Hedge Funds puzzles, solved step by step

Puzzles: 100
Traced to a firm: 38
Topics: 14
Hard: 30

Showing 1–3 of 3 · filtered from 100Clear filters

002The classic Russian roulette puzzle: a six-chamber revolver has two bullets in adjacent chambers. The cylinder is spun once, the trigger is pulled and it clicks empty. You must pull again. Is it safer to spin the cylinder again first, or not?Conditional probability and BayesCoreSchonfeldCentral · 2022
Try it first
Which gives the better chance of surviving the second pull?
Show the worked solution
Do not spin: you survive 75% of the time, against 66.7% if you spin. The empty click puts the cylinder on one of the four empty chambers, each equally likely. Because the two bullets sit together, three of those four empties are followed by another empty and only one is followed by a bullet. A fresh spin throws that information away and gives four chances in six.
What does the empty click tell you?
Picture six people in a queue where two friends always stand together. Pick someone at random who is not one of the friends and ask whether the person behind them is a friend. Only one of the four has a friend behind them: the one standing just in front of the pair. The empty click is information: it tells you which chambers you could be on, and using it is the whole puzzle. The cylinder has no memory, but you do.
After an empty click the cylinder sits on chamber 3, 4, 5 or 6; three of those are followed by an empty chamber and only chamber 6 is followed by a bullet, so not spinning survives 75% of the time against 66.7% for a fresh spin.
The relationship
$P(\text{safe} \mid \text{no spin}) = \frac{3}{4} = 75\% \qquad P(\text{safe} \mid \text{spin}) = \frac{4}{6} \approx 66.7\%$
3/4 three of the four empty chambers are followed by another empty
4/6 four of six chambers are empty after a fresh spin
What it says in wordsConditioning on the empty click beats resetting to the base rate when the bullets sit together.
Why does the answer flip if the bullets are not adjacent?
Separate the bullets, say into chambers 1 and 4. The four empties are now 2, 3, 5 and 6, and the chambers after them are 3, 4, 6 and 1: two empties and two bullets. Not spinning now survives only 2 times in 4, 50%, so spinning, at 66.7%, becomes the better choice. Adjacency is what bunches both bullets behind a single empty chamber. Ask where the bullets sit before you answer, and say that the answer depends on it.
Where does this reasoning show up on a desk?
The same move, updating on what you have just observed instead of resetting to the base rate, is how a trader reads a fill. Getting filled on your bid tells you something about who was selling, just as the empty click tells you which chamber you are on. Ignoring it is the equivalent of spinning the cylinder: it feels neutral, but it throws away an edge you were handed for free.
Where candidates lose it
Candidates say it makes no difference, because a spin feels like a clean reset and the cylinder has no memory. The trap is treating no memory in the device as no information for you: the click has ruled out the two loaded chambers as your position.
The second loss is answering without checking the layout. The case for not spinning rests entirely on the bullets being adjacent; with the bullets apart, the answer reverses. Name that condition in your answer.
What the interviewer asks next
You survive the second pull without spinning. Should you spin before a third?
Three bullets in adjacent chambers: spin or not?
What if the two bullets are in chambers 1 and 4?
Asked at Schonfeld, Quantitative Research, Central, 2022 (Wall Street Oasis): Coding, requires to know DP and divde and conquer., Russian Roulette
027A trade surveillance screen flags 90% of genuinely suspicious trades and wrongly flags 5% of clean ones. One trade in a hundred is suspicious. A trade has just been flagged. What is the chance it is actually suspicious?Conditional probability and BayesCoreCitadelMiami · 2022
Try it first
Gut answer first: a flagged trade is suspicious with probability about
Show the worked solution
About 15%, not 90%. Picture 10,000 trades. 100 are suspicious and the screen flags 90 of them. 9,900 are clean and it wrongly flags 5%, which is 495. So 585 trades are flagged and only 90 are suspicious: 90 over 585 is 15.4%. The false flags swamp the true ones because clean trades are so common.
Why is 90% the wrong number?
A smoke alarm that goes off whenever there is a fire is good. But if it also goes off every time someone makes toast, most of its alarms are toast. The 90% tells you how often a suspicious trade gets flagged; the question asks how often a flag is suspicious, and those run in opposite directions. Mixing them up is called the {term('base rate', 'How common something is before any test is run. Here, one trade in a hundred is suspicious.')} fallacy, and it is exactly what this question is built to catch.
How do you get the number without the formula?
Use counts, not percentages. Start with 10,000 trades because it makes every number whole. Split by the truth first, then by what the screen says, and then read only the flagged column. 1% of 10,000 is 100 suspicious trades; 90% of those, 90, are flagged. 9,900 are clean; 5% of those, 495, are flagged anyway. The flagged column holds 585 trades, and 90 of them are the real thing.
Of 10,000 trades, 100 are suspicious and 90 of those are flagged, while 495 of the 9,900 clean trades are flagged by mistake, so only 90 of the 585 flags, 15.4%, point at a suspicious trade.
The relationship
$P(S\mid F) = \frac{0.90 \times 0.01}{0.90 \times 0.01 + 0.05 \times 0.99} = \frac{0.009}{0.0585} \approx 15.4\%$
S the trade is suspicious
F the screen flags the trade
0.01 the base rate: one trade in a hundred is suspicious
What it says in wordsThe chance a flag is right equals true flags divided by all flags.
Add the desk point after the number. A second, independent check changes things fast: run the 585 flagged trades through a second screen with the same error rates and the 15.4% prior becomes about 77%. A weak test is still useful as a first filter; it is only misleading when its hit rate is read as its accuracy.
Where candidates lose it
Most candidates say 90% within a second, because the question hands them that number. It is the chance of a flag given a suspicious trade, and the interviewer asked the reverse.
The second loss is starting on Bayes' formula with decimals and getting tangled. Say you will use 10,000 trades, draw the two splits, and the answer reads straight off the flagged column.
What the interviewer asks next
What false flag rate would make a flag right half the time?
The flagged trades go through a second independent screen and are flagged again. What is the chance now?
Compliance wants to catch 99% of suspicious trades. What does that usually do to the false flag rate?
Asked at Citadel, Sales and Trading, Miami, 2022 (Wall Street Oasis): I got a question about Bayes' theorem applied to a practical scenario
077In a Monty Hall game you pick door 1. This host does not know where the car is: he opens one of the other two doors at random, and it happens to show a goat. Should you switch, and why does the usual two-thirds answer no longer hold?Conditional probability and BayesCoreSquarepoint CapitalLondon · 2026
Try it first
The host opened a door at random and it happened to show a goat. What is your chance of winning if you switch?
Show the worked solution
It makes no difference: switching and sticking each win half the time. List the six equally likely cases of car position and the host's random pick. Two of them reveal the car, and you have seen that they did not happen. The four that remain split two and two. The knowing host gives two thirds only because he never risks the car, which pushes those two cases into the switch column.
Where does the usual two-thirds answer come from?
In the standard game the host knows where the car is and always opens a goat door. Your first pick is right one time in three, and nothing the knowing host does can change that, so the other two thirds sit on the remaining closed door. His choice carries information because it is forced: when the car is behind door 2, he must open door 3, and when it is behind door 3, he must open door 2.
What changes when the host picks at random?
Picture a friend who does not know the answer to a quiz question and strikes out one option on a whim. If that option happens to be wrong, you have learned less than if someone who knew had struck it. The random host is that friend. Write out six cases: the car behind door 1, 2 or 3, each with the host's coin choosing door 2 or door 3. In two of the six the random host opens the car door, and the goat you saw rules those two out, which removes switch wins rather than stick wins.
In the same six equally likely cases, a knowing host redirects the two where his coin points at the car, so switching wins 4 of 6; a random host shows the car in those two, they are ruled out, and switching wins 2 of the 4 that remain, one half.
Count what is left. The car behind door 1 survives both host choices: two cases where sticking wins. The car behind door 2 survives only when the host opened door 3, and the car behind door 3 only when he opened door 2: two cases where switching wins. Two against two.
The relationship
$P(\text{switch wins} \mid \text{goat shown}) = \frac{2/6}{4/6} = \frac{1}{2}$
2/6 cases where the car is behind the other closed door and the host showed a goat
4/6 all cases where the host showed a goat, which is what you observed
What it says in wordsCondition on what you saw: of the cases where a goat appears, half have the car behind the door you would switch to.
Why would an interviewer want the intuitive answer broken rather than recited?
Because the lesson travels to every desk. The same observation carries different information depending on the process that produced it. A strong track record shown by a manager who launched ten funds and closed the nine that did badly is a host choosing which door to open for you. Before you update on evidence, ask whether the source could have shown you something else, and whether it chose what to show.
Where candidates lose it
Candidates who know the classic puzzle answer two thirds on reflex. The interviewer changed one fact, that the host knows, and is checking whether you notice that the host's knowledge is exactly what made switching better.
The other loss is saying one half without a reason, which sounds like the naive answer to the classic game. Name the two ruled-out cases, where the car would have been shown, and show that both come out of the switch column.
What the interviewer asks next
With 100 doors and a knowing host who opens 98 goat doors, what is your chance if you switch?
With 100 doors and a random host who happens to open 98 goat doors, what is it now?
Where does the same logic show up when you read a fund family's track record?
Asked at Squarepoint Capital, Quant Research Intern Interview, London, 2026 (Wall Street Oasis): notably I was asked why the 'intuitive answer' was not true rather than just what the correct answer was, related to the Monty Hall problem

3/4	three of the four empty chambers are followed by another empty
4/6	four of six chambers are empty after a fresh spin

S	the trade is suspicious
F	the screen flags the trade
0.01	the base rate: one trade in a hundred is suspicious

2/6	cases where the car is behind the other closed door and the host showed a goat
4/6	all cases where the host showed a goat, which is what you observed

Try it first

What does the empty click tell you?

Why does the answer flip if the bullets are not adjacent?

Where does this reasoning show up on a desk?

Where candidates lose it

What the interviewer asks next

Try it first

Why is 90% the wrong number?

How do you get the number without the formula?

Where candidates lose it

What the interviewer asks next

Try it first

Where does the usual two-thirds answer come from?

What changes when the host picks at random?

Why would an interviewer want the intuitive answer broken rather than recited?

Where candidates lose it

What the interviewer asks next