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  1. 040A bag holds four stones, each black or white. Before you look, every count of black stones from 0 to 4 is equally likely. You draw two stones without replacement and both are black. What is the chance the next stone is black, and at what price would you bet on it?Conditional probability and BayesHardCitadelNew York · 2025

    Try it first

    Chance the third stone is black:

    Show the worked solution

    3/4, so fair odds are 3 to 1 on black. Two black draws rule out bags with 0 or 1 black. The ways to draw two blacks in order are 2, 6 and 12 for bags with 2, 3 and 4 black, so those bags now carry 10%, 30% and 60%. The next stone is black with chance 0, 1/2 and 1 in them, which averages to 3/4. A contract paying 100 if black is worth 75.

    How do the two black draws change your view of the bag?

    If a friend pulls two red sweets from a jar you have never seen, you start to suspect it is mostly red. Each possible bag is reweighted by how likely it was to produce what you saw: equal priors times the chance of two blacks. Drawing two blacks in order has 0 ways from bags with 0 or 1 black, 2 x 1 = 2 ways with 2 black, 3 x 2 = 6 with 3 black and 4 x 3 = 12 with 4 black. Out of 20 in total, that is 10%, 30% and 60%: the all-black bag is now the favourite.

    Two blacks drawn: the evidence shifts weight toward the all-black bag0%0 black0%1 black10%2 blackweight 2next black 030%3 blackweight 6next black 1/260%4 blackweight 12next black 1prior 20% eachbefore the drawafter two blacksChance the next is black10% x 0+ 30% x 1/2+ 60% x 13/4Fair odds: 3 to 1 onFair price: 75 per 100
    Before the draw each bag is 20% likely; after two black stones the bags with 2, 3 and 4 black carry 10%, 30% and 60%, and since the next stone is black with probability 0, one half and 1 in those bags, the chance it is black is 3/4.

    How do you turn the posterior into a price?

    Average the chance of black over the bags you still believe in. In the 2-black bag both remaining stones are white; in the 3-black bag one of two is black; in the 4-black bag both are, so the answer is 0.1 x 0 + 0.3 x 1/2 + 0.6 x 1 = 3/4. A contract paying 100 if the next stone is black is worth 75. You would buy it below 75 and sell it above; as a market maker you might quote 70 at 80. Offered even money on black, your expected profit per rupee staked is 0.75 minus 0.25, which is 50 paise.

    The relationship
    P(B3∣B1B2)=∑kP(k∣B1B2) P(B3∣k,B1B2)=220⋅0+620⋅12+1220⋅1=34P(B_3 \mid B_1 B_2) = \sum_{k} P(k \mid B_1 B_2)\,P(B_3 \mid k, B_1 B_2) = \tfrac{2}{20}\cdot 0 + \tfrac{6}{20}\cdot\tfrac{1}{2} + \tfrac{12}{20}\cdot 1 = \tfrac{3}{4}
    kthe number of black stones in the bag
    B1 B2the event that the first two draws are black
    2, 6, 12the ordered ways to draw two blacks from bags with 2, 3 and 4 black
    What it says in wordsThe chance of another black is the average of each bag's chance, weighted by how much the evidence now favours that bag.

    Check it with Laplace rule of successionWith a uniform prior, after s successes in n trials, the chance the next trial succeeds is (s + 1) / (n + 2).: after 2 blacks in 2 draws the next is black with chance (2 + 1)/(2 + 2) = 3/4, and the rule holds exactly for this finite bag. Two routes to 3/4 is what separates a solid answer from a lucky one. The limitation: everything rests on the flat prior. If you had reason to think mixed bags were more common, 3/4 would fall.

    Where candidates lose it

    The common loss is saying 1/2 because the remaining stones are unknown, which throws away the information in the two draws. The question is about updating, and the interviewer wants to see the reweighting.

    The second loss is weighting the surviving bags equally, a third each, which gives 1/2. Each bag must be weighted by how likely it made two blacks: 2, 6 and 12. Then price it: a probability without a bet is half the answer at a trading firm.

    What the interviewer asks next

    • The third stone is black too. What is the chance the fourth is black?
    • You quote 70 at 80 on a contract paying 100 if black and someone who has seen the bag lifts your offer. What now?
    • How does the answer change if the prior is that each stone is black with probability one half, independently?

    Asked at Citadel, Quantitative Trading, New York, 2025 (Wall Street Oasis): Extended bayes derivative question about four stones in a bag (black and white stones).

  2. 066You roll a fair die repeatedly until you have seen every even number, 2, 4 and 6. Given that the last new even number to appear is a 2, what is the probability that your first roll was a 1? Why is the intuitive answer of 1/5 wrong?Conditional probability and BayesHardSquarepoint CapitalLondon · 2026

    Try it first

    Given the game ends on a 2, what is the chance the first roll was a 1?

    Show the worked solution

    The probability is 1/6, not 1/5. Ending on a 2 rules out a first roll of 2, but it does not leave the other five faces equally likely. A first roll of 4 or 6 has already cleared one rival, so 2 then finishes last half the time; after an odd first roll it finishes last only a third of the time. Weighting the faces by those chances gives 1/6 for a 1.

    Why does ruling out one face not spread its weight evenly?

    Suppose you hear that a friend arrived late to a meeting. Before, the bus, the train and the car were equally likely ways she travelled. Learning how things ended shifts weight towards the starts that make that ending more likely, in proportion to how strongly each one leads to it. If the bus is late twice as often as the train, the bus now carries twice the train's weight. The 1/5 answer treats the ending as if it only ruled a face out and said nothing else.

    How likely is a 2 to finish last after each first roll?

    Odd rolls never change the order in which the even numbers first appear, so after an odd first roll the three evens are still symmetric and 2 is last with chance 1/3; after a first roll of 4 or 6 only two evens remain and 2 is last with chance 1/2; after a first roll of 2 it can never be the last new even. Multiply each by the 1/6 chance of that first roll: the joint chances are 1/18 for each odd face, 0 for a 2 and 1/12 for each of 4 and 6. They add to 1/3.

    Knowing how it ends reweights how it beganFirst rolleach face 1/6First roll 1, 3 or 5chance 1/22 last: 1/3joint 1/6First roll 2chance 1/62 last: 0joint 0First roll 4 or 6chance 1/32 last: 1/2joint 1/6Total chance 2 ends it: 1/6 + 0 + 1/6 = 1/3Given the game ends on a 21/61021/631/441/651/46naive 1/5 each (dashed) against the truechances: 1/6 for each odd face
    An odd first roll leaves 2 last among the evens a third of the time, a 4 or 6 leaves it last half the time and a 2 never does, so given the game ends on a 2 each odd face has chance 1/6 and each of 4 and 6 has chance 1/4, not 1/5 each.
    The relationship
    P(first=1∣2 last)=P(first=1) P(2 last∣first=1)P(2 last)=16⋅1313=16P(\text{first}=1 \mid 2 \text{ last}) = \frac{P(\text{first}=1)\,P(2 \text{ last} \mid \text{first}=1)}{P(2 \text{ last})} = \frac{\tfrac16 \cdot \tfrac13}{\tfrac13} = \frac16
    P(first = 1)the chance of rolling a 1 first, 1/6
    P(2 last | first = 1)the chance 2 is the last even to appear after an odd first roll, 1/3
    P(2 last)the overall chance the game ends on a 2, 1/3 by symmetry
    What it says in wordsBayes' rule: the prior chance of a 1, times how strongly a 1 leads to ending on a 2, divided by the overall chance of ending on a 2.

    How do you check the answer?

    Make the six posterior chances add up. Three odd faces at 1/6 each and two even faces, 4 and 6, at 1/4 each give 1/2 plus 1/2, which is 1, with nothing left for a 2. The odd faces keep exactly their starting weight because an odd roll tells you nothing about the evens; all the weight removed from the 2 goes to 4 and 6. Saying that sentence shows the interviewer you understand Bayes ruleThe rule for updating a probability after new information: the prior times the likelihood of the information, divided by the overall chance of the information. rather than recite it.

    Where candidates lose it

    The trap is to condition only by elimination: the game ends on a 2, so the first roll was not a 2, so the five other faces share the weight at 1/5 each. That treats the ending as a filter when it is also evidence about the start.

    The second loss is getting 1/6 and being unable to say where the missing weight went. Name it: 4 and 6 each rise to 1/4, because they make ending on a 2 more likely.

    What the interviewer asks next

    • Given the game ends on a 2, what is the chance the first roll was a 4?
    • What is the expected number of rolls in this game?
    • Given the game ends on a 2, what is the chance the second roll was a 1?

    Asked at Squarepoint Capital, Quant Research Intern Interview, London, 2026 (Wall Street Oasis): why is the probability of seeing a 1 on our first roll, given that we end on a 2, not 1/5

  3. 090Each of your analysts calls the direction of a stock correctly 70% of the time, independently of the others, and your prior is 50/50. Two analysts disagree. What is your probability now that the stock goes up? What if a third analyst then sides with the one who said up?Conditional probability and BayesHardQuant and systematic fundsProp and quant trading firms

    Try it first

    Where are you after the disagreement, and after the third call?

    Show the worked solution

    50% after the disagreement, and 70% once the third analyst sides with up. Work in odds. Each analyst's call multiplies the odds by 0.7/0.3 = 7/3 in the direction called. One up and one down multiply by 7/3 and 3/7, which cancel, leaving the prior of 1:1. The third call multiplies by 7/3 again: odds of 7:3, a probability of 70%. A two-to-one split is worth one analyst, not two thirds.

    Why switch from probabilities to odds?

    Think of two friends who read the weather equally well and disagree about rain: you are back where you started, however good they are. With independent signals, each one multiplies your odds by its likelihood ratioHow much more likely a piece of evidence is if the claim is true than if it is false., so in odds form Bayes' rule is just multiplication. An analyst who is right 70% of the time says up with chance 0.7 if the stock will rise and 0.3 if it will fall, so an up call multiplies the odds by 7/3 and a down call by 3/7.

    Each call multiplies the odds: a disagreement cancels, a 2 to 1 split is one call50% up1 : 1Prior70% up7 : 3A says upx 7/350% up1 : 1B says downx 3/770% up7 : 3C says upx 7/3vote count: 67%(wrong)Three up, none down: odds 343 : 27, a 92.7% chance of up
    Starting from 50%, analyst A's up call moves the chance of up to 70%, analyst B's down call returns it to 50%, and analyst C's up call moves it to 70% again, not to the 67% a vote count suggests; three up calls and none down would give 92.7%.

    How do the three calls combine?

    Start at 1:1. The first analyst says up: 7:3, or 70%. The second says down: 7:3 times 3:7 is 1:1, back to 50%. The third says up: 7:3 again, 70%. Only the net count of calls matters, not the total. Three for up and none against would give 343:27, about 92.7%, which shows how much the one dissenter costs.

    The relationship
    oddspost=1×73×37×73=73⇒p=77+3=70%\text{odds}_{\text{post}} = 1 \times \tfrac{7}{3} \times \tfrac{3}{7} \times \tfrac{7}{3} = \tfrac{7}{3} \quad\Rightarrow\quad p = \frac{7}{7+3} = 70\%
    1the prior odds of up against down, 50/50
    7/3the likelihood ratio of an up call from a 70% accurate analyst
    3/7the likelihood ratio of a down call
    What it says in wordsMultiply the prior odds by one likelihood ratio per independent call, then turn the odds back into a probability.

    What assumption carries all of this?

    Independence. If the analysts read the same research and speak to the same management teams, their errors are correlated, and a second agreeing call adds much less than a factor of 7/3. In the extreme where the second analyst simply copies the first, it adds nothing at all. The rule also assumes each analyst is right 70% of the time whichever way the stock moves; an analyst who calls up too often tells you more when calling down. State independence as the assumption, then say how you would test it: by checking how often the analysts' past calls agreed with each other.

    Where candidates lose it

    The vote-counting answer, two out of three so 67%, is the usual loss. It treats a majority as a probability, when a two-to-one split carries exactly one net call of evidence, 70%.

    The second loss comes after the disagreement: candidates reach for something like 58%, feeling that two good analysts must add something. With equal accuracy and opposite calls, the evidence cancels exactly.

    What the interviewer asks next

    • One analyst is right 80% of the time and the other 60%. The better one says up, the other down. Where are you?
    • Five analysts split three to two. What is your probability?
    • How would you estimate how correlated your analysts' calls are, and how would you adjust for it?