Quant puzzles, solved step by step

Rs 21. Flip n pays 2 to the n with probability 1/2 to the n, so each of flips 1 to 20 contributes exactly 1 rupee: 20 rupees. Beyond flip 20 the casino pays its whole bank, 2 to the 20, and the chance of getting that far is 1/2 to the 20, which adds 1 more. The famous infinite value collapses to 21 as soon as the payer's bank is finite.

Why is the uncapped game worth infinity?

Each extra flip halves the chance and doubles the prize, so each flip adds the same 1 rupee to the average, forever. An expected value is a sum over outcomes of prize times chance, and when every term is 1 and there are infinitely many terms, the sum has no limit. That is the St Petersburg paradox, discussed by Daniel Bernoulli in the eighteenth century: the mathematics says pay anything, and nobody would pay more than a modest sum.

What exactly does the cap remove?

Picture a lottery that promises to double your prize every day for ever, run by a shop with a small safe. The later promises are worth nothing because the shop cannot keep them. The cap turns every term after flip 20 from 1 rupee into 2 to the 20 divided by 2 to the n, which is 1/2, 1/4, 1/8 and so on, and those add up to exactly 1. So the tail that made the value infinite is now worth a single rupee. The bank is Rs 10,48,576 and the game is worth Rs 21.

What does this teach about pricing a payoff?

Doubling the bank adds only one rupee to the fair price: a bank of 2 to the 30, about Rs 107 crore, makes the game worth Rs 31. The value grows with the logarithm of what the counterparty can pay, so the realistic price of a lottery-like payoff depends on who stands behind it. The same idea appears in trading as counterparty risk: a contract's promised payout in extreme states is only worth what the other side can deliver in those states.

Keep a 5 or 6 on the first roll, a 4, 5 or 6 on the second, and take whatever the third gives; the game is worth 14/3, about 4.67. Work backwards. The last roll is worth 3.5. With two rolls left, keep anything above 3.5: (4 + 5 + 6)/6 + (1/2)(3.5) = 4.25. With three, keep anything above 4.25: (5 + 6)/6 + (2/3)(4.25) = 14/3.

Why start from the last roll?

Think of house hunting with three viewings booked: whether to accept the first flat depends on what the remaining viewings are likely to offer, and you only know that once you know how you would behave at the last one. The value of continuing at any point is defined by what you would do later, so the only roll whose value you know outright is the last one, and every earlier decision is built on it. That is backward induction, and the interviewer wants to hear the words before any numbers.

How do the values build up?

On the last roll you take the face, worth 3.5. On the second roll, stop if the face beats 3.5, which means 4, 5 or 6; otherwise you get 3.5 from the last roll. The value is the average of the faces you keep plus the chance you continue times the value of continuing: (4 + 5 + 6)/6 + (3/6)(3.5) = 4.25. On the first roll, the bar to beat is now 4.25, so only 5 and 6 are kept: (5 + 6)/6 + (4/6)(4.25) = 11/6 + 17/6 = 14/3, about 4.667.

What does the common shortcut cost, and where does this lead?

The shortcut is to keep anything above the single-roll average of 3.5 at every stage. On the first roll that keeps a 4, which gives 4.625 instead of 4.667. The threshold rises with the number of rolls left, because each spare roll is an option, and an option is worth more the longer it lives. With six rolls the value is 5.27, and with many rolls it approaches 6, since you can wait for a six. The same structure prices an American option: exercise early only when the payoff beats the value of holding on.

Stop at the first draw of 0.8 or more; the game is worth 0.8 after all fees. Holding x, one more draw improves you by (1 - x)^2 / 2 on average, which equals the 0.02 fee at x = 0.8. Check: you expect 5 draws costing 0.10 in total, and the draw you keep averages 0.90, so the net value is 0.80.

How do you decide whether one more draw is worth it?

Think of hunting for a flat. Each viewing costs you an evening. If the flat in hand is already good, another viewing rarely beats it, and when it does it only beats it by a little. The value of one more look is the chance of beating what you hold times the average margin when you do, and you stop when that falls below the price of looking. With a uniform draw and a current value x, the chance of beating x is 1 - x and the average margin is (1 - x)/2, so the expected gain is (1 - x)^2 / 2.

Why are the threshold and the value the same number?

Because the game has no memory. After a disappointing draw you are back where you started, facing the same game worth V. You should accept a draw exactly when it beats what the fresh game is worth, so the threshold equals the value. This is the same logic as a reservation price: you walk away from any offer below what the next round is worth to you. The arithmetic check is worth saying: with threshold 0.8 a draw succeeds with probability 0.2, so you expect 5 draws and 0.10 of fees, and an accepted draw is uniform on 0.8 to 1 with mean 0.9. The net is 0.9 - 0.10 = 0.8. A seeded simulation of 200,000 games gives 0.800.

What happens as the cost changes?

The threshold is 1 - sqrt(2c), so it responds to the square root of the cost. A fee of 0.005, four times cheaper, only moves the threshold from 0.8 to 0.9. At a fee of 0.5 or more, the threshold hits zero and you take the first draw, because even the first draw's average of 0.5 barely covers what you paid. The limitation of the model is that draws are independent and the distribution is known; if you were learning the distribution as you drew, the first few draws would be worth more than this rule says.

10.6 cards, which is 53/5. Each of the 48 non-aces comes before all four aces with probability 1/5, because among that card and the four aces each is equally likely to be first. So on average 48/5 = 9.6 non-aces precede the first ace, and counting the ace itself gives 10.6. For n cards with m special ones the rule is (n + 1)/(m + 1).

Why isn't the answer 13?

Drop four red counters at random into a line of 48 white ones. Nothing distinguishes the stretch of white before the first red from the stretch between the second and third reds, or the stretch after the last red. The four reds cut the whites into five stretches, and by symmetry each stretch holds the same number on average, 48/5 = 9.6. The answer 13 imagines the aces spread evenly from the top of the deck, but that leaves no room for the stretch after the last ace, which is just as long on average as the others.

How do you make the symmetry argument rigorous?

Use indicators. Pick any non-ace, say the seven of clubs. It is turned before the first ace exactly when it comes first among five cards: itself and the four aces. Those five cards sit in a random order, so that happens with probability 1/5. An expected count is the sum of the probabilities, even when the events depend on each other, so 48 x 1/5 = 9.6 non-aces come first on average and the ace makes it 10.6. The exact sum of the chances that the first ace is later than card k, over k from 0 to 48, gives 53/5 as well, and a seeded simulation of 100,000 shuffles gives 10.63.

The mean hides a skewed shape. The first ace is in the top five cards 34.1% of the time, and the median position is 9, below the mean of 10.6, because a few shuffles bury all four aces deep in the deck and drag the average up. If the interviewer asks for the most likely position, the answer is the very first card: each later position needs every earlier card to be a non-ace, so the chances fall card by card.

How does it generalise, and where is it useful?

With n cards and m special ones, the m specials cut the n - m others into m + 1 gaps, so the first special arrives at (n - m)/(m + 1) + 1 = (n + 1)/(m + 1). Check it with one ace: (52 + 1)/2 = 26.5, the middle of the deck, which is what you would expect. The first spade, with 13 specials, arrives at card 53/14, about 3.79. The same gap symmetry gives the expected wait for the first default in a pool or the first fill among queued orders, provided every ordering is equally likely. That proviso is the limitation: if defaults cluster in time or the deck is stacked, the gaps stop being exchangeable and the rule fails.