Mike Caro recently posted an interesting riddle on his blog: "What's the weakest hold'em hand you can have and still end up on the river with an absolute certainty of victory?"
Saturday, February 12, 2011
I got the answer right, but it took some thinking. Unfortunately, Caro doesn't explain the answer in detail, so I thought I would. Spoiler below, so don't scroll down until you've thought about it all you want to.
The answer is three queens--specifically, two queens in your hand on a board that has these four characteristics: Queen high, unpaired, no three of any one suit, and no straight possible. E.g., Qs-Jd-7c-5c-2h.
Here's why. The answer can't be any one-pair or two-pair hand, because three of a kind is always possible on any hold'em board, so you can never be sure that even top two pair is good. There are some boards on which nothing better than three of a kind is possible--if there is no pair (hence no quads or full house possible), no three of any suit (hence no flush possible), and the ranks are far enough apart that no two hole cards can bridge the gaps so as to make a straight. So the correct answer must be trips of some sort.
Why queens, specifically? Well, whatever set we're talking about has to be made with the highest card on the board; otherwise, obviously, an opponent could have the higher set. So why not three jacks, say? This is the tricky bit. It's because on a jack-high board, the remaining cards will always have a possible straight somewhere. You can try constructing a board to avoid it, but you will fail. Start with a jack, then put in a ten. Now we can't add a 9, because that makes three different straights possible. We can't add an 8 because either 7-9 or Q-9 would make a straight. We can't add a 7 because 8-9 would make a straight. So our next card on the board has to be a 6. We now have J-10-6. But you'll quickly discover that any two final baby cards we put down will make a straight possible.
Incidentally, once when I apparently had too much spare time on my hands, I sat down with a deck of cards and worked out how many different 5-card, unpaired hold'em boards could be made on which no straight would be possible. I counted 81 of them, which is only about 6% of the 1287 possible unpaired hold'em boards. See here.
Here's another hold'em riddle that I heard from a guy named Steve that I used to play with frequently at the Hilton (though I'm sure it was not original with him): If you start with a 5-10 in your hand, there is one kind of straight that you will never be able to make, no matter what cards come on the board. What is it?
I'll put the answer in as the first comment.