 108 rounds of drawing a die and rolling it twice, 36 rounds with each die. The fair die gives two sixes once, the loaded die 9 times, the no-six die never. Of the 10 double sixes, 9 came from the loaded die, and the prior of one third cancels because every die got the same number of rounds. The loaded die's other faces, 1 in 10 each, never enter, because only sixes were seen.
Where candidates lose it
The quick wrong answer is 1/2: two dice can roll a six, so it must be one or the other. That ignores how differently they produce two sixes in a row, a gap of nine to one.
The other loss is getting tangled in the loaded die's other faces, or leaving the no-six die in the denominator with some weight. Neither belongs: only the chance of the observed rolls counts, and for the no-six die that chance is zero.
What the interviewer asks next
A third roll is also a six. What is the probability of the loaded die now? (It rises to 27/28.)
The rolls were a six and then a two. Which die is most likely now?
What is the chance the next roll is a six? (It is 7/15.)
Asked at Belvedere Trading, Capital Markets, Chicago, 2022 (Wall Street Oasis): The technical portion of the interview consisted of probability questions including one questions relating to Bayes' theorem
039A surveillance screen flags suspicious trades. One order in 100 is genuinely manipulative. Alert A fires with a likelihood ratio of 9, and an independent alert B with a likelihood ratio of 4. Both fire on the same order: what is the probability it is manipulative?Conditional probability and BayesCoreCitadelMiami · 2022
Try it first
Both alerts fire. Roughly how likely is the order manipulative?
Show the worked solution
About 26.7%. Work in odds. The prior odds are 1 to 99. Independent evidence multiplies the odds by each likelihood ratio: 1 x 9 x 4 = 36, so the posterior odds are 36 to 99. As a probability that is 36/135, about 26.7%. Even with both alerts, roughly three flagged orders in four are clean, because manipulation is rare to begin with.
Why is odds form the fast way to combine alerts?
Think of two smoke detectors in a kitchen where real fires are rare. Each beep makes a fire more likely, but toast sets both off far more often than fire does. In odds form, Bayes' rule is one multiplication per piece of independent evidence: posterior odds equal prior odds times each likelihood ratioHow much more often the evidence appears when the hypothesis is true than when it is false.. A ratio of 9 means alert A fires nine times as often on manipulative orders as on clean ones, for example on 90% of manipulative orders and 10% of clean ones.
Starting from odds of 1 to 99, alert A multiplies the odds by 9 to reach 9 to 99, an 8.3% chance, and alert B multiplies by 4 to reach 36 to 99, which is only 26.7% because the prior was so low.
How do you check 26.7% by counting?
Take 10,000 orders: 100 manipulative and 9,900 clean. Suppose A fires on 90% of manipulative orders and 10% of clean ones, and B on 80% and 20%, which gives the stated ratios of 9 and 4. Both fire on 100 x 0.9 x 0.8 = 72 manipulative orders and on 9,900 x 0.1 x 0.2 = 198 clean ones. Of the 270 orders where both fire, 72 are manipulative: 26.7%, the same as the odds route.
The relationship
$\frac{P(M\mid A,B)}{P(\bar M\mid A,B)} = \frac{1}{99}\times 9\times 4 = \frac{36}{99} \;\Rightarrow\; P = \frac{36}{135} \approx 26.7\%$
M the order is manipulative
1/99 prior odds: 1 manipulative order per 99 clean
9, 4 likelihood ratios of alerts A and B
What it says in wordsMultiply the prior odds by each alert's likelihood ratio, then turn odds back into a probability.
State the assumption that made multiplication legal: the alerts are independent given the truth. If both alerts key off the same feature, say order size, the second adds little new information and multiplying by 4 overstates the case. With one alert alone the chance is 8.3% for A and 3.9% for B, which is why a desk reviews orders on combined evidence rather than a single flag.
Where candidates lose it
The common loss is treating a likelihood ratio of 36 as odds of 36 to 1 and answering about 97%. That throws away the base rate: the evidence multiplies the prior odds of 1 to 99, not even odds.
The second is adding the ratios, 9 + 4 = 13, instead of multiplying. Independent evidence compounds, and odds form makes that one line of arithmetic.
What the interviewer asks next
How many independent alerts with a ratio of 4 would you need to pass 50%?
Alert B is triggered by the same feature as alert A. How does that change your answer?
A third alert has a likelihood ratio of 0.5 and does not fire. What does that do?
Asked at Citadel, Sales and Trading, Miami, 2022 (Wall Street Oasis): I got a question about Bayes' theorem applied to a practical scenario

R, D	rain and dry
YYY	all three friends say yes
(2/3)^3 and (1/3)^3	the chance of three yeses when it rains, and when it is dry

S, U	skilled and unskilled
0.6^8 0.4^2	the chance of one particular sequence of 8 wins and 2 losses for a skilled manager
0.5^{10}	the same for an unskilled manager

L	the loaded die was drawn
66	the evidence: two sixes in two rolls
1/3	the prior for each die
1/36, 1/4, 0	the chance of two sixes from the fair, loaded and no-six dice

M	the order is manipulative
1/99	prior odds: 1 manipulative order per 99 clean
9, 4	likelihood ratios of alerts A and B

Try it first

Why is the answer not simply 8/9?

How do you set it up so the arithmetic stays small?

What assumption is doing the work, and should you say it?

Where candidates lose it

What the interviewer asks next

Try it first

Why does an impressive record move the needle so little?

How do you set it up quickly?

What does this say about picking managers?

Where candidates lose it

What the interviewer asks next

Try it first

What does the die that never shows six do to the answer?

How much does each surviving die's likelihood count?

Where candidates lose it

What the interviewer asks next

Try it first

Why is odds form the fast way to combine alerts?

How do you check 26.7% by counting?

Where candidates lose it

What the interviewer asks next