This is a brilliant solution.whatshisbucket wrote: ↑24 March 2021, 22:15The exact computation here is actually fairly simple, since we only need to consider the placement of three cards. We are placing them in 3 of 42 buckets, so there are a total of (42 choose 3) = 11480 possible placements.
14 of these buckets are in your partner's hand, so the chance of them having all three aces is (14 choose 3)/11480 = 13/410 ≈ .032
28 buckets are not. Then there are (14 choose 2)(28 choose 1) ways to put 2 aces in your partner's hand and 1 elsewhere, for a probability of 91/410 ≈ .222
Similarly there are (14 choose 1)(28 choose 2) ways to put 1 ace in your partner's hand, for a probability of 189/410 ≈ .461
And of course the probability of your partner having none is (28 choose 3)/11480 = 117/410 ≈ .285
The answer should be 4*choose(52,13)*3*choose(39,13)/choose(56,14)/choose(42,14)/(4*choose(52,13)/choose(56,14)). Far from being as elegant as yours, but it is more like a straight forward thinking.
People who knows classical probability well, either super smart, or read and think a lot. Hope you are both