Quant puzzles, solved step by step

About 3.99. Give each face an indicator that equals 1 if that face appears at least once. A given face is missed on all six rolls with probability (5/6) to the 6th, about 0.335, so it appears with probability 0.665. The expected count of distinct faces is the sum of the six indicators' expectations: 6 x 0.665 = 3.99. No case listing is needed.

Why not list the cases?

You could work out the chance of exactly one, two, up to six distinct faces and average them. It works, but it needs Stirling numbersCounts of the ways to split a set of items into a given number of non-empty groups; they appear when counting surjections. or a lot of careful counting, and it is easy to slip. Linearity of expectation lets you ignore how the faces interact: the expected total of several indicators is the sum of their expectations, whether or not they are independent. Here the six indicators are clearly dependent, since seeing many faces leaves fewer rolls for the others, and it does not matter at all.

How does the indicator trick work step by step?

Think of a teacher counting how many of six friends turn up to a party. Instead of listing every guest list, she asks for each friend separately how likely that friend is to come, then adds. Write the count as I1 + I2 + ... + I6, where I_k is 1 if face k appears; take expectations; each E[I_k] is just the probability face k appears. Face k is missed on one roll with probability 5/6, on all six with (5/6) to the 6th, 0.335. So each indicator averages 0.665 and the total averages 3.99.

Where does this pattern reappear?

The same shape answers how many distinct birthdays a group of n people has, how many of n hash buckets get used, and how many different stocks a random sample of trades touches. For n faces and n rolls, the expected share of faces seen is 1 - (1 - 1/n) to the n, which tends to 1 - 1/e, about 63.2%, as n grows. Six faces give 66.5%, already close. The exact enumeration of all 46,656 rolls gives a mean of 3.9906, the same number.

Rs 4.20. Every roll pays its face, averaging 3.5, and with probability 1/6 the game then starts again, worth the same V. So V = 3.5 + V/6, which gives (5/6)V = 3.5 and V = 4.2. A second route agrees: the number of rolls averages 1/(5/6) = 1.2, and each averages 3.5, so 1.2 x 3.5 = 4.2.

Why write the game in terms of itself?

A pass that gets a free renewal each time you use it on a lucky day is worth its first use plus, on lucky days, another pass exactly like it. When a game can repeat with no memory, the value after the repeat is the value of the whole game, so one equation replaces an infinite sum. Here, after a six you are paid 6 and then face precisely the game you started with, worth V.

How do you set up the equation without slipping?

Condition on the first roll. With probability 5/6 it shows 1 to 5, averaging 3, and you stop. With probability 1/6 it shows 6: you collect 6 and then expect V more. So V = (5/6) x 3 + (1/6) x (6 + V), which simplifies to V = 3.5 + V/6, and V = 3.5 x 6/5 = 4.2. The form 3.5 + V/6 is worth saying: every roll pays 3.5 on average, and one time in six you get another go at the whole game.

Check by counting rolls. The chance a roll triggers another is 1/6, so the number of rolls averages 1/(1 - 1/6) = 1.2, and each roll averages 3.5 whatever came before, giving 1.2 x 3.5 = 4.2. Summing the series directly, k sixes and then a stop, also lands on 4.2, but the recursion gets there in two lines. The limitation to name if asked to price it: 4.2 is a fair value for one play; the payout has a long right tail, since two sixes in a row, one time in 36, already pay at least 13.

15. The first six takes 6 rolls on average, so 5 rolls come before it. Each of those rolls is known not to be a six, so it is uniform on 1 to 5 and averages 3. Expected count times expected size gives 5 x 3 = 15. The check: all rolls including the six average 6 x 3.5 = 21, and taking off the final six leaves 15.

How many rolls come before the six?

Picture waiting at a stop where each minute a bus arrives with chance 1 in 6. On average you wait 6 minutes, and the sixth is the one where it comes. The number of rolls up to and including the first six is geometric with mean 1/p = 6, so the number strictly before it is 5. That is the first factor. Most candidates get this far; the loss comes in the second factor.

Why is each of those rolls worth 3 and not 3.5?

Because you are told something about them. Every roll before the stopping six is, by definition, not a six, so its distribution is the die conditioned on 1 to 5, which averages exactly 3. Using 3.5 gives 17.5, the most common wrong answer. It is the same slip as averaging the income of people who did not win a prize with everyone's income, prize winners included.

How do you check 15 a second way in the room?

Two checks, both fast. The recursion: with chance 5/6 the next roll is not a six, adds 3 on average and you are back where you started, so E = (5/6)(3 + E), which solves to 15. The full sum: Wald's identityFor a stopping time N that does not look into the future, the expected sum of N independent identical draws equals E[N] times the mean of one draw. applied to every roll including the six gives 6 x 3.5 = 21, and the last roll is always exactly 6, so the rest must average 15. Say both; the second one shows you understand why the conditional mean is 3.

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.