Hedge Funds puzzles, solved step by step

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.