Hedge Funds puzzles, solved step by step

Puzzles: 100
Traced to a firm: 38
Topics: 14
Hard: 30

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006You owe exactly Rs pi, that is Rs 3.14159..., and can only pay in whole paise. How do you pay a fair amount on average, and what is the chance you end up paying Rs 3.15?Expected value and dice gamesCoreMillennium ManagementSheung Wan · 2025
Try it first
Under the fair scheme, what is the chance you pay Rs 3.15?
Show the worked solution
Randomise: pay Rs 3.15 with probability 0.159 and Rs 3.14 otherwise. Pi is 3.14159..., which sits 0.159 of the way from 3.14 to 3.15. Paying the higher amount with exactly that probability makes the expected payment 3.14 + 0.01 x 0.1593..., which is pi. So the chance you pay Rs 3.15 is about 15.9%, and over many meals nobody is short-changed.
Why can no fixed amount be fair?
Always round to Rs 3.14 and the restaurant loses 0.159 paise every time; always pay 3.15 and you overpay 0.841 paise. Any fixed amount is unfair to one side, so the only way to be exactly fair is to be fair on average. Two friends who split a Rs 101 bill by taking turns to pay the odd rupee are doing the same thing: neither is exact on any one night, both are exact over time.
Pi sits 0.159 paise above Rs 3.14 and 0.841 paise below Rs 3.15, so paying Rs 3.15 with probability 15.9% and Rs 3.14 otherwise balances exactly at pi, which makes the expected payment fair.
The relationship
$E[\text{pay}] = 3.14\,(1-p) + 3.15\,p = \pi \iff p = \frac{\pi - 3.14}{0.01} \approx 0.1593$
p the probability of paying Rs 3.15
\pi - 3.14 how far pi sits above the lower whole-paisa amount
What it says in wordsThe chance of paying the higher amount equals how far along the gap pi lies.
How do you actually draw a probability of 0.159?
Use any randomness you can split finely. Draw a uniform number between 0 and 1 and pay Rs 3.15 if it falls below 0.1593. With only a die, paying 3.15 on a six gives 1/6, an expected payment of Rs 3.141667: close, not exact. With only a fair coin you can be exact: toss it to generate the binary digits of a uniform number one at a time and stop as soon as the digits so far settle which side of 0.1593 it falls. Each toss settles it with probability one half, so on average it takes two tosses.
Where does randomised rounding show up in a fund?
Whenever a quantity has to be split in whole units. A fund allocating 1,003 shares across three accounts cannot give each 334.33; handing the odd share out by lottery, or in rotation, keeps each account fair on average. The principle is the same: when the exact amount is impossible, make the expected amount exact and keep the error unbiased. The limitation is that fair on average is not fair every time, which is why allocation policies also cap how far any account can drift.
Where candidates lose it
Candidates round to Rs 3.14 and argue the gap is too small to matter. The interviewer is not asking about a sixth of a paisa; the question is whether you see that a fair expected value can be built from amounts that are each individually wrong.
The second loss is the coin flip. Fifty-fifty between 3.14 and 3.15 feels even-handed but averages 3.145, overpaying by nearly half a paisa every time. The probability has to match where pi sits in the gap.
What the interviewer asks next
How would you hit the probability exactly using only a fair coin?
How many coin tosses does that take on average?
What if you owe Rs e, 2.71828...?
Asked at Millennium Management, Quantitative Research, Sheung Wan, 2025 (Wall Street Oasis): How to pay the restaurant fairly if I owe pi dollars. Need to pay with usual dollars and cents.
017You roll a fair die until the first 6 appears and are paid the sum of every roll, including the final 6. What is the expected payout?Expected value and dice gamesHardQuant and systematic fundsProp and quant trading firms
Try it first
What is the expected payout?
Show the worked solution
21. The number of rolls until the first 6 averages 6, so there are on average 5 non-six rolls plus the 6. A roll known not to be a 6 is equally likely to be 1 to 5, so it averages 3, not 3.5, giving 5 x 3 + 6 = 21. Wald's identity confirms it: 6 expected rolls times 3.5 a roll is also 21, because a stop that looks only at past rolls does not bias the total.
Why is it tempting to get 23.5?
Suppose you keep buying scratch cards until one wins. Every card before the winner is, by definition, a loser, so those cards are worth less than an average card. The stopping rule changes the rolls before the stop: each one is known not to be a 6, so it averages 3, and valuing them at 3.5 overpays by 0.5 a roll, 2.5 in all. Five non-sixes at 3.5 plus a 6 is 23.5, which counts the high side twice: once in the 3.5 and again in the final 6.
In a typical run of 4, 1, 5, 2, 3 and then 6, the five rolls before the stop are non-sixes averaging 3 and the total is 21; 5 x 3 + 6 and 6 x 3.5 both give 21, while 5 x 3.5 + 6 = 23.5 wrongly treats the early rolls as ordinary rolls.
How do the two routes agree?
Route one splits the sum: the expected number of non-six rolls times their average, plus the final 6. The count of rolls is a geometric wait with success chance 1/6, so it averages 6, of which 5 are non-sixes: 5 x 3 + 6 = 21. Route two is Wald's identityFor a stopping rule that uses only rolls already seen, the expected total equals the expected number of rolls times the average roll.: the expected total is the expected number of rolls times the average roll, 6 x 3.5 = 21. The low early rolls and the high final roll balance exactly.
The relationship
$E[S] = E[N]\cdot E[X] = 6 \times 3.5 = 21 \qquad E[S] = (E[N]-1)\times 3 + 6 = 5\times 3 + 6 = 21$
S the total paid
N the number of rolls, including the 6
X a single roll, averaging 3.5 before any conditioning
What it says in wordsCounted either as all rolls at 3.5 or as non-sixes at 3 plus a 6, the expected payout is 21.
Why would a trading firm ask this?
Stopping rules are everywhere on a desk: exit at the first stop-loss hit, rebalance at the first breach of a band. The question checks whether you can tell when a stopping rule biases what you observe, as it does for the early rolls, and when it does not, as for the total. Say the condition too: Wald's identity needs the decision to stop to use only rolls already seen, and the expected number of rolls to be finite. A rule that could peek at the next roll would break it.
Where candidates lose it
The slip is 5 x 3.5 + 6 = 23.5. It treats the rolls before the 6 as ordinary rolls, when the stopping rule guarantees none of them is a 6, which pulls their average down to 3.
The opposite slip is to distrust 6 x 3.5 because stopping at a 6 seems to bias it. It does not: the total is unbiased for any stopping rule that looks only at the past. Give both routes and say why they agree.
What the interviewer asks next
What is the expected payout if the final 6 is not paid?
You stop at the first 5 or 6 instead. What is the expected payout?
If you could choose to stop whenever you like, what would you pay to play?
031You roll two fair dice and are paid the higher of the two faces, in rupees. What is the expected payout?Expected value and dice gamesWarm upWolverine TradingChicago · 2025
Try it first
Your quick estimate:
Show the worked solution
161/36, about Rs 4.47. The higher face equals k in 2k minus 1 of the 36 equally likely rolls: 1, 3, 5, 7, 9 and 11 rolls for k from 1 to 6. Multiply each value by its count, add to 161, and divide by 36. The lower face averages 91/36, about 2.53, and the two add to 7, the average total of two dice, which is a quick check.
How many of the 36 rolls give each maximum?
Think of two runners and a prize for the faster time: the winning time is better than a typical single runner's because you always keep the better of two. The higher face is at most k in k x k of the 36 rolls, so it equals exactly k in k squared minus (k minus 1) squared, which is 2k minus 1 rolls. That gives 1 roll with a maximum of 1, 3 with a maximum of 2, and on up to 11 with a maximum of 6. On the grid those cells form L shapes that grow as you move towards the corner.
Of the 36 equally likely rolls, the higher face is 1 in just one roll and 6 in eleven rolls, so the expected maximum is 161/36, about 4.47, well above the 3.5 of a single die.
The relationship
$E[\max] = \sum_{k=1}^{6} k \cdot \frac{2k-1}{36} = \frac{161}{36} \approx 4.47$
k the value of the higher face
2k - 1 the number of rolls, out of 36, whose higher face is exactly k
What it says in wordsWeight each possible maximum by how many of the 36 rolls produce it.
How do you check 4.47 in ten seconds?
Use the pair. The higher face plus the lower face always equals the total of the two dice, so their averages must add to 7. The lower face is at least k in (7 minus k) squared rolls, which gives an average of 91/36, about 2.53. 4.47 plus 2.53 is 7.00. A second method that lands exactly is what makes an interviewer stop checking your arithmetic and move on to the follow-up.
The follow-up is usually a game. If you could pay to roll one die or to roll two and keep the higher, the second is worth about Rs 0.97 more. That gap, the value of a free second look, is the same idea as an option: the right to choose after seeing the outcome is worth paying for.
Where candidates lose it
The common loss is answering 3.5 plus something vague, or 5, from instinct. Both skip the count of how often each maximum occurs, which is the whole question.
The second is listing all 36 rolls one by one under time pressure. Say the 2k minus 1 rule, give 161 over 36, and use the lower face check to show the number is right.
What the interviewer asks next
What is the expected higher face with three dice?
What would you pay to roll two dice and keep the higher, if you could reroll both once?
What is the expected value of the lower face, and why do the two add to 7?
Asked at Wolverine Trading, Equity Hedge, Chicago, 2025 (Wall Street Oasis): Typical dice questions that you can find in most probability textbooks
043You may roll a fair die up to three times. After each roll you either stop and are paid the face in rupees, or throw that roll away and roll again; if you reach the third roll you must take it. What is the best stopping rule, and what is the game worth?Expected value and dice gamesHardSquarepoint CapitalLondon · 2026
Try it first
On the first roll you get a 4. What do you do?
Show the worked solution
Stop on the first roll only with a 5 or 6, on the second with a 4 or more, and the game is worth 14/3, about Rs 4.67. Work backwards. The last roll is worth 3.5. With two rolls left, keep 4, 5 or 6 and reroll otherwise: worth (4 + 5 + 6)/6 + 3/6 x 3.5 = 4.25. With three rolls left, keep only what beats 4.25, a 5 or 6: worth (5 + 6)/6 + 4/6 x 4.25 = 14/3.
Why start from the last roll?
Deciding whether to take a job offer is easier if you know what your fallback is worth. Each keep-or-reroll decision compares the roll in hand with the value of the rolls still to come, so you need the value of the future first, and the only stage with no future is the last one. On the last roll you must take whatever comes, which is worth 3.5 on average. That single number lets you solve the stage before it, and so on backwards. This is {term('backward induction', 'Solving a sequence of decisions from the last one to the first, using the value of each later stage to make the earlier decision.')}, the core of dynamic programming.
With one roll left the game is worth 3.5, so with two left you keep 4 or more and the game is worth 4.25; with three left you keep only 5 or 6, and the whole game is worth 14/3, about 4.67.
How do the thresholds come out?
With two rolls left, a roll of 4, 5 or 6 beats the 3.5 you expect from rerolling, and 1, 2 or 3 does not. So the two-roll game is worth the average of the kept faces times their chance, plus the chance of rerolling times 3.5: 15/6 + 1.75 = 4.25. On the first roll the fallback is now 4.25, so a 4 is no longer good enough: only 5 or 6 is kept. That gives 11/6 plus 4/6 x 4.25, which is 1.833 plus 2.833, or 14/3.
The relationship
$V_n = \frac{1}{6}\sum_{f=1}^{6} \max\left(f,\, V_{n-1}\right), \quad V_1 = 3.5,\; V_2 = 4.25,\; V_3 = \tfrac{14}{3}$
V_n the value of the game with n rolls left
f the face you just rolled
max(f, V_{n-1}) keep the roll or throw it away, whichever is worth more
What it says in wordsEach stage is worth the average, over the six faces, of the better of keeping the face or playing on.
Add the pattern. More rolls always raise the value, but by less each time: 3.5, 4.25, 4.67, then about 4.94 with four rolls. An extra option is always worth something and never worth more than what it can still improve. That is the same logic as valuing a trade you can exit early: the right to wait is priced by what the future is worth, not by the average outcome.
Where candidates lose it
The common loss is using 3.5 as the threshold at every stage, which keeps a 4 on the first roll. The fallback on the first roll is the two-roll game, 4.25, not a single roll.
The second loss is solving forwards and getting lost. Say you will start from the last roll, compute 3.5, 4.25 and 14/3 in that order, and the thresholds fall out.
What the interviewer asks next
What is the game worth with four rolls, and what is the first-roll threshold?
You must pay Rs 1 for every reroll. How do the thresholds change?
You are paid the square of the face instead. What is the optimal rule?
Asked at Squarepoint Capital, Quantitative Research, London, 2026 (Wall Street Oasis): Dynamic programming questions with focus on probability at the end.
056Game A: roll two fair dice and be paid the product of the faces in rupees. Game B: roll one fair die and be paid the square of its face. Which game is worth more, and by exactly how much?Expected value and dice gamesCoreCitadelMiami · 2022
Try it first
Which game would you rather play, and by roughly how much?
Show the worked solution
The one-die game is worth more: Rs 15.17 against Rs 12.25, a gap of 35/12, about Rs 2.92. With two independent dice the average product is the product of the averages, 3.5 x 3.5 = 12.25. With one die the average square is (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6. The gap between the two is exactly the variance of one die.
Why is averaging a square not the same as squaring an average?
Take two students who score 2 and 8 in a test. Their average is 5, and 5 squared is 25. Square each score first and then average, and you get (4 + 64)/2 = 34. Squaring rewards the high value more than it penalises the low one, so the average of the squares is always at least the square of the average, and the gap is the spread. Here the gap, 9, is exactly the variance of the two scores.
How do the two games come out?
In game A the dice are independent, so the average of the product is the product of the averages. Two independent dice give 3.5 x 3.5 = Rs 12.25, because a high roll on one die is as likely to meet a low roll on the other as a high one. In game B one number is multiplied by itself, so a 6 always meets a 6 and a 1 always meets a 1. The average of the six squares is 91/6, about Rs 15.17.
The relationship
$E[X^2] = \mathrm{Var}(X) + E[X]^2 = \tfrac{35}{12} + 12.25 = \tfrac{91}{6} \approx 15.17, \qquad E[XY] = E[X]\,E[Y] = 12.25$
X, Y the faces of two independent dice
Var(X) the variance of one die, 35/12
E[X] the average face, 3.5
What it says in wordsThe average square is the square of the average plus the variance; the average product of independent dice has no variance term.
The six squares from 1 to 36 average 15.17, above the 12.25 that squaring the average roll gives; the two-dice product game is worth 12.25, so the one-die square game is worth 2.92 more, exactly the variance of one die.
What is the desk lesson?
A payoff that is the square of a move gains from dispersion; one built from two independent pieces does not. Correlation is what turns a product into a square: if the second die always copied the first, game A would be game B. Flip it and make the second die show 7 minus the first, and the average product falls to 9.33. Say that range out loud and the interviewer knows you see the payoff as a bet on varianceThe average squared distance of an outcome from its mean; for one fair die it is 35/12, about 2.92. as well as on the average.
Where candidates lose it
The quick wrong answer is that the games are worth the same, because both feel like three and a half times three and a half. That is true only for independent dice. Squaring one die ties the factors together.
The second loss is getting 15.17 and 12.25 and not naming the gap. Say that 2.92 is the variance of a die; that one sentence is what the interviewer is waiting for.
What the interviewer asks next
What is game A worth if you are paid the sum of the two dice instead of the product?
The second die always shows 7 minus the first. What is the average product now?
What is the expected value of the larger of two dice?
Asked at Citadel, Sales and Trading, Miami, 2022 (Wall Street Oasis): No, pretty typical interview questions, dice questions, etc.
068You and an opponent each secretly show heads or tails. If both show heads you win Rs 3, if both show tails you win Rs 1, and if they differ you pay Rs 2. The payoffs look balanced. What mix should each player use, and what is the game worth to you?Expected value and dice gamesHardCitadelsydney · 2025
Try it first
If both of you play well, what is the game worth to you per round?
Show the worked solution
Both players should show heads 3/8 of the time, and the game is worth minus Rs 0.125 a round to you. Choose your mix so the opponent gains nothing by switching: 3p - 2(1 - p) = -2p + (1 - p) gives p = 3/8. The opponent's mix solves the same balance from your side, also 3/8. At those mixes you lose an eighth of a rupee a round, although the payoffs look even.
Why is a fair coin the wrong strategy?
Think of a penalty taker and a goalkeeper. If the taker always shoots left, the keeper dives left; the taker's only defence is to mix so the keeper cannot profit from guessing. A mix is right only if it leaves the other side indifferent, and a fair coin here does not: against it the opponent's tails pays you 0.5 x (-2) + 0.5 x 1 = -0.50 a round. So the opponent always shows tails, and you lose Rs 0.50 a round rather than breaking even.
How do you find the equilibrium mix?
Set up your payoff for each of your choices as the opponent's chance of heads, q, varies. Showing heads pays 3q - 2(1 - q) = 5q - 2; showing tails pays -2q + (1 - q) = 1 - 3q; they are equal at q = 3/8, where both pay -1/8. The opponent plays 3/8 heads so that nothing you do beats -1/8. By the same balance from the opponent's side, you play 3/8 heads so that nothing the opponent does pushes you below -1/8. That pair is the Nash equilibriumA pair of strategies in which neither player can do better by changing only their own choice..
The relationship
$5q - 2 = 1 - 3q \;\Rightarrow\; q = \tfrac{3}{8}, \qquad V = 5 \cdot \tfrac38 - 2 = -\tfrac18$
q the opponent's chance of showing heads
5q - 2 your expected payoff if you show heads
1 - 3q your expected payoff if you show tails
V the value of the game to you
What it says in wordsThe opponent's mix makes your two choices pay the same, and that common payoff is what the game is worth.
Your payoff from showing heads rises with the opponent's chance of heads and your payoff from tails falls; the lines cross at 3/8, where either choice pays minus Rs 0.125, so the opponent plays 3/8 heads and the game is worth minus Rs 0.125 a round to you.
Why is the game negative when the payoffs look even?
Because the balance is in the totals, not in the play. Your two winning cells need coordination the opponent will not give you, while their winning cells, the mismatches, pay the same 2 either way. The opponent can lean towards tails, starving your big Rs 3 cell, and the only price is feeding your small Rs 1 cell. Say what a desk would do with it: ask to be paid about 13 paise a round to play, or ask to swap sides.
Where candidates lose it
The trap is adding up the payoffs, 3 and 1 against 2 and 2, calling the game fair and playing a fair coin. That ignores that the opponent chooses too, and against a fair coin their best reply costs you Rs 0.50 a round.
The second slip is solving for your own mix by making yourself indifferent. Each player's mix is chosen to make the other player indifferent; set up the equation from the opponent's payoffs.
What the interviewer asks next
Change the tails-tails payoff to Rs 2. What are the mixes and the value now?
What would you pay per round to play this game from the opponent's side?
The opponent is known to show heads half the time. What do you do, and what do you earn?
Asked at Citadel, Quantitative Trading, sydney, 2025 (Wall Street Oasis): many probability questions for OA. mix of prob and game theory for technical
081Which is more likely: at least one six in four rolls of a single die, or at least one double six in twenty-four rolls of a pair of dice?Expected value and dice gamesCoreQuant and systematic fundsProp and quant trading firms
Try it first
Before any arithmetic, which do you back?
Show the worked solution
The single six in four rolls: 51.8% against 49.1%. Use the complement for each. Four rolls with no six happen (5/6)^4 = 48.2% of the time, so at least one six is 51.8%. Twenty-four rolls of two dice with no double six happen (35/36)^24 = 50.9% of the time, so at least one is 49.1%. Only the first is better than even money.
Why does the proportional argument give the same answer for both?
The old gamblers' rule, in the problem usually credited to the Chevalier de Méré, went like this: a six comes up one time in six and you get four tries, so 4/6; a double six comes up one time in thirty-six and you get twenty-four tries, so 24/36, also 4/6. Think of phoning a friend four times: four tries do not give four times the chance of getting through, because once they pick up, the later calls add nothing. Adding the chance per try counts the runs with two or more hits more than once, so it overstates the chance of at least one hit, and the overstatement grows with the number of tries.
At least one six in four rolls comes up 51.8% of the time and at least one double six in twenty-four rolls only 49.1%, on opposite sides of even money, while the proportional rule wrongly puts both at 66.7%.
How does the complement settle it?
Ask how likely it is that nothing happens. Four rolls with no six: (5/6)^4 = 625/1,296 = 48.2%, so at least one six is 51.8%. Twenty-four rolls with no double six: (35/36)^24 = 50.9%, so at least one is 49.1%. The rare event with many tries falls short, because its misses compound over six times as many rolls.
The relationship
$1 - \left(\tfrac{5}{6}\right)^{4} = 0.518 \qquad 1 - \left(\tfrac{35}{36}\right)^{24} = 0.491$
5/6 the chance one roll of a die is not a six
35/36 the chance one roll of two dice is not a double six
4, 24 the number of tries in each bet
What it says in wordsThe chance of at least one hit is one minus the chance that every single try misses.
Why is the gap so small, and what is it worth as a bet?
The two probabilities are only 2.6 points apart, which is why the question needed a careful calculation to settle and why it still tests method rather than intuition. As an even-money bet, the first earns 3.5 paise per rupee staked on average: 51.8% of winning a rupee less 48.2% of losing one. The second loses 1.7 paise per rupee. A small edge repeated many times is the whole business of a casino, and of many trading strategies. Give the second bet one more roll and it tips over: 1 - (35/36)^25 = 50.6%.
Where candidates lose it
The trap is the proportional argument: 4 x 1/6 = 24 x 1/36 = 2/3, so the bets are equal. It is the reasoning the question was built to catch, and any answer that makes the chance grow in a straight line with the tries fails the moment the tries pass six.
The second loss is getting 51.8% for the first bet and assuming the second is also above half. Compute both; the point of the question is that they fall on opposite sides of 50%.
What the interviewer asks next
How many rolls of two dice do you need before a double six is more likely than not?
You are offered even money on the second bet. What is your expected result per Rs 100 staked?
Why is 1 - (1 - p)^n close to 1 - e^(-np) when p is small, and what does that give for the second bet?
092A casino offers the St Petersburg game: a fair coin is tossed until the first tail, and if that takes n tosses you are paid Rs 2 to the power n. The casino can pay out at most Rs 1 crore. What is a fair price to play?Expected value and dice gamesHardQuant and systematic fundsProp and quant trading firms
Try it first
Roughly what is the capped game worth?
Show the worked solution
About Rs 24.19. Round n pays Rs 2^n with probability 1/2^n, so each round adds exactly Rs 1 of expected value. That holds up to n = 23, since 2^23 is about Rs 84 lakh and 2^24 is above the Rs 1 crore cap. From round 24 on, the payout is stuck at Rs 1 crore, with total probability 1/2^23, which adds about Rs 1.19. So 23 + 1.19, about Rs 24.
Why is the uncapped game worth an infinite amount?
Each round is a doubling bet: the payout doubles while the chance of reaching it halves. Think of a raffle where a ticket twice as valuable is half as likely to win: every prize tier is worth the same to you. Round n pays 2^n with probability 1/2^n, so every round contributes exactly Rs 1 to the expected value, and there are infinitely many rounds. That is the famous paradox: few people would pay even Rs 100, yet the expected value has no bound. The resolution that matters on a desk is not psychology; it is that nobody can pay out an unlimited amount.
Rounds 1 to 23 each add exactly Rs 1 to the expected value, and once the Rs 1 crore cap binds the later rounds add 0.60, 0.30, 0.15 and so on, Rs 1.19 in all, so the capped game is worth about Rs 24.19.
How does the cap change the sum?
Find the round where the cap starts to bind. 2^23 is Rs 83,88,608, under Rs 1 crore; 2^24 is Rs 1,67,77,216, over it. So rounds 1 to 23 each add Rs 1, and every round from 24 on pays the capped Rs 1 crore, which together happen with probability 1/2^23 and add 1,00,00,000 / 83,88,608, about Rs 1.19. The fair price is about Rs 24.19. Nearly all of the textbook infinity lives in outcomes the casino cannot pay.
The relationship
$E = \sum_{n=1}^{23} \frac{2^n}{2^n} + \frac{10^7}{2^{23}} = 23 + 1.19 = 24.19$
2^n the payout if the first tail arrives on toss n
1/2^n the chance the first tail arrives on toss n
10^7 / 2^23 the capped Rs 1 crore times the chance of reaching round 24 or later
What it says in wordsEvery uncapped round is worth one rupee; the capped tail is worth the cap times the chance of getting that far.
What does a bigger casino buy you?
Very little. The value grows only with the logarithm of the cap: each doubling of the casino's bankroll adds about Rs 1. A cap of Rs 1,000 crore, a thousand times larger, lifts the fair price only to about Rs 34. That is the lesson a risk manager takes away: a payoff whose expected value rests on rare, enormous outcomes is worth what the other side can actually pay, and any estimate built on the tail should be checked against who stands behind it.
Where candidates lose it
Answering infinity is the trap for anyone who knows the textbook game. The interviewer added the cap to see whether you can find where it binds and redo the sum, not recite the paradox.
The other loss is dropping the tail beyond the cap and saying Rs 23, or valuing it crudely at the full Rs 1 crore times a guessed chance. The tail is a clean sum: probability 1/2^23 of receiving Rs 1 crore.
What the interviewer asks next
The cap rises to Rs 1,000 crore. What is the fair price now?
With logarithmic utility and wealth of Rs 1 lakh, roughly what would you pay for the uncapped game?
How is a book short deep out-of-the-money options like the casino in this game?

p	the probability of paying Rs 3.15
\pi - 3.14	how far pi sits above the lower whole-paisa amount

S	the total paid
N	the number of rolls, including the 6
X	a single roll, averaging 3.5 before any conditioning

k	the value of the higher face
2k - 1	the number of rolls, out of 36, whose higher face is exactly k

V_n	the value of the game with n rolls left
f	the face you just rolled
max(f, V_{n-1})	keep the roll or throw it away, whichever is worth more

Try it first

Why can no fixed amount be fair?

How do you actually draw a probability of 0.159?

Where does randomised rounding show up in a fund?

Where candidates lose it

What the interviewer asks next

Try it first

Why is it tempting to get 23.5?

How do the two routes agree?

Why would a trading firm ask this?

Where candidates lose it

What the interviewer asks next

Try it first

How many of the 36 rolls give each maximum?

How do you check 4.47 in ten seconds?

Where candidates lose it

What the interviewer asks next

Try it first

Why start from the last roll?

How do the thresholds come out?

Where candidates lose it

What the interviewer asks next

Try it first

Why is averaging a square not the same as squaring an average?

How do the two games come out?

What is the desk lesson?

Where candidates lose it

What the interviewer asks next

Try it first

Why is a fair coin the wrong strategy?

How do you find the equilibrium mix?

Why is the game negative when the payoffs look even?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the proportional argument give the same answer for both?

How does the complement settle it?

Why is the gap so small, and what is it worth as a bet?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the uncapped game worth an infinite amount?

How does the cap change the sum?

What does a bigger casino buy you?

Where candidates lose it

What the interviewer asks next