Quant puzzles, solved step by step

10 is more likely: 27 ways out of 216 against 25. The equally likely outcomes are the 216 ordered rolls, not the unordered combinations. A combination of three different faces covers six ordered rolls, one with a pair covers three, and a triple covers one. Total 9 includes 3 + 3 + 3, which counts once, and one fewer all-different combination, so it loses two orderings to 10.

Why are combinations the wrong thing to count?

Think of dealing two cards and asking whether a pair of kings or a king with a queen is more likely. There is one combination of each, but king and queen can arrive in either order while two kings are just two kings. Probability comes from counting outcomes that are equally likely, and with dice those are the ordered rolls: first die, second die, third die, 6 x 6 x 6 = 216 of them. Unordered combinations bundle different numbers of those outcomes, so counting combinations gives the wrong weights.

How do you count the orderings fast?

Classify each combination by its repeats. Three different faces give 3! = 6 orders, a pair gives 3 orders, one for each position of the odd die, and a triple gives 1. For 9: 1 2 6, 1 3 5 and 2 3 4 are all different, 18; 1 4 4 and 2 2 5 are pairs, 6; 3 3 3 is a triple, 1; total 25. For 10: 1 3 6, 1 4 5 and 2 3 5 give 18; 2 2 6, 2 4 4 and 3 3 4 give 9; total 27. So 10 comes up 12.5% of the time and 9 only 11.6%.

Is there a shortcut that avoids listing?

Yes: symmetry. Replacing each face x by 7 - x maps a total of t to 21 - t, so the distribution of three dice is symmetric about 10.5, and 10 and 11 are the two most likely totals, each 27/216. Anything further from 10.5, including 9, must be less likely or equal; a quick count confirms it is 25. Historically this is the question gamblers put to Galileo, who answered it by counting ordered outcomes, which is still the method.

8 cubes have 3 painted faces, 12 have 2, 6 have 1 and 1 has none; the chance the top face is painted is exactly 1/3. Corners carry three, edge middles two, face centres one, and the core none. Picking a random cube and rolling it picks a random small face out of 27 x 6 = 162. The painted ones are the big cube's surface, 6 x 9 = 54, so the probability is 54/162 = 1/3.

Where do the 8, 12, 6 and 1 come from?

Think of a Rubik's cube: its pieces are corners, edges and centres, plus a hidden core. A small cube's painted faces equal the number of outer walls it touches: a corner touches three, an edge middle two, a face centre one, the core none. A cube has 8 corners, 12 edges with one middle piece each, and 6 faces with one centre each. That accounts for 8 + 12 + 6 = 26 cubes; the 27th is the core.

Why is the roll probability exactly one third?

Do it the long way first: weight each cube type by its share of cubes and its share of painted faces. 8/27 x 3/6 + 12/27 x 2/6 + 6/27 x 1/6 + 1/27 x 0 = (24 + 24 + 6)/162 = 54/162. Then notice the shortcut: a random cube with a random face up is a uniform pick from all 162 small faces, and the painted small faces are exactly the big cube's surface, 6 x 9 = 54. That gives 1/3 without any case split, and it works for any size: an n x n x n cube gives 6n squared over 6n cubed, which is 1/n.

Interviewers use the second part to see whether you look for the structure before the arithmetic. Counting faces instead of cubes turns a four-case weighted average into one division. Say both routes: the case split proves you can count, the face count proves you can see.

110 paths avoid the block, so the probability is 110/210 = 11/21, about 52.4%. All shortest paths use 6 rights and 4 ups in some order: C(10,4) = 210. Paths through (3,2) combine 10 ways in with 10 ways out, 100 in all. Subtract, and 110 survive.

How do you count all the shortest paths?

A shortest path is a string of 10 moves with exactly 6 rights and 4 ups, like a ten-letter word made of R and U. Choosing which 4 of the 10 slots are ups fixes the path completely, so there are C(10,4) = 210 shortest paths. That is the whole sample space, and every one of those strings is equally likely by the question's rule.

Why multiply for the paths through the blocked point?

Think of a trip from home to the office with a stop at a coffee shop. If there are 10 routes to the shop and 10 routes from the shop to the office, there are 10 x 10 = 100 full trips, because every first half pairs with every second half. Paths through (3,2) split the same way: C(5,2) = 10 in and C(5,2) = 10 out, so 100 of the 210 pass through the block. Subtract and 110 remain.

The grid method in the figure is the check, and it is also what you would code. Each point's count is the count from the left plus the count from below, because the last step into any point came from one of those two neighbours. Setting the block to zero removes every path through it automatically, and the same method handles several blocks, where the subtraction formula needs inclusion and exclusion.

Does the answer change if the walker flips a coin at each step?

Yes, and interviewers often ask this next. If the walker flips a fair coin for right or up at each step, any visit to (3,2) happens on move five, and a walker still able to get there has not yet touched the top or right edge, so all five of those moves were free coin flips. The chance of standing on (3,2) after five flips is C(5,2)/2^5 = 10/32 = 5/16, so the coin-flip walker avoids the block with probability 11/16, about 68.8%, well above 11/21. Choosing uniformly among complete paths is not the same as flipping coins: conditioning on the end point (6,4) pulls paths towards the diagonal that leads there, and (3,2) sits on it. Say which model the question means before you answer.