A poker brainteaser

Playing at the Rio tonight, I saw a final board of Qh-Qd-Qs-9s-Ks. It occurred to me that almost every rank of poker hand was available with this unusual set of community cards. That made me wonder whether it is the maximum possible for a hold'em board.

What no player could have with this board are the lowest hand rankings: high card (i.e., no pair), one pair, or two pairs. Because of the trip queens, everybody will have a minimum of three of a kind. But every possible type of hand three of a kind or better is possible here. A 10-J in the hole yields a straight. Two spades makes a flush. There are a whole bunch of possible full houses (13, if I counted right, but don't quote me on that). A queen in the hole gives four of a kind. And 10-J in spades produces a straight flush. A royal flush is not possible here, but I don't consider that actually to be a different hand ranking; it is merely one type of straight flush.

So that makes six different ranks of hands possible with this board. The question is whether that is the most that any hold'em board could generate.

Here's another candidate board: Ah-Ad-Kh-Kd-10h. Here the worst hand any player could achieve would be two pairs, so at first glance I thought this might beat the previous board and yield seven different hand rankings. But as I scrutinized it more closely, I realized that three of a kind is not available here; any player who ends up with three cards of the same rank actually has a full house. So as with the previous board, six hand rankings are possible, though this time not all consecutive ones (because of skipping over trips).

After thinking about it for a little while, I came up with a board that surpasses both of these. If you enjoy puzzles, take some time and work it out for yourself, before you scroll down for my solution.


["Jeopardy" theme music plays here.]



















There are many similar possibilities, but the one I wrote down was Ah-Ad-Kd-Qd-Jd. If you held, say, 2-3 of clubs, you would have one pair. If you held J-2 of clubs, you would have two pairs. If you held A-2 of clubs, you would have three of a kind. If you held 10-2 of clubs, you would have a straight. If you held any diamond other than the 10, you would have a flush. There are six full houses possible. The other two aces in the hole gives you quads. And the 10d in your hand gives you a straight flush (the royal, in this case, though that's an unimportant distinction for this little game). That's every possible hand ranking except the lowest, a high card (no pair). This has to be the maximum, because in order to make a bare high card possible, you would have to have an unpaired board, but then you lose both the full house and quad possibilities, for a net loss of one ranking.

So eight is the maximum number of different hand ranks that might be available from one hold'em board.

If somebody wants to tackle the question of whether the answer is any different for Omaha, be my guest in the comments. Just thinking of that this late at night makes my head hurt.