Join Date: Aug 2011
Posts: 5,086
For now I'll give a quick upper bound:
4(4C2) * {(26C3)(13C5)² + 26(13C6)² + 2[(26C2)(13C6)(13C5) + 26(13C7)(13C5) + (13C7)(13C6) + (13C8)(13C5)]} / (52C13) = 22.46%
That overcounts the chance of more than one shared flush, which is small of course, so the estimate is already good (I say call it 22% and call it a day). Edit to add: I can picture an exact solution (eg starting with the above and correcting it via inclusion-exclusion), but it would require a lot of work which I probably won't feel like doing. So instead, I'll add some explanation to the above.
The part in { } is the number of ways a specific two players can share a specific flush, say clubs. The denominator is all 52C13 ways the clubs can be distributed. In the numerator, the term (26C3)(13C5)² is the number of ways to pick 5 places for the clubs within one player's 13 cards, 5 places in the other player's, and 3 places within the other two players' cards. We have to separately add the case of a 6-flush vs 5-flush, 6&6 and so on. Any asymmetrical matchup must be multiplied by 2 because either player can have the 6-flush or 5-flush.
Since any two players can have the flushes, we multiply all that by 4C2. If the question were about a specific suit, that would give the exact answer. Since it can be any suit, we also must multiply everything by 4 (but now we've overcounted).
Last edited by heehaww; 11-27-2024 at 04:17 PM.