Quant puzzles, solved step by step

8/11, about 72.7%. If it is raining, all three say yes with probability (2/3)^3 = 8/27. If it is dry, all three must be lying, (1/3)^3 = 1/27. Weight each by how often it happens: 1/4 x 8/27 against 3/4 x 1/27, which is 8 parts to 3. Three agreeing witnesses move a 25% prior a long way, but not to certainty.

Why is the answer not simply 8/9?

Picture a clinic where a test is quite reliable but the illness is uncommon. A positive result makes the illness more likely, but how much more depends on how rare it was to begin with. The friends' agreement tells you how much more likely rain makes their answer than dry does, eight times, but it does not erase the fact that dry days are three times as common. 8/9 is the answer you get if rain and dry start level. Here they do not.

How do you set it up so the arithmetic stays small?

Use odds rather than probabilities. Posterior odds are prior odds times the likelihood ratioHow many times more likely the evidence is if the hypothesis is true than if it is false., and both are easy numbers here. Prior odds of rain are 1 to 3. The likelihood ratio of three yeses is (2/3)^3 over (1/3)^3, which is 2 cubed, 8. So the posterior odds are 8 to 3, and the probability is 8 over 8 plus 3, 8/11. Each additional agreeing friend would double the odds again.

What assumption is doing the work, and should you say it?

The calculation needs the friends to lie independently. If they could be coordinating a joke, three yeses are really one piece of evidence, and the answer falls back towards the one-friend figure of 2/5. Say the independence assumption out loud, then give 8/11. Interviewers often follow up by making one friend unreliable or by letting them talk to each other.

About 12.7%. A skilled manager wins exactly 8 of 10 with probability 0.1209; an unskilled one with 0.0439, a likelihood ratio of about 2.75. Prior odds of skill are 5 to 95, 1 to 19. Posterior odds are 2.75 to 19, so the probability is 0.05 x 0.1209 / (0.05 x 0.1209 + 0.95 x 0.0439) = 0.127. The record helps, but luck has far more players.

Why does an impressive record move the needle so little?

Imagine a thousand people each tossing a coin ten times. About 55 of them will get eight heads or better with a fair coin. If a handful of the thousand had slightly biased coins, you still could not pick them out from the lucky crowd by one run of ten. Evidence moves a belief in proportion to how much more likely it is under one explanation than the other, and 8 wins in 10 is not much more likely from a 60% manager than from a 50% one. The ratio is about 2.75.

How do you set it up quickly?

Use odds. Prior odds of skill are 1 to 19; the likelihood ratioHow many times more likely the evidence is under one hypothesis than under the other. of the record is (0.6/0.5) to the 8 times (0.4/0.5) squared, which is 1.2 to the 8 times 0.64, about 2.75; multiply to get posterior odds of about 0.145. Converting, 2.75 over 2.75 + 19 is 12.7%. The binomial coefficient, 45, is the same in both likelihoods and cancels, so you never need it.

What does this say about picking managers?

When skill is rare and its edge is small, even a long, strong track record leaves luck as the likelier explanation. Using 8 or more wins instead of exactly 8 barely changes things: the answer becomes 13.9%. The honest limitation is that the model is stylised: real skill is not a fixed 60%, and survivorship means the managers you hear about were already filtered for good records, which pushes the true figure lower still.

90%. Each die starts at one in three. The chance of two sixes is 1/36 for the fair die, 1/4 for the loaded die and zero for the die with no six. Weight each by its prior: 1/108 for the fair die, 1/12 for the loaded die, nothing for the third. The loaded die's weight is nine times the fair die's, so its probability is 9/10.

What does the die that never shows six do to the answer?

A neighbour tells you a red car blocked the gate this morning. If one of your suspects owns only a blue scooter, that suspect is out, however likely they looked before. A hypothesis that cannot produce the evidence gets zero weight afterwards, no matter what its prior was. The no-six die could never give two sixes, so it drops out, and the question becomes a contest between the fair die and the loaded die, which started level at one third each.

How much does each surviving die's likelihood count?

Now compare how easily each remaining die produces what you saw. The fair die gives two sixes 1 time in 36. The loaded die gives a six half the time, so two in a row 1 time in 4. With equal priors, the posterior odds are just the ratio of the likelihoodsThe probability of the observed evidence under each hypothesis, before any prior is applied.: 1/4 against 1/36, which is 9 to 1. Nine parts in ten is 90%.

Check it by counting, the safer habit under pressure. Imagine 108 rounds of drawing a die and rolling it twice, 36 rounds with each die. The fair die gives two sixes once, the loaded die 9 times, the no-six die never. Of the 10 double sixes, 9 came from the loaded die, and the prior of one third cancels because every die got the same number of rounds. The loaded die's other faces, 1 in 10 each, never enter, because only sixes were seen.

About 26.7%. Work in odds. The prior odds are 1 to 99. Independent evidence multiplies the odds by each likelihood ratio: 1 x 9 x 4 = 36, so the posterior odds are 36 to 99. As a probability that is 36/135, about 26.7%. Even with both alerts, roughly three flagged orders in four are clean, because manipulation is rare to begin with.

Why is odds form the fast way to combine alerts?

Think of two smoke detectors in a kitchen where real fires are rare. Each beep makes a fire more likely, but toast sets both off far more often than fire does. In odds form, Bayes' rule is one multiplication per piece of independent evidence: posterior odds equal prior odds times each likelihood ratioHow much more often the evidence appears when the hypothesis is true than when it is false.. A ratio of 9 means alert A fires nine times as often on manipulative orders as on clean ones, for example on 90% of manipulative orders and 10% of clean ones.

How do you check 26.7% by counting?

Take 10,000 orders: 100 manipulative and 9,900 clean. Suppose A fires on 90% of manipulative orders and 10% of clean ones, and B on 80% and 20%, which gives the stated ratios of 9 and 4. Both fire on 100 x 0.9 x 0.8 = 72 manipulative orders and on 9,900 x 0.1 x 0.2 = 198 clean ones. Of the 270 orders where both fire, 72 are manipulative: 26.7%, the same as the odds route.

State the assumption that made multiplication legal: the alerts are independent given the truth. If both alerts key off the same feature, say order size, the second adds little new information and multiplying by 4 overstates the case. With one alert alone the chance is 8.3% for A and 3.9% for B, which is why a desk reviews orders on combined evidence rather than a single flag.

1/6, the same as with no information. An odd first roll says nothing about the order in which 2, 4 and 6 first appear, so it is independent of 2 finishing last. A first roll of 4 or 6 raises the chance 2 is last from 1/3 to 1/2, so conditioning on that ending shifts weight onto 4 and 6, which rise to 1/4 each. The odd faces keep 1/6 each; 1/5 wrongly spreads the weight evenly.

Why does the ending tell you anything about the start?

Suppose you hear that a friend reached a party last. That makes it a little more likely they left home late, because leaving late and arriving last go together. It says nothing about whether they wore a blue shirt, which has no bearing on arrival order. Conditioning on an outcome reweights every starting state by how likely that state makes the outcome, and a state that does not affect the outcome keeps its original probability. Here the outcome is 2 finishing last among the evens; the question is which first rolls make that more or less likely.

How do the numbers work out with Bayes?

Odd rolls never change which new even appears next, so only the order of first appearances matters, and without information it is a random ordering of three: 2 is last with chance 1/3. If the first roll is 4, then 2 and 6 are left to race, and each is equally likely to show first, so 2 ends last with chance 1/2. Now weigh: each face has prior 1/6. The joint chance of first roll 1 and 2 last is 1/6 x 1/3 = 1/18; of first roll 4 and 2 last, 1/6 x 1/2 = 1/12. The total is 1/3, so first roll 1 has posterior (1/18)/(1/3) = 1/6 and first roll 4 has (1/12)/(1/3) = 1/4.

Where does the 1/5 intuition go wrong?

It treats the information as simply ruling out one face and renormalising the rest. Ruling out an outcome and conditioning on an event are the same thing only when every remaining outcome makes the event equally likely, and here they do not. A check: the posteriors 1/6, 1/6, 1/6, 1/4, 1/4 and 0 add to 1, while five faces at 1/5 would give 4 and 6 the same weight as 1. On a desk this is the error of reading a trade's outcome as if it said nothing about which signal triggered it.

32/41, about 78%. Before flipping, the odds are 1 to 9 against the double-headed coin. Five heads happen for certain with it and with chance 1/32 with a fair coin, a likelihood ratio of 32. Multiply: posterior odds are 32 to 9, which is 32/41. Each extra head doubles the odds, so a sixth head would take it to 64/73, about 88%.

Why is the answer not close to certain?

Imagine a rare illness and a decent test. A positive result raises the chance you have it, but if the illness is rare enough, most positives still come from healthy people. Evidence is weighed against how common each explanation was to begin with, so five heads, which a fair coin produces only once in 32 tries, still has to overcome nine fair coins for every double-headed one. The 97% instinct takes 1 minus 1/32 and forgets the nine-to-one start.

How do you run Bayes in odds form?

Odds form is the fastest way to say it in the room. Posterior odds equal prior odds times the likelihood ratio: (1 to 9) times 32 gives 32 to 9. Converting back, 32 out of 32 + 9 is 32/41, about 78%. The long form gives the same thing: the joint chance of picking the special coin and seeing five heads is 1/10, the joint chance of a fair coin and five heads is 9/10 x 1/32 = 9/320, and the posterior is (32/320) / (41/320).

What does the odds picture tell you about more flips?

Each head is twice as likely under the double-headed coin, so every head doubles the odds and every tail ends the question, since the special coin never shows tails. After 0 to 5 heads the chance runs 10%, 18%, 31%, 47%, 64% and 78%; it passes 50% only after the fourth head. The useful follow-up is the next flip: it lands heads with chance 32/41 + (9/41)(1/2) = 73/82, about 89%. On a desk, the same arithmetic tells you how many winning days it takes before a new strategy's record says anything about skill.

10/12, about 0.833. A uniform prior on the win rate, updated with 9 wins and 1 loss, gives a Beta(10, 2) posterior. The chance of winning tomorrow is the posterior mean, (9 + 1)/(10 + 2). The answer is below 0.9 because ten days cannot rule out a lower true win rate, and averaging over that uncertainty pulls the estimate towards one half.

Why is 0.9 too confident?

A new restaurant with nine five-star reviews out of ten looks excellent, but you would not bet that the next diner rates it five stars with 90% certainty. Ten reviews is a small sample, and a restaurant that truly earns five stars 70% of the time could easily post nine out of ten. The right forecast averages over every win rate the evidence still allows, and with only ten days that includes plenty of rates below 0.9. The average of that spread of possibilities is what you should quote for tomorrow.

How does the Bayesian update give 10/12?

Start with every win rate p between 0 and 1 equally likely. The chance of the observed record is proportional to p^9 (1 - p), so the posterior is proportional to that, which is the Beta(10, 2) distribution. The chance of a win tomorrow is the average of p over the posterior, and the mean of a Beta(a, b) is a/(a + b), here 10/12. The shortcut is Laplace's rule of succession: add one imaginary win and one imaginary loss to the record, then divide.

When does the pull towards one half stop mattering?

When the record is long. At 90 wins out of 100 the rule gives 91/102, about 0.892, almost exactly the raw 0.9, because a hundred days of data swamp the one imaginary win and loss. The limitation is the prior itself. A uniform prior says a 99% win rate was as plausible as a 50% one before you saw any data, which no trader believes about a new strategy. With a sceptical prior centred near one half, ten days would pull the estimate even further below 0.9. The posterior also tells you more than one number: the chance of winning both of the next two days is (10 x 11)/(12 x 13), about 0.705, not 0.833 squared.