Equity Research puzzles, solved step by step

The answer is 1/6. After a first roll of 1, 3 or 5 the game ends on 2 with chance 1/3; after a 2 it cannot; after a 4 or 6 it ends on 2 with chance 1/2. Overall the game ends on 2 with chance 1/3, so a first roll of 1 carries (1/6 x 1/3) / (1/3) = 1/6. The 4 and 6 absorb the share the 2 lost.

Why does 1/5 feel right, and where does it go wrong?

Think of a cricket team told that the match was won off the last ball. That news does not just rule out one scenario; it makes the close games far more likely than the easy ones. Conditioning works the same way here. Being told the game ended on 2 is evidence, and evidence reweights every starting point by how well it explains what you saw. The 1/5 answer treats the news as if it only deleted the 2 and left the other five faces equally likely.

So ask, for each first roll, how likely it is that 2 ends up the last even number. After a 1, 3 or 5, all three evens are still missing and each is equally likely to come last, so 1/3. After a 2, the 2 is already seen and cannot come last, so 0. After a 4, only 2 and 6 are missing, and 2 comes last exactly when 6 shows first, so 1/2. The same holds after a 6.

How do you turn those chances into the answer?

Apply Bayes ruleThe chance of a cause given what you saw equals its prior chance times how likely it made the observation, divided by the total chance of the observation.. Each face starts at 1/6. Multiply by its chance of ending on 2 and add them up: three faces at 1/18, one at 0, two at 1/12, which totals 1/3. The 1 contributes 1/18 of that 1/3, which is 1/6, exactly its starting weight. The 4 and the 6 rise from 1/6 to 1/4 each, and the 2 falls to zero.

The research version of this is reading a data point. A company that beats estimates is more likely to be one that guided low, not just one that is doing well. Before you update, ask which starting stories make the thing you observed more likely, and shift weight towards them.

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.

135 over 211, about 64%. Let P(i) be the chance of reaching Rs 5 from Rs i. Each P(i) is 0.6 times P(i+1) plus 0.4 times P(i-1), with P(0) = 0 and P(5) = 1. The solution is P(i) = (1 - (2/3)^i) / (1 - (2/3)^5), and at i = 2 that is (5/9) divided by (211/243), or 135/211. A fair coin would give 40%.

How do you set the problem up without drawing every path?

Think of a cricket team chasing a small target with few wickets in hand. What matters is not the path of each ball but where the game stands now: runs needed and wickets left. The only thing that matters at any moment is how much money you hold, so write one unknown for each amount and one equation linking it to its neighbours. From Rs i you move to Rs i+1 with chance 0.6 and to Rs i-1 with chance 0.4, so P(i) = 0.6 P(i+1) + 0.4 P(i-1). The two ends are fixed: P(0) = 0 because you are broke, and P(5) = 1 because you have won.

How do you check 64% without trusting the formula?

Solve the chain by hand, starting from the bottom. Write P(1) = a. Then P(2) = (a - 0.4 x 0) / 0.6 = a/0.6 = 1.667a, and each later step follows the same rule until P(5) = 211/81 x a. Setting P(5) = 1 gives a = 81/211, or 38.4%, and P(2) = 135/211. Two routes landing on the same fraction is the check an interviewer wants to hear. It also tells you the shape: the edge matters most in the middle, where the fair line and the edge curve are furthest apart.

Say the desk version in one line. A small, repeatable edge does not make ruin impossible when you start thin: from Rs 2 you still go broke about 36 times in 100. More capital, not a bigger edge, is what pushes the ruin chance towards zero.

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.

With three cars the last car reaches about 1,833 miles; with n cars it reaches 1,000 x (1 + 1/2 + ... + 1/n). The cars drive together until they have burned one tankful between them, then one car refills the rest and stops. That happens after 1,000/n miles, then 1,000/(n - 1), down to the last car's 1,000. The sum has no ceiling, but it grows only like the logarithm of n.

Why does a car drop out after exactly 1,000/n miles?

Picture a group of hikers sharing water on a long desert walk. Once the group has drunk one person's worth between them, that person can pour what is left into the others' bottles, fill every one back up, and wait by the path. Carrying that hiker any further only costs water. The cars work the same way. After d miles each car has burned d miles of fuel and has room for d more. One car can refill the other n - 1 cars exactly when (n - 1) x d equals what it has left, 1,000 - d, which solves to d = 1,000/n.

Why do three cars buy 1,833 miles and not 3,000?

Follow the road left to right. Three cars run 333 miles together and burn 1,000 miles of fuel between them, one full tank. Car 3 has 667 miles of fuel left and hands 333 to each of the other two, which fills them. Two cars then run 500 miles, burning another tankful, and car 2 refills car 1. Car 1 runs its own 1,000. Every stretch burns exactly one tank, but each extra car buys a shorter stretch, because its tank has to move more cars.

What happens as n goes to infinity?

The harmonic series never converges, so enough cars can reach any distance at all. But the growth is logarithmic: ten cars reach about 2,929 miles and a hundred cars about 5,187, so every tenfold increase in cars adds only about 2,300 miles. When asked for the asymptotic answer, say both halves: unbounded, and slow. State the assumptions too: fuel moves between cars without loss, and a car that drops out is simply left behind. If the helper cars had to drive home, the answer would shrink sharply.

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.

200 miles. Ignore the zigzag and ask how long the bird is in the air. The car takes 100 / 50 = 2 hours to reach B, and the bird flies the whole time at 100 miles an hour, so it covers 2 x 100 = 200 miles. Summing the legs gives the same answer: 133.3 miles in the first round trip, then a third as much each round, and 133.3 / (1 - 1/3) = 200.

What is the shortcut, and why does it work?

Picture walking home while your dog runs back and forth between you and the front door. Ask how far the dog ran and the back and forth looks like the hard part, but the dog simply ran at its own speed for exactly as long as your walk took. When something moves at a constant speed for a known time, the distance is speed times time, however many times it turns around. The car's trip fixes the time at 2 hours, and the bird's speed does the rest.

How do you check it the long way?

The first meeting comes when the gap of 100 miles closes at 150 miles an hour, after 40 minutes, 33.3 miles from A. The bird has flown 66.7 miles and flies 66.7 back to B, a first round trip of 133.3 miles. By then the car is two thirds of the way, and every later round trip covers a third of the remaining gap, so each is a third of the last. A geometric series with first term 133.3 and ratio one third sums to 133.3 / (2/3) = 200, the same answer the shortcut gives in one line.

Why would an investing firm ask a puzzle like this?

Because choosing the quantity that stays simple is most of the skill. Here time is simple and distance is messy, so solving in time turns an infinite sum into one multiplication. The same habit finds total cash generated over a period from an average rate instead of adding every quarter, or total interest paid on a loan from the average balance. Say the frame you chose before you give the number.

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.

Stickers: 6n squared. Cubes on the surface: n cubed minus (n minus 2) cubed. Each of the six faces is an n by n grid of stickers, so there are 6n squared, which is 54 for n of 3. For the cubes, count the hidden core: peeling one layer off every side leaves (n minus 2) cubed, so 27 less 1 is 26 on show. Both formulas need n to be at least 2.

Why are stickers and surface cubes different counts?

A corner cube carries three stickers, an edge cube two and a face-centre cube one. The sticker count measures surface area, while the cube count measures how many pieces touch the surface, and the corners and edges make the first larger than the second. The surface area the question asks for is the sticker count: six faces, each n by n, so 6n squared unit squares.

What is the fast way to count the cubes on show?

Picture a box of laddoos packed three by three by three: to find how many touch the box, count the ones that do not. Only a core with one layer peeled from every side is hidden, so the hidden cubes form an (n minus 2) by (n minus 2) by (n minus 2) block, and everything else shows. For n of 3 that core is a single cube, so 26 show. The subtraction is quicker and harder to get wrong than adding corners, edges and faces.

How do you check the formula before you commit?

Test the smallest case you can picture. A 2 by 2 by 2 cube has no hidden centre, so all 8 pieces show, and the formula gives 8 minus 0, which is 8. Then look at the expanded form: 6n squared counts every face-cube once per face, minus 12n plus 8 removes the double and triple counts along the 12 edges and at the 8 corners. The formula breaks for n of 1, where it gives 2 for a single cube, so state that it holds from n of 2 upwards.