Probability of a bigger pair

Ed Miller's column in the June 27 issue of Card Player magazine is about bet-sizing tells. He walks the reader through a hand that he played while holding Q-Q, showing how opponents' bet sizes helped him deduce what hands he was up against. I have no quibbles with his reasoning. In fact, it's a highly worthwhile piece, and the sort of insight that makes me never want to miss his columns.

I do, however, have a quibble with an incidental point tossed in along the way. In explaining why he elected to just call rather than reraise when faced with an opening raise from a nit under the gun, he says:

[T]here are still five unknown hands behind me. There's about a one percent chance each player has either A-A or K-K, making about a five percent chance in total that one of the two hands I'm most afraid of is lurking behind me.

(Note: this is just one of the reasons he cites for not reraising, not the only one.)

At first I wondered why he was making a distinction between the players who have already acted and those yet to act. After all, A-A or K-K is an ugly problem for Q-Q no matter where on the table it is being held. But I think I see what he means. If he were in the big blind, and thus last to act on the raise, and nobody had reraised yet, then he could quite confidently assume that nobody was holding A-A or K-K, unless it is the UTG nit.

Or put it another way: Before there is any action, there is a fixed probability that any other player has been dealt a pocket pair higher than Q-Q. But people so reliably raise and reraise with A-A or K-K that when they fail to do so, we can safely revise downward that estimated probability.

What is the a priori probability that somebody at a ten-handed table has been dealt A-A or K-K when I'm holding Q-Q? It turns out to be a ferociously complicated mathematical problem, one that I could never solve on my own. Fortunately, though, somebody a lot smarter than me has already worked it out, and you can see both the process and the result here:

http://people.math.sfu.ca/~alspach/comp35/

The table containing the solution is at the bottom of the page. The answer for Q-Q is 0.0841. That is, if I see Q-Q in my hand, and I have nine opponents, about 8.4% of the time one of them will be holding a pair bigger than mine. That does indeed work out to a round estimate of about a 1% probability that any particular one of them has the goods. And that's a useful rule of thumb. Before there is any action (action constituting additional information to throw into our estimation math), each player in a full ring game has about a 1% chance of having been dealt a pair higher than my queens.

If I'm in the big blind and nobody has raised, then probably nobody has a bigger pair, unless they're trying the ol' limp/reraise thing. Also, if I open-raise from under the gun with my Q-Q and get only callers with no reraise, I can go to the flop reasonably confident that nobody has A-A or K-K. But really, that fact has everything to do with what we know from experience about how people play, and little to do with the math. The conclusion would be equally true if the a priori probability of somebody holding a bigger pair were 80%, or 0.8%, rather than the actual 8%.

But I think there's a problem with how Ed applies the math to the situation he describes. Once a nitty player has opened for a raise from under the gun, we no longer have the baseline 8% chance that somebody at the table has A-A or K-K. The probability now has to be estimated to be vastly greater than that. Exactly how much greater depends, obviously, on exactly how nitty this particular nit is. If he is the extreme case and will only raise UTG with exactly A-A or K-K, then our previous 8% probability goes to 100%. Of course, it's not often you'll have a player with a range that narrowly definable, so it usually won't go up all the way to 100%. But it's surely a damn sight higher than 8%.

If the nit does, in fact, have A-A or K-K, then the probability that any other player at the table also has A-A or K-K is much lower than the roughly 1% apiece that we would otherwise be using as our rule of thumb. Using the next column in Alspach's table, we see that the a priori probability that two players have been dealt pocket pairs bigger than my queens is only about 0.2%--a negligible quantity. The implication is this: The fact that a nit put in an UTG raise means that the probability of a bigger pocket pair being out there is much greater than the baseline 8% (as just discussed), but also that nearly all of that probability resides in the nit, not in the players who will be acting behind me.

It is thus no longer true that each player behind us still has a 1% chance of holding aces or kings. Our estimate of the probability that any one of them has A-A or K-K must be greatly reduced from its baseline ~1% by the fact of having observed that UTG raise from the nit. Put most simply: As the probability that the UTG nit has aces or kings goes up, the probability that anybody else at the table has aces or kings goes down.

Now, this is where it gets a little weird and mind-bendy. Because two things are simultaneously true: (A) The probability that any player is dealt A-A is independent of whether any player is dealt K-K.* (B) Alspach's calculation of a 0.2% probability of two players having A-A or K-K is no longer valid once we actually know that one player has one of those hands. The 0.2% is valid only when we have no information except for having seen our own Q-Q.

This is analogous to questions such as the probability of a roulette wheel hitting 00 twice in a row. It is 1/38 x 1/38 = 0.026 x 0.026 = 0.0007. But if we spin and the ball falls in that 00 slot the first time, now the probability of seeing 00 twice in a row greatly increases, to 0.026. Of course, the probability that the second spin will be 00 has not changed one smidgen by virtue of the result of the first spin, yet the probability of seeing two 00s in a row has changed.

So it is with the A-A and K-K. If we know that the UTG player surely or almost surely has A-A or K-K, we have to throw out the window the Alspach table numbers. They no longer apply. The 8% probability (that exactly one player has A-A or K-K when I have Q-Q) is no longer valid because we have strong reason to believe that this specific shuffle of the deck did, in fact, result in one of those 8% events. And the 0.2% probability (that two players have each been dealt A-A or K-K when I have Q-Q) is no longer valid both for that reason, and for the reason that once the UTG nit is known to have A-A or K-K, two of those eight cards are no longer available to be given to other players.

The knowledge (or at least high suspicion) that UTG nit has one of the two premium pairs is akin to opening the sealed box in which Schrodinger's poor cat has been confined. It is now known to be either alive or dead, not in some hybrid or probabilistically indeterminate state.

Mind you, I don't claim to have worked out the math so as to be able to tell you exactly what the new, revised probability is that each of the five players yet to act has a pair bigger than Q-Q, in addition to the one that we presume is hiding under the nit's card protector, but it is definitely not well estimated by the 1% rule of thumb. It must be substantially lower than that. Exactly how much lower depends entirely on the degree of confidence with which we can assign A-A or K-K to the open-raising nit.

*Note: I realize that that is not quite true, because as soon as two aces are in a player's hand and no longer in the deck from which the dealer is pitching cards, the concentration of kings in the deck rises slightly by virtue of card subtraction, so the probability of K-K goes up slightly. But it's a tiny difference.