Quant puzzles, solved step by step

Puzzles: 100
Traced to a firm: 71
Topics: 12
Hard: 30

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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?Games and strategic reasoningHardQuant 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 relationship
$E[\pi(b)] = \underbrace{\frac{b}{100}}_{\text{accepted}}\;\Big(1.5\cdot\frac{b}{2} - b\Big) = -\frac{b^2}{400} < 0 \text{ for all } b > 0$
b 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?Games and strategic reasoningHardQuant 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 relationship
$P_{\text{air}} = \tfrac23\cdot\tfrac37 + \tfrac13\cdot\tfrac13 = \tfrac{25}{63} \qquad P_{C} = \tfrac13\cdot\tfrac17 + \tfrac23\cdot\tfrac{25}{63} = \tfrac{59}{189} \qquad P_{B} = \tfrac23\cdot\tfrac{25}{63} = \tfrac{50}{189}$
3/7 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?
045Simplified poker with three cards, A, K and Q: each player antes 1, you are dealt one card and I am dealt another. You may bet 1 or check; if you bet, I call or fold, and if you check the higher card wins the antes. How often should you bluff with the Q, and how often should I call with the K, in equilibrium?Games and strategic reasoningHardOld Mission CapitalNew York · 2022
Try it first
How often should I call a bet when I hold the K?
Show the worked solution
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.
In equilibrium the bettor always bets the A, always checks the K and bluffs the Q one time in three, while the caller always calls the A, folds the Q and calls with the K one time in three; together the A and the K calls defend two thirds of hands facing a bluff.
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.
The relationship
$\underbrace{\tfrac12(-2) + \tfrac12\big(c(-2) + (1-c)(1)\big)}_{\text{Q bluffs}} = -1 \Rightarrow c = \tfrac13 \qquad \underbrace{\frac{-2 + 2b}{1 + b}}_{\text{K calls}} = -1 \Rightarrow b = \tfrac13$
c how often the caller calls with the K
b how often the bettor bluffs with the Q
-1 the payoff of the alternative: checking the Q or folding the K, losing the ante
What it says in wordsEach frequency is set so the opponent's two choices are worth the same.
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.
Where candidates lose it
The common loss is setting the K's calling rate to two thirds. Two thirds is the total defence the caller needs against a bluff, and the A already provides half of it, so the K calls only one time in three.
The second is never bluffing because the Q cannot win a showdown. A player who never bluffs lets the caller fold every K to a bet, and the A's bets stop earning. The bluff is what gets the A paid.
What the interviewer asks next
What is the value of the game to each player?
The bet size doubles to 2. How do the bluffing and calling frequencies change?
Now the caller may also bet after a check. What changes?
Asked at Old Mission Capital, Quantitative Research, New York, 2022 (Wall Street Oasis): Asking to find the game theory optimal strategy in a simplified poker game
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?Games and strategic reasoningHardQuant 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?

b	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

3/7	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

c	how often the caller calls with the K
b	how often the bettor bluffs with the Q
-1	the payoff of the alternative: checking the Q or folding the K, losing the ante

Try it first

Why does 75 look right and fail?

How do you set up the expected profit?

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

Where candidates lose it

What the interviewer asks next

Try it first

Who does everyone else aim at?

How do you work out the two-player duels?

What is the general lesson?

Where candidates lose it

What the interviewer asks next

Try it first

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

How do the indifference conditions fix both frequencies?

Where candidates lose it

What the interviewer asks next

Try it first

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

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

What is the general lesson for the second mover?

Where candidates lose it

What the interviewer asks next