Derivatives Foundation puzzles, solved step by step

Puzzles: 100
Traced to a firm: 66
Topics: 12
Hard: 29

Showing 1–7 of 7 · filtered from 100Clear filters

003A logger stamps every event to the nanosecond, nine decimal places, and whenever the timestamp is missing it writes nine zeros instead. In 100,000 records you find 15 whose fractional part is exactly nine zeros. What is the probability that at least one of those 15 is a filled-in missing value?Conditional probability and BayesHardJump TradingAnonymous interview candidate in · 2022

Try it first

First instinct: roughly how many genuine timestamps, out of 100,000, should end in nine zeros by chance?

Show the worked solution

Essentially 1; the 15 are gaps. A genuine stamp ends in nine zeros with probability 10^-9, so in 100,000 records you expect 0.0001 such endings. The chance that 15 or more arise genuinely is about 10^-72, which no reasonable prior on missing data can overcome: even a missing rate of one in ten thousand would produce about 10 filled-in endings. So the probability that at least one of the 15 is a filled-in value is 1 to every decimal place you could print.

What is the question really asking you to compare?

A shopkeeper who finds 15 notes with the same serial number does not ask what the chance of a coincidence is; she asks which explanation makes 15 identical notes likely. This is a Bayes question in disguise: compare how likely 15 all-zero endings are if nothing is missing against how likely they are if some values are missing, then weight by a prior. The first likelihood is astronomically small; the second is ordinary. The prior would need to be more extreme than anything a real system justifies to change the answer.

Genuine nanosecond stamps should produce 0.0001 all-zero endings in 100,000 records, while 15 were observed, five orders of magnitude more, and the chance of 15 or more genuine ones is about 10^-72, so the observation is explained only by filled-in missing values.

How do you put a number on the genuine case?

Each of the 100,000 stamps ends in a specific nine-digit string with probability one in a billion, so the count of genuine all-zero endings is Poisson with mean 0.0001. The probability of exactly 15 is e^(-0.0001) times 0.0001^15 over 15 factorial, which is about 10^-72. You do not need the exact figure in the room; say that 0.0001 to the fifteenth power is 10^-60 before dividing by 15 factorial, and the interviewer has what they need. The point is to show you can set up the count, not to print 72 zeros.

The relationship

P(\geq 1 \text{ missing} \mid 15) = 1 - \frac{P(15 \mid \text{none}) \, P(\text{none})}{P(15)} \approx 1 - \frac{10^{-72} \, P(\text{none})}{P(15)} \approx 1

P(15 \| none)	the chance of 15 genuine all-zero endings when no value is missing, Poisson with mean 0.0001
P(none)	your prior that the data set has no missing values at all
P(15)	the overall chance of seeing 15, dominated by the missing-value explanation

What it says in wordsThe probability that none are missing is the genuine likelihood times its prior, divided by the total, and the genuine likelihood is so small that the result rounds to 1 whatever prior you hold.

What does the interviewer want to hear about the prior?

The honest answer is that the question is underspecified: without a prior on how often values go missing, you cannot write a single number. Say that, then show it does not matter: for the posterior to drop even to 99.9% you would need a prior of no missing values more than 10^69 times stronger than the alternative, and no logging system earns that confidence. For contrast, a modest missing rate of one record in ten thousand would give an expected 10 filled-in endings, right where the observed 15 sits. A limitation worth adding: the argument assumes the genuine fractional digits are uniform, which breaks if the clock quantises to microseconds and pads with zeros itself.

Where candidates lose it

Candidates reach for the binomial probability of 15 genuine zeros and stop, reporting a tiny number as if it were the answer. The question asks for the probability of a missing value given the data, which needs the comparison with the alternative, not a single likelihood.

The second loss is freezing because no prior is given. The strong move is to name the missing input, then show that the likelihood ratio is so lopsided that the prior cannot matter. That is what a desk wants: a conclusion that survives the unknown.

What the interviewer asks next

Now the logger stamps to the microsecond, six digits, and pads with three zeros. Does the argument survive?
Suppose only 1 record ends in nine zeros. What would you conclude then, and what would you need to know?
How would you check the data itself rather than reason about it?

Asked at Jump Trading, Prop Trading, Anonymous interview candidate in, 2022 (Wall Street Oasis): What is the probability of at least 1 missing value given that we see 15 data points with 0's in the end

006A trade surveillance system raises an alert on 95% of genuinely suspicious trades and, wrongly, on 2% of normal trades. One trade in a thousand is genuinely suspicious. An alert has just fired on a trade. What is the probability the trade is suspicious?Conditional probability and BayesWarm upCitadelMiami · 2022
Try it first
Gut answer before you count anything.
Show the worked solution
About 4.5%. Count 100,000 trades. One in a thousand is suspicious, so 100 are, and 95 of those alert. The other 99,900 are normal, and 2% of them, 1,998, alert anyway. Alerts total 2,093, of which 95 are genuine, so the probability that an alerted trade is suspicious is 95 over 2,093, about 4.5%. The 95% hit rate is not the answer; the base rate is what decides it.
Why does a 95% accurate system give a 4.5% answer?
A smoke alarm that goes off for 2% of toast is a fine alarm in a house that is never on fire; nearly every ring will be toast. When the thing you are looking for is rare, even a small false-alarm rate applied to the huge normal pile produces more alerts than the true cases produce. Here 2% of 99,900 normal trades is 1,998, twenty times the 95 genuine alerts. The system is not bad; the base rate is low, and that is what the question is testing.
Of 100,000 trades, 100 are suspicious and raise 95 true alerts, while the 99,900 normal trades raise 1,998 false ones, so alerts total 2,093 and a trade that alerts is genuinely suspicious only 4.5% of the time.
The relationship
$P(S \mid A) = \frac{P(A \mid S)\,P(S)}{P(A \mid S)\,P(S) + P(A \mid N)\,P(N)} = \frac{0.95 \times 0.001}{0.95 \times 0.001 + 0.02 \times 0.999} = \frac{95}{2{,}093} \approx 4.5\%$
S, N a suspicious trade, a normal trade
A an alert fires
P(A | S) = 0.95 the hit rate
P(A | N) = 0.02 the false-alarm rate
P(S) = 0.001 the base rate
What it says in wordsTrue alerts divided by all alerts, where all alerts are the true ones plus the false ones from the normal pile.
What is the fastest way to say it in the room?
Do not write Bayes' formula; count a round number of trades. Say: in 100,000 trades, 100 are suspicious and 95 alert; 99,900 are normal and 1,998 alert; 95 over 2,093 is about 4.5%. Three sentences, no algebra, and every number is checkable by the person listening. The odds form is just as quick: prior odds 1 to 999, likelihood ratio 0.95 over 0.02, about 47.5, so posterior odds 47.5 to 999, roughly 1 to 21.
What does the desk do with a 4.5% answer?
It decides what the alert is for. A 4.5% hit rate is fine for a filter that sends trades to a human for a second look, and useless for an automatic block, because 95% of blocked trades would be legitimate business. That is the trade-off every surveillance, fraud and risk-limit system lives with: a lower threshold catches more of the 100 but drags in more of the 99,900. Say the limitation too: the 2% and 95% are themselves estimates from past data, and a system tuned on last year's patterns can drift.
Where candidates lose it
The whole trap is answering 95%, confusing the probability of an alert given a suspicious trade with the probability of a suspicious trade given an alert. Interviewers ask this precisely because the two sound the same and are twenty times apart.
The second loss is reaching for the formula and tangling the denominator. Count 100,000 trades and the denominator builds itself: 95 plus 1,998.
What the interviewer asks next
The false-alarm rate is cut to 0.5%. What is the probability now?
Two independent systems both alert on the same trade. What is the probability it is suspicious?
What base rate would make an alert a coin flip, and what does that tell you about where surveillance is worth running?
Asked at Citadel, Sales and Trading, Miami, 2022 (Wall Street Oasis): I got a question about Bayes' theorem applied to a practical scenario, which I handled decently
038Box A holds 3 red and 1 black ball, box B holds 1 red and 3 black. You pick a box at random and draw a red ball. You then draw again from the same box without replacement. What is the probability the second ball is red?Conditional probability and BayesCoreProp trading firmsMarket making
Try it first
Pick the probability of a second red before you compute.
Show the worked solution
Exactly 1/2. The first red is evidence: box A produces a red with probability 3/4, box B with 1/4, so after seeing red the odds are 3 to 1 for A, a posterior of 3/4. Then update the contents. Box A has 2 red of 3 left, so a second red has probability 2/3; box B has 0 red of 3, probability 0. Average over the updated boxes: 3/4 x 2/3 + 1/4 x 0 = 1/2. Forgetting the update gives 1/3, which is wrong.
Why does the first red change which box you think you hold?
If a friend's cooking is spicy three times out of four and yours one time in four, and a dish picked at random from one of you turns out spicy, you would guess it was your friend's, and you would be right three times in four. The first red ball is the same clue: it comes from box A three times as easily as from box B, so the box you are holding is A with probability 3/4, not 1/2. The arithmetic is Bayes. The weight for A is 1/2 x 3/4 = 3/8, for B it is 1/2 x 1/4 = 1/8, and A's share of the total 1/2 is 3/4. The question is built so that candidates who skip this update get a clean wrong answer.
After a red first draw the box is A with posterior 3/4 and B with 1/4, box A has 2 red of 3 left and box B none, so the chance of a second red is 3/4 x 2/3 plus 1/4 x 0, which is exactly one half.
What has to be updated besides the belief about the box?
The box itself. One red ball has left it. Box A now holds 2 red and 1 black, so a red is 2/3; box B now holds 0 red and 3 black, so a red is impossible. Both updates matter and they pull in opposite directions: believing more in A pushes the answer up, removing a red from A pushes it down, and here they land on exactly 1/2. Say that out loud, then check it by counting. Imagine 8 rounds, 4 with each box. Box A yields a first red in 3 of its 4 rounds and box B in 1 of its 4: four red first draws in all. In the three from A the second draw is red 2/3 of the time, 2 rounds; in the one from B, never. Two of four is 1/2.
The relationship
$P(R_2 \mid R_1) = P(A \mid R_1)\,P(R_2 \mid A, R_1) + P(B \mid R_1)\,P(R_2 \mid B, R_1) = \tfrac34 \cdot \tfrac23 + \tfrac14 \cdot 0 = \tfrac12$
R1, R2 the first and second draws are red
P(A | R1) the posterior that the box is A after one red, 3/4
P(R2 | A, R1) a second red from box A with one red already gone, 2/3
What it says in wordsWeight each box's chance of a second red by how likely that box is after the first red.
How do the variants separate candidates who understand from those who memorised?
Put the first ball back and the box contents do not change but the belief still does: 3/4 x 3/4 + 1/4 x 1/4 = 5/8, which is above 1/2 because the posterior leans to A and A stays rich in red. Forget the belief update and keep the contents update, and you get 1/2 x 2/3 + 1/2 x 0 = 1/3. The gap between 1/3, 1/2 and 5/8 is the whole lesson: the first draw tells you about the box and changes the box, and you need both. If a second red does appear, the posterior for A becomes 1, since B has no reds left, and a third red is then 1/2. The limitation is the usual one for Bayes: the 1/2 prior for each box is given here, and in a trading setting the prior is the thing you are least sure of.
Where candidates lose it
The common loss is 1/3: keeping the 1/2 prior on each box and only updating the contents. The first red is evidence about the box, and leaving it out throws away half the information in the question.
The second is 2/3: updating to box A and then forgetting that B is still possible with probability 1/4, or forgetting that A has lost a red. Carry both boxes through to the end, then average.
What the interviewer asks next
The first ball is put back before the second draw. What is the probability now?
The second ball is also red. What is the probability the box is A, and that a third draw is red?
Box B held 0 red and 4 black instead. What changes, and what is the answer?
A trader sees one winning trade from a new signal. What is the analogue of the box update?
053It rains on 20% of days. Two forecasters work independently: each calls rain on 90% of the days it actually rains, and also calls rain on 15% of the dry days. Both call rain for tomorrow. What is the probability it rains? And if one calls rain and the other does not?Conditional probability and BayesCoreJane StreetNew York · 2025
Try it first
Two independent forecasters both say rain. Instinct first: how likely is rain?
Show the worked solution
90% if both call rain, and 15% if they disagree. Over 1,000 days, 200 are rainy and 800 are dry. Both call rain on 200 x 0.9 x 0.9 = 162 rainy days and on 800 x 0.15 x 0.15 = 18 dry days, so rain is 162 of 180. A split call happens on 36 rainy days and 204 dry days, so rain is 36 of 240, which is below the 20% base rate.
Why count days instead of multiplying probabilities?
A doctor reading two test results does the same thing: imagine a thousand patients, split them by who is actually ill, then split each group by what the tests said. Natural frequencies keep the base rate in the picture, where the probability form hides it, so you never confuse the chance of the evidence given rain with the chance of rain given the evidence. The 200 rainy days produce two rain calls 162 times; the 800 dry days, despite each forecaster being wrong only 15% of the time, produce two rain calls 18 times because 800 is a big base. The ratio is 162 to 18, so 90%.
Of 1,000 days, 200 rainy days give 162 double rain calls, 36 split calls and 2 with no call, while 800 dry days give 18, 204 and 578, so two rain calls make rain 90% likely and a split call makes it only 15% likely.
Why does a split call push the chance below the base rate?
Because a forecaster saying dry is itself evidence, and it is strong evidence. Each forecaster calls dry on 85% of dry days but only 10% of rainy days, so a dry call divides the odds of rain by 8.5, while a rain call multiplies them by only 6. Put together, one rain call and one dry call multiply the odds by 6 x (1/8.5), about 0.71, so the odds fall from 1 to 4 to about 1 to 5.67, which is 15%. The common feeling that the two calls cancel is wrong because the two kinds of error are not symmetric.
The relationship
$\frac{P(R\mid\text{both})}{P(D\mid\text{both})} = \frac{0.2}{0.8}\times\frac{0.9}{0.15}\times\frac{0.9}{0.15} = \frac{1}{4}\times 36 = 9 \;\Rightarrow\; P(R\mid\text{both}) = \tfrac{9}{10}$
0.2/0.8 the prior odds of rain, 1 to 4
0.9/0.15 the likelihood ratio of one rain call, 6
9 the posterior odds of rain after two independent calls
What it says in wordsPrior odds of one to four, multiplied by six for each independent rain call, give odds of nine to one, which is ninety per cent.
What assumption does the answer lean on, and when would it fail?
Independence given the weather. If both forecasters read the same satellite feed, two rain calls are closer to one piece of evidence than two, and the odds should be multiplied by something well short of 36. On a desk the same mistake appears when three models trained on the same data all flag a trade: the signals agree because they share inputs, not because the evidence is three times as strong. The answer also treats the 90% and 15% rates as known; with rates estimated from a short record, the posterior is softer than it looks.
Where candidates lose it
The fast wrong answer is 81%, from 0.9 times 0.9, which is the likelihood of the evidence rather than the probability of rain. Candidates who write Bayes' formula from memory often lose the base rate and give the same number.
The second trap is the split call: most people say about 50% or go back to 20%. Say that a dry call is strong evidence, divide the odds by 8.5 and multiply by 6, and the 15% follows.
What the interviewer asks next
Both forecasters say dry. What is the chance of rain now?
The two forecasters share the same data and always agree. What does two rain calls tell you?
A third independent forecaster with the same accuracy calls rain. Update the 90%.
How accurate would one forecaster need to be for a single rain call to make rain more likely than not?
Asked at Jane Street, Generalist, New York, 2025 (Wall Street Oasis): a question about the probability of rain the next day that relied on a very in depth understanding of bayes theorem
064A bag holds 9 fair coins and 1 coin with heads on both sides. You pull one out at random and flip it 5 times, getting 5 heads. What is the probability it is the two-headed coin, and what is the probability the next flip is heads?Conditional probability and BayesCoreQuant tradingHedge funds
Try it first
Five heads in a row from a coin that is two-headed one time in ten. How likely is it the two-headed one?
Show the worked solution
78.0% that it is the two-headed coin, and 89.0% that the next flip is heads. Prior odds are 1 to 9. The two-headed coin gives a head with certainty and a fair coin with chance 1/2, so each head multiplies the odds by 2; after five heads the odds are 32 to 9, which is 32/41. The next flip is heads with chance 32/41 x 1 + 9/41 x 1/2 = 73/82.
Why work in odds rather than probabilities?
A doctor who sees the same symptom five mornings running does not recompute the whole diagnosis each day; each new observation multiplies the odds of the condition by one fixed factor. In odds form, Bayes' rule is a multiplication: posterior odds equal prior odds times the likelihood ratio of each observation, and here every head has the same ratio of 1 to 1/2, which is 2. So the odds on the two-headed coin go 1:9, 2:9, 4:9, 8:9, 16:9, 32:9. Convert at the end: 32 divided by 32 plus 9 is 32/41. Doing it in probabilities means dividing by a different normaliser five times, which is where people slip.
Each head doubles the odds on the two-headed coin, lifting the probability from 10% to 78% after five heads, and the chance the next flip is heads, 89%, mixes a certain head from the two-headed coin with a coin flip from a fair one.
Why is the next flip not simply 78% heads?
Because the fair coin also produces heads. The next flip is heads if the coin is two-headed, with chance 32/41, or if the coin is fair and lands heads, with chance 9/41 times 1/2, and the two routes add to 73/82, about 89%. Candidates who answer 78% have confused the probability of the hypothesis with the probability of the outcome. The gap between the two is the fair coin's half chance of a head, weighted by the 22% chance you are holding a fair coin.
The relationship
$\frac{P(\text{two-headed}\mid 5H)}{P(\text{fair}\mid 5H)} = \frac{1}{9}\times\left(\frac{1}{1/2}\right)^5 = \frac{32}{9} \;\Rightarrow\; P = \frac{32}{41}, \qquad P(\text{next } H) = \frac{32}{41}\cdot 1 + \frac{9}{41}\cdot\frac{1}{2} = \frac{73}{82}$
1/9 the prior odds of drawing the two-headed coin
(1 / (1/2))^5 the likelihood ratio of five heads, two to the fifth
73/82 the chance of a sixth head, mixing both coins
What it says in wordsFive heads multiply the prior odds by thirty-two, giving thirty-two to nine, and the next flip mixes a sure head with a fair flip in those proportions.
How many heads would it take to be nearly sure, and what does a desk take from this?
Each head doubles the odds, so after 10 heads the odds are 1,024 to 9, about 99.1%, and after 5 you are only at 78%. Evidence that is merely consistent with a hypothesis moves you slowly when the alternative also produces it often, which is why five good months from a new trading strategy prove far less than people feel they do. The limitation is the prior: if the bag held 99 fair coins and one two-headed, five heads would leave you at 32 to 99, still under 25%, and no amount of looking at the flips alone tells you the composition of the bag.
Where candidates lose it
The fast wrong answer is 1 minus 1/32, about 97%, which is the chance a fair coin would not have done this. That number ignores that there are nine fair coins for every two-headed one. Start from the prior odds and double.
The second loss is giving 78% for the next flip. The next flip is a mixture: a certain head from the two-headed coin and a half chance from a fair one.
What the interviewer asks next
After how many heads does the probability it is the two-headed coin pass 99%?
The sixth flip is tails. What is the probability it is the two-headed coin now?
The bag has 99 fair coins and one two-headed coin. What are the two answers after five heads?
Why is the probability of the next head always between the fair coin's 1/2 and 1?
089Four doors hide one prize. You pick a door. The host, who knows where the prize is, opens one of the other three that is empty and offers you a switch to either of the two remaining closed doors. Should you switch, and what are your chances each way?Conditional probability and BayesCoreSquarepoint CapitalLondon · 2026
Try it first
After the host opens an empty door, what is the chance your original door hides the prize?
Show the worked solution
Switch: either other closed door wins with probability 3/8, against 1/4 for staying. Your door had a 1/4 chance and the host's action cannot change it, because he can always open an empty door among the other three. The remaining 3/4 sat on three doors and now sits on two, so each carries 3/8. Switching does not make you a favourite, it only beats staying, 37.5% against 25%.
Why does your own door keep its 1/4?
A friend who knows the answers to a quiz looks at three questions you did not attempt and crosses out one wrong option. She would have done that whether your own answer was right or wrong, so her crossing out says nothing about your answer. Information only moves probability onto or off your door if the action could have depended on what is behind it, and the host can open an empty door among the other three every time. So your door stays at 1/4, exactly where it started.
Before the host acts each of the four doors carries 1/4, and after he opens empty door 4 your door 1 still carries 1/4 while the 3/4 that sat on doors 2, 3 and 4 now sits on doors 2 and 3, 3/8 each, so switching wins 37.5% of the time against 25% for staying.
How does Bayes give 3/8 for each of the other doors?
Say the host opens door 4. If the prize is behind your door 1, he chooses among doors 2, 3 and 4 at random, so he opens door 4 with chance 1/3. If it is behind door 2, he must choose between 3 and 4, chance 1/2; the same for door 3. If it is behind door 4, he never opens it. Multiply each by the prior of 1/4: 1/12, 1/8, 1/8 and 0, which add to 1/3. Divide through and the posterior is 1/4 for door 1, 3/8 for door 2, 3/8 for door 3 and nothing for door 4. The host is more likely to open door 4 when the prize sits behind door 2 or 3 than when it sits behind yours, and that difference is the information.
The relationship
$P(\text{door 2} \mid \text{opens 4}) = \frac{\tfrac{1}{4}\cdot\tfrac{1}{2}}{\tfrac{1}{4}\cdot\tfrac{1}{3} + \tfrac{1}{4}\cdot\tfrac{1}{2} + \tfrac{1}{4}\cdot\tfrac{1}{2} + 0} = \frac{1/8}{1/3} = \frac{3}{8}$
1/4 the prior on each door
1/2 the chance the host opens door 4 when the prize is behind door 2
1/3 the chance he opens door 4 when the prize is behind your door
What it says in wordsEach closed door you did not pick ends up with three eighths, half again as much as your own quarter.
Then generalise, which is what the question tests. With n doors and one opened, staying wins 1/n and each other closed door wins (n - 1)/(n(n - 2)). With three doors that is 2/3, the familiar answer; with 100 doors it is 99/9,800, barely above the 1/100 for staying. The edge from switching shrinks as the host reveals a smaller share of the doors. The limitation is the host's rule: if he opened a door at random and it merely happened to be empty, he would carry no information about the other doors and switching would gain nothing.
Where candidates lose it
The intuitive loss is 1/3 each: three closed doors, so a third apiece. It treats the host's choice as if he had picked blindly, which he did not. The interviewer asks this version precisely to see whether you say why the intuitive answer is wrong rather than recite 2/3.
The second loss is reciting the three-door logic and saying switching wins 3/4. Switching moves you to one of two doors, not to both, so you collect only half of the 3/4.
What the interviewer asks next
Now the host opens two empty doors and offers you the last closed one. What are the chances?
If the host opens a door at random and it happens to be empty, should you switch?
With 100 doors and one opened, by how much does switching beat staying?
Asked at Squarepoint Capital, Quant Research Intern Interview, London, 2026 (Wall Street Oasis): I was asked why the 'intuitive answer' was not true rather than just what the correct answer was, related to the Monty Hall problem
098One trader in ten has a real edge and wins 60% of days; the rest win 50%. A new hire wins 8 of her first 10 days. What is the probability she has an edge?Conditional probability and BayesCoreBelvedere TradingChicago · 2022
Try it first
Roughly how likely is it that she has an edge?
Show the worked solution
About 23%. With an edge, 8 wins in 10 has probability 45 x 0.6^8 x 0.4^2 = 12.1%; without, 45/1,024 = 4.4%. Out of 1,000 traders, 100 have an edge and about 12.1 of them go 8 of 10; 900 do not and about 39.6 of them go 8 of 10 by luck. So 12.1 of 51.6, about 23.4%, have an edge. A strong start raises the chance from 10% but leaves luck the likelier story.
Why is the answer so much lower than her win rate?
If a test for a rare condition comes back positive, most positives can still be false, simply because there are so many more people without the condition to produce them. Trading records work the same way. Ten days is a weak test and skill is rare, so most traders with a strong start are lucky members of the large unskilled group, not members of the small skilled one. The win rate of 80% is what she did, not what she is; the question is which kind of trader is likelier to produce that record, weighted by how many of each kind there are.
Of 1,000 new traders, the 100 with an edge produce about 12.1 who go 8 of 10 and the 900 without produce about 39.6, so only about 23% of traders with that start have an edge, even though the record is 2.75 times likelier with one.
How do you run the numbers in the room?
Use natural frequencies. Of 1,000 hires, 100 have an edge. Each goes 8 of 10 with probability C(10,8) x 0.6^8 x 0.4^2 = 45 x 0.01680 x 0.16 = 0.1209, so about 12.1 of them. Of the 900 without, each goes 8 of 10 with probability 45/1,024 = 0.0439, so about 39.6. Among the 52 or so traders with 8 wins, about 12 have an edge, a posterior of 23.4%, because a likelihood ratio of 2.75 cannot overcome prior odds of 1 to 9. In odds form: 1/9 x 2.75 = 0.306, and 0.306/(1 + 0.306) = 23.4%.
The relationship
$P(\text{edge} \mid 8/10) = \frac{0.1 \times 0.1209}{0.1 \times 0.1209 + 0.9 \times 0.0439} \approx 0.234$
0.1 the share of traders with an edge
0.1209 the chance of exactly 8 wins in 10 at 60% a day
0.0439 the chance of exactly 8 wins in 10 at 50% a day
What it says in wordsWeight each explanation by how common it is and how well it fits the record, then take the edge's share.
Two refinements the interviewer may ask for. If the record were 8 or more wins rather than exactly 8, the answer rises a little, to about 25%. And the practical point: a desk that promotes on a ten-day record mostly promotes luck. It takes many more days to separate a 60% trader from a 50% one, because the gap in win rate is small next to the day-to-day noise. The limitation of the model is the two-type world; real skill comes in shades, which spreads the posterior out but does not change the lesson about rare skill and short records.
Where candidates lose it
The common loss is answering 80% or 60%, confusing the trader's record with the probability that she is skilled. The interviewer is testing whether you start from the base rate.
The second loss is computing the likelihoods correctly and forgetting the prior, which gives 12.1/(12.1 + 4.4) = 73%. That is the answer for a world where half of all traders have an edge, which is not the world described.
What the interviewer asks next
How many days would she need at an 80% win rate before you were 90% sure she had an edge?
If one trader in three had an edge, what would 8 of 10 imply?
Her next 10 days are 5 wins and 5 losses. Update the probability.
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

S, N	a suspicious trade, a normal trade
A	an alert fires
P(A \| S) = 0.95	the hit rate
P(A \| N) = 0.02	the false-alarm rate
P(S) = 0.001	the base rate

R1, R2	the first and second draws are red
P(A \| R1)	the posterior that the box is A after one red, 3/4
P(R2 \| A, R1)	a second red from box A with one red already gone, 2/3

0.2/0.8	the prior odds of rain, 1 to 4
0.9/0.15	the likelihood ratio of one rain call, 6
9	the posterior odds of rain after two independent calls

1/9	the prior odds of drawing the two-headed coin
(1 / (1/2))^5	the likelihood ratio of five heads, two to the fifth
73/82	the chance of a sixth head, mixing both coins

Try it first

What is the question really asking you to compare?

How do you put a number on the genuine case?

What does the interviewer want to hear about the prior?

Where candidates lose it

What the interviewer asks next

Try it first

Why does a 95% accurate system give a 4.5% answer?

What is the fastest way to say it in the room?

What does the desk do with a 4.5% answer?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the first red change which box you think you hold?

What has to be updated besides the belief about the box?

How do the variants separate candidates who understand from those who memorised?

Where candidates lose it

What the interviewer asks next

Try it first

Why count days instead of multiplying probabilities?

Why does a split call push the chance below the base rate?

What assumption does the answer lean on, and when would it fail?

Where candidates lose it

What the interviewer asks next

Try it first

Why work in odds rather than probabilities?

Why is the next flip not simply 78% heads?

How many heads would it take to be nearly sure, and what does a desk take from this?

Where candidates lose it

What the interviewer asks next

Try it first

Why does your own door keep its 1/4?

How does Bayes give 3/8 for each of the other doors?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the answer so much lower than her win rate?

How do you run the numbers in the room?

Where candidates lose it

What the interviewer asks next