I played this afternoon at Binion's with Cliff Clavin. No, I didn't play with John Ratzenberger. I played with Cliff Clavin--or at least somebody who acted a whole lot like him. He knew everything, or thought he did. Of course, he was no more likely than the real Cliff to be right about anything, but that didn't stop him from spouting alleged "facts" about whatever topic happened to under discussion at the moment. Naturally, this included poker.
In one hand, I was in the big blind with 9-2 of clubs, which I would have thrown away if anybody had so much as coughed. But I got to see the flop for free, and it brought two more clubs, with a non-club ace. Nobody bet at it. The turn made my flush, so I led out with a $7 bet, roughly 2/3 of the pot. Cliff, two seats to my left, was the only person to call. The river neither paired the board nor put a four-flush out there, so I bet again, $20 this time. He quickly called with two pairs--aces and fives. He had paired his ace on the flop, made the nut flush draw on the turn, and caught his second pair on the river. He may not have played his hand optimally (e.g., playing A-5 offsuit from early position was a bad idea to begin with), but he certainly didn't do anything monumentally stupid. I might have gotten trapped into making those calls, too, if I had made the initial error of playing the hand at all.
Anyway, as the pot was being pushed my way, he said, "That was gutsy to be betting a flush that small. You know, when you have a flush, there is a 33% chance that somebody else has one, too."
I'm pretty conversant with the most common probability facts of hold'em, but this is not one that I can recall having heard before. Nevertheless, I know it's wrong. It's wrong because the real number is unknowable. It is unknowable because it depends on all sorts of things that can't be calculated--for example, the likelihood that any particular player likes chasing flushes enough that he will enter the pot with suited cards. It also matters whether you make your flush with three of the suit on the board or four, since it is obviously easier for an opponent to have a flush in the latter situation, needing only one card. It would matter further when you make your flush. If, e.g., you flop top pair but then happen to back into a runner-runner flush, it's presumably less likely that somebody else did the same, whereas if you flop a flush, it's more likely that an opponent with either a higher or lower flush will stick around. In real-world situations, you would also adjust your estimate of the probability of an opponent having a flush based on the betting pattern, though I suppose technically that can't be said to affect the probability of another flush being out there or not.
So I repeat: for any actual table of real opponents, the probability of an opponent also holding a flush when you do is highly dependent on (1) how and when you made the flush, and (2) the other players' propensities, which are, for all practical purposes, unknowable. It is therefore just silly to make any sort of blanket pronouncement about the likelihood of an opponent holding a flush.
We can, however, take a decent stab at knowing how likely it is that somebody else was initially dealt two cards of the flush suit. This is, I suspect, what Cliff heard at some point, and he then warped it to the very different "fact" that he so kindly shared with the table.
OK, I have 9-2 of clubs, so there are 50 cards left, including 11 more clubs. In order to get at the probability that somebody else at the table was also dealt two clubs, we're going to have to use a binomial calculation. First, let's figure out the probability that any one randomly selected opponent was given two clubs. His first card has a 11/50 chance of being a club, and if it is, his second card has a 10/49 chance of being a club. Combine those, and the probability is 0.0449.
Here we come to a wrinkle in the math. It would be easier if the probability of every opponent being dealt two clubs were independent of that of every other opponent--but it is not. If, say, the first six players all get two clubs each, it is impossible for anybody else to start with two clubs, because there is only one left in the deck. A full-out attack on this question would require a series of calculations far more complex than I think it's worth, adding up the probabilities with each of a bunch of different conditions for who does or does not get clubs. But the binomial calculation should give us a good estimate because, as you'll soon see, the numbers drop off so fast that we really don't need to consider the tiny odds of more than two opponents also having been dealt two clubs each.
When dealing with binomials, this is my favorite site for both explaining how they work and providing a nifty, easy-to-use calculator. I'm using p=0.0449 and q=0.9551. Here is a table of the results:
Interpretation: At a ten-handed table (nine opponents), the probability of exactly one opponent also having been dealt two clubs, given that I was, is about 28%. The probability that one or more opponents were all dealt two clubs is about 34% (though that's where the estimation error comes in; the real number would be lower, because once we know that one opponent has two clubs, the probability of another player also getting two clubs drops, because there are fewer clubs left in the deck). As you can see, the probability of four players (i.e., you and three opponents) all having been dealt two cards of the same suit is negligibly small.
But we're not done analyzing Cliff's claim yet. The above calculations were done as if we were still before the flop, when all that we knew was that we had suited cards. Once we're to the river in a hand such as the one that I played this afternoon, we have more information to factor in. Now I know that three more clubs and two more non-clubs were unavailable to be dealt to the players at the start of the hand. That changes the p in the binomial equation. Specifically, we can now say that the probability of a randomly selected opponent having been dealt two clubs is 8/45 (probability of his first card being a club) x 7/44 (conditional probability of his second card being a club, given that his first card was a club), which is 0.0283, and, correspondingly, q (which is just 1-p) becomes 0.9717.
Plugging that value into the binomial calculator, the same table as above changes to this:
So here's the conclusion: At a nine-handed table (like Binion's uses), if on the river you have a flush made with exactly three board cards plus the suited cards in your hand, the probability that one or more opponents started the hand with two cards of the same suit is about 21% (and remember that this is a slight overestimate, because once one opponent has been given two clubs, there are fewer left for other players to be dealt).
Given that information, it is up to you to determine whether any such opponent actually stayed in the hand until the end, given the betting sequence, and whether his flush is higher or lower than yours. (Of course, those questions are not independent of each other, since players are more likely to stick with higher suited cards than lower ones.) As a gross ballpark figure, it seems reasonable to guess that even in a limped pot, half of all suited starting hands got thrown away before the flop. If that is approximately correct, then the probability of there being one or more opponents who caught the flush draw on the flop falls to the vicinity of 10%.
But Cliff insisted that I was bold to be betting such a relatively weak made flush, because of his inflexible and incorrect conviction that there was a 33% chance that somebody else also had a flush when I put my money in. I doubt that the guy is capable of grasping in how many ways he was wrong.