Equity Research puzzles, solved step by step

Flip twice: call heads on HT, tails on TH, and flip two more times on HH or TT. HT and TH each have probability 1/3 x 2/3 = 2/9, so they are equally likely whatever the bias. A pair is kept with chance 4/9, so on average you need 9/4 pairs, which is 4.5 flips per fair result.

Why are HT and TH always equally likely?

Think of two friends who each toss the same lopsided coin once. The chance the first gets heads and the second tails is the same as the chance the first gets tails and the second heads, because the same two numbers are multiplied, just in a different order. For any bias p, HT and TH both have probability p times (1 minus p), so the order of two different outcomes is a perfectly fair coin. This is known as the von Neumann trick.

How do you get the 4.5 flips, and can you do better?

Each pair succeeds with chance 4/9. The number of pairs until the first success averages one over the success chance, 9/4, and each pair is two flips. So the fair coin costs 4.5 biased flips on average, a steep price for throwing away five pairs in every nine.

The interviewer's follow-up is efficiency. The discarded pairs still carry information: HH and TT are themselves a biased coin, and you can run the same trick on them, and on the pattern of which pairs matched. Iterating like this recovers most of the waste. The ceiling is set by how unpredictable the coin is: about 0.918 fair bits per biased flip here, or roughly 1.09 flips per fair result at best. Naming that limit is what separates a good answer from a complete one.

The probability is 0.875. The sum of two uniform numbers has a triangular density, rising from 0 to a peak at 1 and falling back to 0 at 2. The region above 1.5 is a small triangle with base 0.5 and height 0.5, area 0.125. So the chance the sum is below 1.5 is 1 minus 0.125, which is 0.875.

Why is the sum not spread evenly from 0 to 2?

Think of rolling two dice: a total of 7 turns up far more often than 2 or 12, because many pairs make 7 and only one pair makes 2. Continuous numbers behave the same way. Adding two flat distributions gives a triangle, because middle values can be reached in many ways and extreme values in very few. Combining two densities like this is called convolutionThe operation that gives the distribution of a sum of independent random quantities by sliding one density across the other and adding up the overlap..

How do you get 0.875 quickly?

Use geometry rather than integrals. Every pair of draws is a point in the unit square, all points equally likely, so probability is area. The line x + y = 1.5 cuts off a corner triangle whose legs are each 0.5 long, so its area is half of 0.5 times 0.5, which is 0.125. Everything else, 0.875, is where the sum is below 1.5. The triangle density on the right shows the same tail from the other side.

The general rule for a threshold t between 1 and 2 is 1 minus half of (2 minus t) squared; for t below 1 it is half of t squared. The research version is diversification of two independent return sources: the average of two is less likely to be extreme than either alone, which is the whole argument for combining them.

6 tosses. Call S the expected tosses from the start and H the expected tosses once you have one head. From the start, one toss either gives a head (move to H) or a tail (stay at S): S = 1 + H/2 + S/2. From one head, one toss either finishes or sends you back: H = 1 + S/2. Substituting, S = 6 and H = 4.

Why is the answer not four?

Think of climbing two steps on a slippery stair where any slip sends you back to the bottom. Reaching the second step is not just two climbs; every slip costs you what you had gained. The chance of heads-heads on a single pair is one in four, but a tail after a head wipes out your progress, so the expected wait is longer than four. The clean way to count the cost of those slips is to name each position you can be in.

How do you check 6 another way?

Count sequences directly. The first HH lands on toss n in exactly F(n-1) of the 2ⁿ equally likely sequences, where F is the Fibonacci sequence: 1 way on toss 2, 1 on toss 3, 2 on toss 4, 3 on toss 5. Summing n times F(n-1) over 2ⁿ across all n gives 6.0000, the same 6. The same series shows the spread: you finish within six tosses 67% of the time, and the other third of runs can go on for a long while, which is what pulls the average up to 6.

0.648. At least two means exactly two or all three. Exactly two can happen three ways, each with chance 0.6 x 0.6 x 0.4 = 0.144, so 0.432 together. All three beat with 0.6 cubed, 0.216. Adding them gives 0.648. The complement checks it: none beat 0.064 plus exactly one 0.288 is 0.352, and 1 minus 0.352 is 0.648.

How do you avoid missing an outcome?

Think of three friends each deciding whether to come to dinner. At least two coming covers four different guest lists: each of the three possible pairs, and all three. List the outcomes by how many succeed, count the ways each count can happen, and multiply by the chance of one such way. With three stocks there are only eight results, so write them down; with more, the count of ways is a binomial coefficientThe number of ways to choose k items out of n, written n choose k; here 3 choose 2 is 3..

Is independence a fair assumption for three stocks you cover?

Usually not, and saying so is the part that sounds like an analyst. Stocks in one sector share demand, input costs and the same consensus-setting habits, so beats tend to cluster: when one beats, the others are more likely to. Clustering fattens both ends. The chance of all three beating and of none beating both rise above 0.216 and 0.064, and whether at least two beat can move either way, so the independent answer is a starting point, not a forecast.

Pull again straight away: you survive with probability 3/4, against 2/3 after a fresh spin. The four empty chambers sit together in a row and the cylinder advances one chamber per pull. The click you heard came from one of those four, each equally likely. Only one of them, the last empty chamber before the bullets, is followed by a bullet, so the chance of dying without a spin is 1/4. A fresh spin puts you back at two bullets in six.

Why does the first click carry information?

Picture a street with four dark houses in a row, then two lit ones, laid out in a circle. You are told you are standing at a dark house and that you will walk to the next door. Three of the four dark houses have another dark house next door. The click tells you that you are somewhere in the run of four empty chambers, and because the bullets sit together, most of that run is followed by another empty chamber. A fresh spin throws that information away.

Would the answer change if the bullets were not next to each other?

Yes, and saying so shows you understand the mechanism rather than the trick. Put the bullets in chambers 1 and 4. The empty chambers are 2, 3, 5 and 6, and two of them, 3 and 6, sit in front of a bullet. With the bullets apart, pulling straight away survives only 50% of the time, so a fresh spin at 66.7% becomes the better choice. The adjacency is what makes the empty chambers cluster, and the clustering is what makes the click informative in your favour.

Why would an investing firm ask this?

Because it is a small version of the question every analyst faces after new information: does what just happened change the odds of what comes next? An outcome you have already seen is data about the state of the world, and ignoring it in favour of the unconditional average is a common and costly habit. Say that link in one line after the numbers.

Switching wins two times in three. Your first pick is right with probability one third, and nothing the host does changes that. In the other two thirds of cases the prize is behind one of the two doors you did not pick, and the host, who must open an empty door, is forced to leave the prize door shut. One half is wrong because it treats the host's reveal as random when it is not.

Why does one half feel right?

Two closed doors, one prize: the picture in front of you looks like a coin toss, and intuition prices the picture rather than the process that produced it. One half would be right only if the two remaining doors were equally likely before the reveal and the reveal told you nothing about either of them. Neither is true. Your door was chosen blind; the other door survived a filter run by someone who knew the answer.

What exactly does the host's knowledge change?

Think of a shopkeeper who knows which of three mangoes is ripe. You pick one; she then removes an unripe one from the two you did not pick, always careful never to remove the ripe one. She has concentrated the two thirds chance that the ripe mango was among the other two onto the single one she left. Your own pick is untouched at one third, because she would have been able to remove an unripe mango whatever you picked. The reveal carries information about the other door, not about yours.

When would one half actually be the right answer?

This is the part the interviewer is asking for when they ask why intuition fails. If the host did not know and opened one of the other doors at random, and it just happened to be empty, then switching would win only one half of the time. A random reveal that could have shown the prize rules out the branches where it would have, and what is left is symmetric. So the answer depends on the host's rule, not on the doors. Scale it up to make the point land: with 100 doors, you pick one, and a knowing host opens 98 empty ones; nobody thinks the last closed door is a coin toss.

The analyst's version is reading a signal from someone who knows more than you do. A management team that chooses which numbers to disclose is running a filter, and what survives the filter is not a random sample.