Quant puzzles, solved step by step
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010A company's value to its current owner is equally likely to be anything from Rs 0 to Rs 100 crore, and only the owner knows the figure. In your hands the company would be worth 1.5 times that value. You may make one take-it-or-leave-it offer, which the owner accepts only if it is at least the company's value to them. What should you bid?Quant tradingQuant research
Try it first
Which bid maximises your expected profit?
Show the worked solution
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.
The naive line values the company at its overall average and shows profit for any bid under 75, but conditioning on the owner accepting gives expected profit of minus b squared over 400, which is below zero for every positive bid, minus 6.25 crore at a bid of 50. 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.
The relationshipb your bid in Rs crore b/100 the chance the owner's value is below b b/2 the owner's average value, given that they accepted What it says in wordsThe chance of winning times the loss when you win is negative for every positive bid.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.
Where candidates lose it
Most candidates bid somewhere between 50 and 75, reasoning from the unconditional average value. That ignores the information in the owner's acceptance, which is the entire point of the question.
The second loss is a partial fix: realising acceptance is informative but then bidding a little lower, such as 60, to leave a margin. Any positive bid loses here. Write the expected profit as a function of b and let the algebra say zero.
What the interviewer asks next
- What multiplier would make you willing to bid, and how much would you then bid?
- What if the owner's value is uniform on 50 to 100 instead?
- How does this relate to a market maker who gets filled on their quotes?
022A three-way duel: you hit your target with probability 1/3, B with 2/3, and C never misses. You shoot first, then B, then C, repeating in that order until one person is left, and everyone aims to maximise their own survival. Where should you aim your first shot?Quant tradingQuant research
Try it first
Which first shot gives you the best chance of surviving?
Show the worked solution
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.
Firing into the air gives you 39.7% survival, against 31.2% for aiming at C and 26.5% for aiming at B, because hitting either rival makes you the survivor's only target while missing on purpose lets B and C shoot at each other first. 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.
The relationship3/7 your survival in a duel with B when you shoot first 1/7 your survival in a duel with B when B shoots first 25/63 your survival after a deliberate miss What it says in wordsMissing on purpose beats both targeted shots: 75/189 against 59/189 and 50/189.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.
Where candidates lose it
The instinctive answer is to shoot at C, the most dangerous player. It ignores what happens after a hit: you have just made yourself B's only target, and B shoots first.
The second loss is assuming that firing into the air is allowed but not checking it is optimal. Candidates who have heard the answer before often cannot produce 25/63, 59/189 and 50/189 when asked. The numbers are the answer; the slogan is not.
What the interviewer asks next
- What if your hit rate were 1/2 instead of 1/3?
- What if C shot first and you shot last?
- What is B's overall survival probability when you fire into the air?
059We play a coin game. I pick a sequence of three heads or tails, you then pick a different sequence after seeing mine, and we flip a fair coin until one of the two sequences appears; whoever's comes first wins. I pick HHH. What do you pick, and how often do you win?Quant tradingProp trading firms
Try it first
Which reply to HHH is best?
Show the worked solution
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.
HHH wins only when the first three flips are heads, a 1/8 chance; in every other run the first HHH is preceded by a tail, so THH is completed one flip earlier and wins the remaining 7/8 of games. 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.
Where candidates lose it
The trap is answering that every sequence has probability 1/8, so the game is fair, or picking TTT because it has nothing in common with HHH. Both treat the race as independent draws of three flips rather than a stream where patterns overlap.
The second trap is reaching for a four-state Markov chain and running out of time. The tail-before-the-run argument settles it in one sentence; set up the chain only if asked about a harder pair.
What the interviewer asks next
- I pick HTH. What do you pick, and how often do you win?
- What is the expected number of flips to see HHH, and to see THH?
- Why can no three-flip sequence be the best first choice?