Derivatives Foundation puzzles, solved step by step

5/9, about 55.6%. Put A's arrival time across and B's up a 60 by 60 square; every pair of times is a point, all equally likely, so probability is area. They meet when the times are within 20 minutes, the band either side of the diagonal. They miss in two corner triangles, each with legs of 40 minutes and area 800 of 3,600, which is 2/9. So the meeting chance is 1 minus 4/9 = 5/9.

Why turn two arrival times into a square?

Throw a dart at a square board without aiming and the chance it lands in any patch is just that patch's share of the board. Two independent arrival times, each spread evenly over the hour, behave exactly like that dart: A's time picks a position across, B's time picks a position up. With two independent uniform times, every pair of arrivals is a point in a 60 by 60 square, equally likely anywhere in it, so a probability becomes an area you can see. The event they meet is the set of points where the two times differ by less than 20 minutes, a band hugging the diagonal.

Why is it easier to compute where they miss?

The band is an awkward six-sided shape; the regions outside it are two clean triangles. In the top-left triangle B arrives more than 20 minutes after A, so A has gone; in the bottom-right one, A is the late one. Each triangle has legs of 60 minus 20 = 40 minutes, so its area is 40 x 40 / 2 = 800 square minutes out of 3,600, which is 2/9, and the two together are 4/9. The clause about leaving at 2 pm changes nothing, because no one can arrive after 2 pm anyway; it only stops the question from looking ambiguous.

How does the answer move with the waiting time?

Not in a straight line, and that is a common follow-up. Doubling the wait from 10 to 20 minutes takes the meeting chance from about 31% to about 56%, not from one third to two thirds, because the miss region shrinks as a square. The table runs the formula for four waits. The trading version is two orders that must arrive within a latency window to match: halving the gap you can tolerate does more than halve the matches, and a picture of the square is the fastest way to see by how much. The limitation to state is the uniform assumption; real arrivals bunch near the hour, which raises the meeting chance.

1/3. Picking a cube at random and then a face at random makes every one of the 27 x 6 = 162 small faces equally likely to end up on top. The painted small faces are exactly the squares on the big cube's surface, 6 faces of 9 each, 54 in all. So the chance is 54/162 = 1/3. The breakdown by cube agrees: corners give 8 x 3, edges 12 x 2, face centres 6 x 1, the core 0, total 54.

Why count faces rather than cubes?

If a bag holds sweets of different sizes and you want the chance a random bite is chocolate, you count chocolate bites, not chocolate sweets. Every small cube is equally likely and every face of it is equally likely to land on top, so every one of the 162 small faces has the same chance, 1/162, and the answer is just the share of small faces that are painted. That share is easy, because the painted small faces are exactly the visible squares of the big cube: 6 faces with 9 squares each, 54. The answer, 54/162 = 1/3, comes in one line without classifying a single cube, which is what the interviewer hopes to see.

How does the cube-by-cube count confirm it?

Classify the 27 cubes by position. The 8 corners have 3 painted faces each, the 12 edge cubes have 2, the 6 face centres have 1, and the single core cube has none: 8 + 12 + 6 + 1 = 27. Weight each type by how often you pick it and by the chance its top is painted: 8/27 x 3/6 + 12/27 x 2/6 + 6/27 x 1/6 + 1/27 x 0 = (24 + 24 + 6) / 162 = 1/3, the same answer by the long road. The two methods are the law of total probability written two ways, once by cube and once by face, and saying that out loud shows you know why they must agree. For an n x n x n cube the face count gives 6n squared painted faces out of 6n cubed, so the answer is 1/n: 1/2 for a 2 x 2 x 2 cube, 1/10 for a 10 x 10 x 10.

What is the natural follow-up, and how do you answer it?

Turn it round: the top face is painted; what is the chance you picked a corner? That is Bayes on the same count. Of the 54 painted faces, 24 belong to corners, so the chance is 24/54 = 4/9, far above the 8/27 a corner has before you look. Seeing paint is evidence for the cubes with more paint, and the face count gives the posterior directly without a formula. A second follow-up asks for the chance that the picked cube has any paint at all, which is 26/27, and the gap between 26/27 and 1/3 is exactly the trap in the original question. The limitation is that the face count relies on every face being equally likely to land on top; a weighted cube, or a rule that picks cubes by size, would need the long route.

110 routes. Without the block there are C(10,4) = 210 routes, one for each way of placing 4 ups among 10 moves. A route through (3,2) is a route from (0,0) to (3,2), C(5,2) = 10 ways, followed by a route from (3,2) to (6,4), another C(5,2) = 10 ways, so 100 routes pass through the block. Subtract: 210 - 100 = 110.

Why is a lattice route a choice of positions rather than a sequence of decisions?

Think of a delivery driver in a city laid out as a grid who only ever drives east or north. Whatever order the turns come in, the trip is six blocks east and four blocks north, and the only freedom is which of the ten blocks are the north ones. A monotone route is fully described by choosing which 4 of its 10 moves go up, so the number of routes is 10 choose 4, which is 210, and no decision tree is needed. The same logic prices any question that asks how many ways a count can reach a level in fixed-size steps.

Why does multiplying the two legs count each bad route exactly once?

Because a monotone route visits a given point at most once; it can never come back. A route through (3,2) splits uniquely into the part before the block and the part after, so the number of such routes is the product of the two leg counts, 10 x 10 = 100, with no double counting to correct. Each leg is three rights and two ups, so C(5,2) = 10. If there were two blocked points, the same idea works but needs inclusion and exclusion: subtract the routes through each, then add back the routes through both.

How do you check 110 another way?

Fill the grid with counts. Each point's count is the sum of the counts to its left and below, with the blocked point set to zero, and the corner comes out at 110. That dynamic-programming check takes a minute on paper and catches arithmetic slips in the binomials. It also answers the probability version the interviewer sometimes asks: if each step is right or up with equal chance, the chance a random walk reaches (6,4) at all is not 1, because it can overshoot, so the probability of avoiding the block among routes that do arrive is 110/210, about 52%, which is a different question from the probability for a free walk.

The probability is 1/6, not 1/5. By symmetry the game ends on a 2 one third of the time. A first roll of 1 happens one sixth of the time and leaves all three evens unseen, so 2 is last with probability 1/3. The joint probability is 1/18, and 1/18 divided by 1/3 is 1/6. The naive 1/5 treats the five possible first rolls as equally likely given the ending, and they are not.

Why does the ending change the odds of the start?

Think of a race where you learn only who finished last. If you are told that runner C came last, the start line was still fair, but some starting arrangements make C last more often than others, and you should update towards those. The die works the same way. A first roll of 4 leaves only two evens unseen, so 2 comes last half the time; a first roll of 1 leaves three evens unseen, so 2 comes last only one third of the time. The ending is twice as consistent with a first-roll 4 as with a first-roll 1, and a first-roll 2 is ruled out entirely, because a 2 that has already appeared cannot complete the set.

How do you set up the Bayes calculation in thirty seconds?

Group the first roll into three cases rather than six, because 1, 3 and 5 are interchangeable and so are 4 and 6. Odd first roll: probability 1/2, and then 2 ends the game with probability 1/3, giving a joint weight of 1/6. First roll 2: joint weight 0. First roll 4 or 6: probability 1/3, then 2 ends the game with probability 1/2, joint weight 1/6. The weights add to 1/3, which is the unconditional chance of ending on a 2, as symmetry says they must. Given the ending, the first roll was odd with probability 1/2 and was 4 or 6 with probability 1/2, so each odd face carries 1/6 and each of 4 and 6 carries 1/4.

The sanity check the interviewer wants to hear: 3 x 1/6 + 2 x 1/4 = 1, so the posterior weights over the five possible first rolls add up. Then say the general point. An odd roll is a wasted roll that tells you nothing about which even finishes last, which is why its posterior weight is simply its prior, 1/6, unchanged. The information in the ending all goes into shifting weight from the 2, which is now impossible, onto 4 and 6.

One half. By the time the last passenger boards, seats 2 to 99 are always taken, because each owner either sat in theirs or found it taken. The free seat is either seat 1 or seat 100. Every passenger who chooses at random, the first and each one displaced after him, faces seat 1 and seat 100 with exactly the same chance, so the game is equally likely to settle on either. The answer is 1/2 for any plane with two or more seats.

Why do only two seats matter?

Think of a cloakroom where one guest hangs his coat on a random hook. Each later guest uses their own hook if it is free and a random free hook if not. Every hook except two has an owner who will turn up and either use it or find it used. The last passenger can only end up in seat 1 or seat 100, because every seat in between has an owner who boards earlier and leaves it occupied. So the whole question is which of those two seats is still empty when the last passenger walks down the aisle.

Why are the two endings equally likely?

Follow the chain of displaced people. Passenger 1 picks at random. If he takes seat 1, nobody is ever displaced and the last passenger gets seat 100. If he takes seat 100, the last passenger is shut out. If he takes some other seat k, everyone up to k sits normally and passenger k inherits the problem: a random choice among the free seats, which still include both seat 1 and seat 100. Every random chooser in the chain picks seat 1 and seat 100 with equal probability, and the chain stops the moment either is taken, so the two endings carry the same total weight, 1/2 each. The other seats only pass the problem along; they never decide it.

Check it on the smallest case aloud: with two seats, passenger 1 takes his own or the other with equal chance, so 1/2. Running the recursion for every plane from 2 to 100 seats returns exactly 1/2 each time. A useful extension the interviewer often asks next: passenger j, for j from 2 to n, gets their own seat with probability (n - j + 1)/(n - j + 2). Passenger 2 is almost always fine; passenger 99 is fine two times in three. The limitation is in the rules: if displaced passengers preferred seats near the front, the symmetry between seat 1 and seat 100 breaks and so does the half.