Hedge Funds puzzles, solved step by step

One round trip makes at most 16; unlimited round trips make 26. For one trade, walk the prices once, carrying the lowest price so far and the best sale against it: buy at 96, sell at 112. For many trades, add every day-on-day rise and skip every fall: 8 + 11 + 7 = 26. Without short selling the falls are simply sat out, never profited from.

How do you find the best single trade without checking every pair?

Imagine walking down a street of shops that all sell the same phone, planning to buy once and sell once further along. You do not need to compare every pair of shops: carry the cheapest price seen so far in your head, and at each shop ask what selling here would make against it. One pass, keeping the running minimum and the best gap found so far, gives the best single trade. Here the running minimum drops to 96 on day 2 and the best gap appears on day 7: 112 - 96 = 16.

Why is the many-trade answer just the sum of the rises?

Any rise from a low to a later high is the sum of the daily steps inside it, and some of those steps may be falls. With no short selling and no costs, the most you can make is the total of every positive day-on-day move, 26 here, because trading only the up steps collects everything a longer trade would and skips its falls. In practice that is three round trips: buy 96, sell 104; buy 99, sell 110; buy 105, sell 112.

What does the interviewer add next?

Costs. Once each round trip costs something, the sum of rises overstates the profit, because small moves stop being worth trading. With a cost of 6 per round trip, the three separate trades net 2 + 5 + 1 = 8, the best two-trade split nets 9, and the single trade from 96 to 112 nets 10, so the single trade now wins. The general version is a short dynamic programme that tracks the best profit on each day while holding and while flat.

Move the mid from 10.5 to 13 and tighten the market by about a fifth. Each die averages 3.5, so three dice average 10.5. Once one die shows 6, the sum is 6 plus two unknown dice averaging 7, which is 13. The variance falls from 3 x 35/12 to 2 x 35/12, so the standard deviation drops from 2.96 to 2.42. If the first market was 9.5 at 11.5, the new one is about 12.2 at 13.8.

Where do you put the first market, and how wide?

Start from the fair value and then decide the width from how uncertain the outcome is. The mid is the expected sum, 3 x 3.5 = 10.5, and the width should scale with the standard deviation of the sum, because that is how far the answer typically lands from the mid. One die has variance 35/12, so three independent dice have 35/4 = 8.75, a standard deviation of 2.96. A market of 9.5 bid, 11.5 offered is a reasonable opening: tight enough to trade, with room for your edge.

What does revealing one die change?

A weather forecast for tomorrow is more precise than one for next week, because fewer things can still change. Once one die is known, it contributes a certain 6 and no uncertainty, so the mid rises by 2.5 and only two dice of variance remain. The mid becomes 6 + 7 = 13. The variance becomes 35/6, a standard deviation of 2.42, down from 2.96. Scale the width by the same ratio, about 0.82, and a 2.0 wide market becomes about 1.6 wide: 12.2 at 13.8.

Say what would make you widen instead. If the person revealing the die can choose which die to show, or picks the moment, the reveal itself carries information and you should be more careful, not less. A 6 chosen as the highest of three tells you the other two are 6 or lower, and they no longer average 7. Interviewers like it when you ask who chose what to reveal before you requote.

The lift says they saw at least two heads, so the fair value is now at least 5.75, not 5; requote around 5.75 bid, 6.5 offered. The other seven flips are worth 3.5 heads, so the buyer's value is heads seen plus 3.5. Paying 5.5 only makes sense with two heads (5.5) or three (6.5). Those are 3 to 1 likely, giving 5.75; a buyer who needs a strict edge saw three heads, worth 6.5.

What is the trader's view before they trade?

A friend offers to buy your raffle ticket after the first few numbers are drawn. The offer itself is the warning. The informed trader values the contract at heads already seen plus 3.5, the expected heads in the seven unseen flips, so their value is 3.5, 4.5, 5.5 or 6.5 with chances 1, 3, 3 and 1 in 8. Against your market of 4.5 bid and 5.5 offered, they buy only if their value is at least 5.5, and sell to you at 4.5 only if it is 4.5 or less. The flat 5 you quoted around is right only for someone who has seen nothing.

How do you turn the trade into a new fair value?

Condition on the fact that they traded. Only the two-head and three-head worlds produce a buy at 5.5, and they are 3/8 and 1/8 likely, so given a lift the value is (3 x 5.5 + 1 x 6.5) / 4 = 5.75. If you assume they would not bother trading at zero edge, only the three-head world is left and the value is 6.50. Either way you sold too cheaply: this is adverse selectionThe tendency of a market maker to trade most with the people who know more, so the trades that happen are the ones that lose money for the market maker., and it is the cost every market maker prices into the spread.

Now requote. Your bid should not be below what you now believe the floor is, and your offer should sit where even the best informed buyer has no edge. Something like 5.75 bid, 6.5 offered does both: a buyer who saw three heads is indifferent at 6.5, and you are no longer selling below value. Cut your size too, because you know someone is trading with more information than you, and say you would ask whether they could see the flips before quoting again.