A-A vs. K-K on hand #1

You may remember a few years ago at the World Series of Poker on the very first hand of the Main Event Sammy Farha had A-10, and another guy (seems to me it was a lesser-known Hollywood actor, but now I'm forgetting who it was; I guess that's because he's lesser known!) had 10-10. When the flop came A-A-10, it was pretty much inevitable that they'd get all the chips in.

This year, Card Player columnist Matt Lessinger lasted only about 25 minutes before running into the bad side of a set-over-set situation against Johnny Chan, and getting knocked out.

This kind of thing has made me wonder many times how likely it is that somebody in the vast field of the Main Event is unlucky enough to be on the ugly side of A-A versus K-K on the very first hand. My inclination, without running any numbers, has been to think that it probably does happen in a typical year. Today, thanks to another bout of insomnia, we're going to find out.

We start with the hardest calculation already done for us. The probability of two specific pairs being dealt turns out to be a much more complicated calculation than one would think, but fortunately for the math-impaired among us, Brian Alspach worked it all out years ago and posted his work here. If you jump all the way to the bottom table, you can just read the answer: If one person is dealt K-K, the probability that he is facing an opponent with A-A at a nine-handed table is 0.0439.

So what is the probability that somebody has K-K? That's not too hard. If you draw two cards at random from the deck, the probability that the first one will be a king is 4/52, or 1/13. If that worked, then the probability that the second one is also a king is 3/51, or 1/17. Combine those, and the answer is 1/221. Using a binomial probability calculator, such as the one here, we can quickly discover that the probability of somebody at a nine-handed table being dealt K-K on any given hand is about 0.04.

Therefore, the combined probability that somebody has K-K and somebody else has A-A is 0.04 x 0.0439, or about 0.00176.

At this year's Main Event there were 7319 contestants, or just over 813 tables with nine players each. Yes, I know that they didn't all play on the same day, and not everybody showed up for the first hand, and they didn't have a perfect arrangement of all tables being exactly nine-handed except for one. But let's pretend that all of those things were true. They're close enough for my purposes, which is just to get an estimate of the overall probability.

Turning again to the binomial calculator, we plug in n=813, k=1, and p=0.00176. The computer instantly tells us that the probability that exactly one table in the field has K-K versus A-A is 0.34. The probability that at least one table has that confrontation is 0.76.

The conclusion, then, is that my hunch was right. In a field of this size, the odds are about 3:1 that at one or more tables, right after "Shuffle up and deal," somebody is facing his worst nightmare. Of course, that doesn't mean that the two players necessarily get all their chips in, nor that the guy with kings loses and goes home. Maybe Mr. K-K can find a fold after about the fifth pre-flop raise, or when an ace flops. Or maybe he gets lucky and hits a set to felt the guy with aces. What happens after the deal is much harder to figure out mathematically.

But I'm setting aside all of the consequential action. All I care about is this fact: In about three years out of four, on average, at least one person will be so unlucky as to be looking down at kings on his very first hand when somebody else at his table is looking down at aces. One way or another, that's gonna leave a mark.