DF. The Best Full House
Peter Winkler brings us a short poker puzzle, from his new collection Mathematical Mind Benders: What is the best full house?
(The answer is not three aces and two kings…)
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jlundell said,
December 29, 2007 at 11:14 am
I arrived at the answer you were looking for (aces over 6..9), but I wonder whether it’s the complete answer to the question you actually asked.
The intended question, I think, was to find the best full house against any random poker hand. The actual question, though, specified draw poker, and it seems to me at least possible that this complicates the situation. I by no means have an analysis, but I’ll suggest one direction the consideration might go.
In choosing the full house, we have two ways to distribute suits. Our aces represent three suits, and the pair can either be from those three suits, or can include the fourth suit, in which we have no ace.
Now consider our opponents, who presumably don’t know that we’ve stacked the deck (or they’d be fools to ante up), so they’re playing draw poker (per the question as asked) and trying to win the hand. It’s reasonable to assume that a flush is going to be a pat hand; that is, our opponents, playing the odds, are not going to risk busting a flush in order to improve their hand by drawing a straight flush.
In that case, knowing that we will beat any plain flush, we’d like to encourage as many pat flush hands as we can. And if we choose our pair to be in the same suit as two of our aces, rather than choosing one in the fourth suit, we leave slightly more flushes to be dealt to our opponents, leading to slightly more pat–but losing–hands (at least according to my back-of-the-envelope calculation).
This is a grossly incomplete analysis, of course, but a complete analysis seems pretty complex. Maybe everything balances out in the end and the pair suits don’t matter, but I’d hate to have to prove it.
The simplest recourse: fix the question: no draw.