Quant puzzles, solved step by step

Bid nothing. If a bid of b is accepted, the owner has told you the company is worth less than b to them, so its value is uniform on 0 to b and averages b/2. In your hands that is 1.5 x b/2 = 0.75b, a quarter less than you paid. Expected profit is (b/100) x (0.75b - b) = -b squared/400, negative for every positive bid. This is the winner's curse in its purest form.

Why does 75 look right and fail?

Picture buying a used car from someone who knows its history while you do not. If they agree to your price at once, that is itself news: sellers of good cars refuse low offers. Acceptance is not random; it happens exactly in the states where the company is worth less than you offered, so the average value you actually receive is the average below your bid, not the average overall. The naive 75 uses the unconditional average of 50 and forgets that you only trade when the owner is happy to sell.

How do you set up the expected profit?

Split it into the chance of a deal and the profit given a deal. A bid of b is accepted with probability b/100; given acceptance the owner's value is uniform on 0 to b, averaging b/2, so your value averages 0.75b and your profit averages minus 0.25b. Multiply: minus 0.25b x b/100, which is minus b squared over 400. At a bid of 50 that is minus 6.25 crore: you win half the time and lose 12.5 crore on average when you do.

When would bidding make sense, and where does this show up on a desk?

The multiplier is the lever. With a multiplier m, the profit given a deal is (m/2 - 1)b, so bidding pays only if you add more than double the owner's value; at exactly 2 you break even, and above 2 you should bid the full 100. On a trading desk the same logic is called adverse selectionThe tendency for the trades you actually get to come from counterparties with better information than you, so they are worse on average than a random trade.: the orders that fill against you are disproportionately the ones from people who know more. A quote that looks profitable against the average counterparty loses against the ones who choose to trade.

Fire into the air. B and C each target the other, the bigger threat, so while both live nobody shoots at you. Aiming in the air gives survival of 2/3 x 3/7 + 1/3 x 1/3 = 25/63, about 39.7%. Aiming at C gives 31.2%, because a hit leaves you in a duel with B shooting first. Aiming at B gives 26.5%, because a hit leaves C, who never misses, to shoot you.

Who does everyone else aim at?

Start with the stronger players, because their choices fix yours. B aims at C, because if B shot you instead, C would kill B next turn for certain; C aims at B, the more dangerous of the two remaining threats. So while all three are alive, nobody is shooting at you. Think of two large firms in a price war while a small competitor stays out of it: the small firm's best move is often to let the giants weaken each other.

How do you work out the two-player duels?

Against B with you shooting first, you win if you hit now, or if both miss and the same duel restarts. Call your survival x: x = 1/3 + (2/3)(1/3)x, so x = 3/7; if B shoots first, you must survive B's first shot, 1/3 of the time, giving 1/7. Against C you get exactly one shot, since C never misses: 1/3 if you shoot first, 0 if C does. Now combine. In the air: B hits C two times in three, giving you the 3/7 duel; otherwise C kills B and you get your one shot at C, 1/3. Total 25/63.

What is the general lesson?

In a game with several players, weakening one rival can hurt you if it frees the strongest remaining player to turn on you. Your best shot is the one that keeps the others focused on each other. Say the limitation too: the answer depends on the hit rates and the order. Change the order of shooting, or let C aim at you, and the tree changes; the interviewer will often change a number or the order to see whether you rebuild the tree or repeat the slogan.

Go second. The player facing 1, 5, 9, 13, 17 or 21 matches loses against good play, and 21 is on that list. Whatever your opponent takes, take 4 minus that, so every round removes exactly 4. They then face 17, 13, 9, 5 and finally 1, and are forced to take the last match.

How do you find the losing positions?

Start from the end of the game, the way you would plan the last few stops of a journey before the first. With 1 match in front of you, you must take it and lose. With 2, 3 or 4, you take enough to leave exactly 1 and win. With 5, every move leaves 2, 3 or 4, each a winning spot for the other player, so 5 loses. A position is losing when every move from it hands your opponent a winning position, and here that happens every 4 matches: 1, 5, 9, 13, 17, 21.

Why does answering with 4 minus k always work?

Your opponent can take 1, 2 or 3; you can always take 3, 2 or 1 in reply. The pair of moves removes exactly 4 matches whatever they chose, so you control the count at the end of every round. From 21 the rounds end at 17, 13, 9 and 5, and then your opponent faces a single match. The number 4 is the maximum take plus one; that is where the period comes from.

A short check you can say aloud: the game has only 21 positions, and marking each as winning or losing from the bottom up, a position wins if any move reaches a losing one, reproduces the list 1, 5, 9, 13, 17, 21. If the pile had been 20, you would go first and take 3 to leave 17. This type of game has a backward inductionSolving a game by working out the best move at the last step first, then the step before, back to the start. solution, and the interviewer mainly wants to hear you build it from the end.

Bluff with the Q one time in three, and call with the K one time in three. You always bet the A and always check the K; I always call with the A and fold the Q. A Q bluff risks 1 more to win the 2 antes, so I must defend two thirds of hands facing it: the A gives half, the K calling one time in three gives the rest. The game is worth 1/18 per hand to you.

Which hands are easy, and where is the real decision?

Clear away the obvious hands first. With the A you always bet, because you win whether I call or fold. With the K, betting only gets called by the A and folds out the Q, which you beat anyway, so you check. As the caller, I always call with the A and always fold the Q. The whole game comes down to two mixed choices: how often you bluff with the Q, and how often I call with the K.

How do the indifference conditions fix both frequencies?

A teacher who spot-checks homework faces the same logic: check every paper and time is wasted, never check and everyone copies, check at the right rate and copying stops paying. In equilibrium each player mixes at the rate that makes the other indifferent between their two options. Your Q loses 1 by checking. Bluffing loses 2 against my A, and against my K loses 2 if I call and wins 1 if I fold. The two are equal only when I call with the K one time in three. My K loses 1 by folding; calling loses 2 against your A and wins 2 against a bluff, which is worth -1 only when you bluff one time in three.

Check against the pot-odds rule. A bluff risks 1 extra to win the 2 antes, so the caller must defend 2/(2 + 1) = 2/3 of the hands facing it; the A already covers half, and the K calling one time in three covers the other sixth. That two thirds is the number people misremember as the K's calling rate. Put the strategies together and the bettor, who acts first with more information about their own hand, earns 1/18 of a chip per hand. The limitation: with a bigger bet or more cards the ratios change, but the method, indifference on both sides, carries over.

Pick THH; you win 7 times in 8. HHH can only win if the first three flips are all heads, which has chance 1/8. In any other run, the first HHH is preceded by a tail, and that tail with the next two heads spells THH, which is completed one flip before HHH. So THH wins every game except the one that opens with three heads.

Why is this not a fair race between two 1/8 sequences?

Think of two runners on the same track where one always starts one step ahead of the other on the only route to the finish. Their speeds are identical but the race is not even. In a race between patterns, what matters is not how often each appears but which one tends to appear first, and that depends on how the patterns overlap. THH is built from HHH's own first two heads with a tail placed in front, so it ambushes HHH whenever HHH has not already won at the start.

How do you prove 7/8 without a Markov chain?

Look at the first time HHH appears. If it does not start at flip 1, the flip immediately before it must be a tail, because otherwise an earlier HHH would already have appeared. That tail plus the first two heads of the HHH is THH, finished one flip before HHH. So HHH wins only if flips 1 to 3 are heads, chance 1/8, and THH wins otherwise: 7/8. The proof is one sentence, and interviewers want to hear it rather than a transition matrix.

What is the general lesson for the second mover?

This is Penney's gameA coin sequence race in which the second player, choosing after seeing the first, can always pick a sequence that wins more than half the time., and the second player always has an edge because the winning relation among three-flip sequences is not transitive: every sequence has another that beats it. The recipe: take the opponent's first two flips, and put in front of them the opposite of the opponent's second flip. Against HHH that gives THH at 7/8; against HTH it gives HHT at 2/3. In trading terms, a strategy that looks as good as any other in isolation can still lose systematically to one designed around it.

I show heads 3/8 of the time; so should you; and the game is worth -1/8 to you per round. At my mix q, your heads pays 3q - 2(1 - q) and your tails pays -2q + (1 - q). Setting them equal gives q = 3/8, where both pay -1/8. By the same algebra your 3/8 mix makes my choices equal, so neither of us can improve: the table looks fair and is not.

Why do I randomise to make you indifferent, not to help myself?

Think of a penalty taker and a goalkeeper. If the taker shoots left more often than he should, the keeper dives left and gains; any pattern is exploited. In a zero-sum game, the mix that protects you is the one that leaves your opponent with nothing to exploit, which means making every one of their choices pay the same. So I do not pick my heads probability by looking at my own payoffs in isolation; I pick it so that your heads and your tails earn you equal amounts. Then it no longer matters what you do.

How do the two equations give 3/8 and -1/8?

Let q be my chance of heads. Your heads earns 3q - 2(1 - q) = 5q - 2 and your tails earns -2q + 1(1 - q) = 1 - 3q, and they are equal when 8q = 3, at q = 3/8. Plug back in: 5(3/8) - 2 = -1/8. For your side, let p be your chance of heads; my payoffs are the negatives of yours, and because the table is symmetric the same algebra gives p = 3/8. At those mixes, each cell's frequency is HH 9/64, TT 25/64 and the two mismatches 15/64 each: 3(9) + 1(25) - 2(30) = -8, over 64, is -1/8.

Why does a table with equal wins and losses favour me?

Because the cells are not played equally often. Your big win needs both of us on heads, and I can make that cell rare; the mismatches, where I win, happen 30 times in 64 at equilibrium. If I played heads half the time you could earn +1/2 by always showing heads, so a naive 50/50 from me would be a gift. In trading the same idea sets a market maker's quotes: they are placed so that the informed side has no choice that beats the others, not so that the market maker profits on any one trade.

Do not spin: pulling straight away survives with probability 3/4, against 2/3 with a spin. The click tells you the hammer sat on one of the four empty chambers. Because the two bullets are adjacent, the four empties form a run, and only the last empty in that run is followed by a bullet. A spin throws that information away and resets you to 4 empties out of 6.

What does the click actually tell you?

Think of a row of six houses where two neighbours keep dogs. You knocked at a random house and no dog barked. If you now try the next house along, you are only in trouble if you had knocked on the one house sitting just before the dogs. The click narrows the hammer's position to the four empty chambers, and the question becomes how many of those four have a bullet immediately after them. With the bullets side by side, the empties run 3, 4, 5, 6 in firing order, and only chamber 6 hands over to a bullet.

Does the answer depend on the bullets being adjacent?

Completely, and that is the follow-up most interviewers ask. If the two bullets are not next to each other, the empties split into two runs, two empties now sit in front of a bullet, and pulling straight away survives only 1/2, worse than the 2/3 of a spin. The same count works for bullets one apart or directly opposite: in both cases two of the four empties are followed by a bullet. So the rule is not spin or do not spin; it is count the empties that border a bullet and compare with a fresh spin.

What is the general lesson for a trading interview?

A random reset destroys information, and information has value only if the structure of the problem lets you use it. Here the structure is clustering: the bullets sit together, so a safe chamber is likely followed by another safe one. Markets have the same feature in volatility: a calm day tends to be followed by a calm day, so conditioning on what just happened beats assuming each day is a fresh draw. Say that link in one line after the arithmetic.

No die is best: red beats green, green beats blue and blue beats red, each with probability 5/9. So who picks first matters a great deal. Let me choose, then take the die that beats mine and win 5/9 of the time. If you are forced to pick first, every choice loses 5/9 of the time against an opponent who knows the cycle.

How do you work out who beats whom?

Each matchup has only nine equally likely pairs of faces, so write the 3 by 3 grid and count. Red against green: red's 2 beats only the 1, while its 6 and 7 each beat the 1 and the 5, for 1 + 2 + 2 = 5 wins out of 9. Do the same for the other two pairs and every matchup comes out 5 to 4: red over green, green over blue, blue over red. It is rock, paper, scissors built out of dice, and in rock, paper, scissors nobody asks which hand shape is best.

Why does green lose to red when green has the higher average?

The averages are red 5, green 6 and blue 5. Winning is about how often, not by how much. Green's 12 wins every time it shows, but it shows only a third of the time, and green's other two faces, 1 and 5, lose to both of red's high faces. A higher mean and a higher chance of winning are different things, and the gap between them is the whole puzzle. Change the rules so the winner collects the difference between the two numbers, and green's expected margin against red is 6 - 5 = +1: now you want green against red, and blue against red is a dead heat at 0.

Where does a trader meet the same thing?

Head-to-head comparisons need not line up into a ranking. Strategy A can beat strategy B on more days than not, B can beat C, and C can beat A, whenever one of them earns its money in rare large wins, as green does. Before choosing between strategies, decide whether you care about how often you win or how much you make, because under the first a cycle like this one means there may be no best choice at all. The limitation of the puzzle is that it is one roll; over many rolls with the total score counted, the mean matters more and green's 12 starts to pay.