Derivatives Foundation puzzles, solved step by step
- Puzzles
- 100
- Traced to a firm
- 66
- Topics
- 12
- Hard
- 29
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?Jane 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 relationship0.2/0.8 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
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?Squarepoint 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 relationship1/4 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?Belvedere 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 relationship0.1 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
