Doyle Brunson is wrong

I've been reading Doyle Brunson's newest book, My 50 Most Memorable Hands. In it he tells one story illustrating how winning became harder after publishing his famous book, Super/System. Doyle has K-K, Lyle Berman has 4-4 in a pot that has been raised and re-raised before the flop. The flop comes A-9-4, rainbow.

I checked. Lyle bet and I moved in on him, a very big bet. Lyle studied his
hand a moment and then called. The turn and river cards were no help; and as
Lyle was dragging in the pot, I said in a sad voice, "Didn't you think I might
have three aces?"

Lyle stood up and screamed, "Doyle, I read your book!"

After a moment I realized what he was talking about. I say in my no-limit
hold'em section, "I never give a free card when an ace comes on the flop because
unless the board is paired, the next card can always make a straight." So Lyle
knew I didn't have three aces. We have laughed about that many times
since.

Good story, but I got hung up on this assertion: "[W]hen an ace comes on the flop...unless the board is paired, the next card can always make a straight."

At first I thought he was claiming that with an unpaired ace-high flop, a straight will always be possible with any turn card that comes. That's clearly not true. For example, the flop could be A-2-6, and if the turn card is anything other than a 3, 4, or 5, it is not possible for anybody to have a straight.

So if I were wording Doyle's assertion to be more unambiguous, I would put it like this: Assuming an unpaired board, when the flop contains an ace, there is always at least one card that could come on the turn that could make somebody a straight. For example, if the flop is A-6-9, it's not possible for anybody to have flopped a straight, but a 10 on the turn will make a straight for a player holding 7-8; an 8 will make a straight for a player holding 5-7 or 7-10, etc. I was skeptical of this at first, but I played around with a deck for a while and convinced myself that it really is true.

Then I wondered whether the same was also true when there was not an ace on the flop. Turns out that it's not--but it's almost true. If I've counted correctly, there are only four possible unpaired flops for which no turn card can make a possible straight for any player: specifically, 2-7-Q, 2-7-K, 2-8-K, and 3-8-K.

Ignoring suits, there are 66 possible unpaired flops containing an ace.* As stated above, for all 66 of those, there is at least one card that could come on the turn that makes a straight possible if a player is holding just the right down cards.

There are 220 possible unpaired flops not containing an ace.** Of those, 216 (i.e., all except for the four I listed above) share that same quality--that there is at least one card that could come on the turn that makes a straight possible if a player is holding just the right down cards.

Put another way, with an ace on the flop, it is 100% certain that an opponent holding just the right hole cards could make a straight with just the right turn card. Without an ace on the flop, it is "only" 98.2% certain (216/220 = 0.982).

In other words, I can't see any reason that Doyle makes a distinction here. It is virtually always the case that the turn card could make an opponent's straight, whether or not there is an ace on the flop.

Now, this isn't to say that one should slow-play a strong hand on an ace-high flop and risk giving an opponent a free card. As with most things in poker, it all depends on the specifics of the situation: the presence of possible flush draws, one's position, the aggressiveness of the opponent(s), one's guess as to the possible range of cards an opponent might be playing, the size of the pot, the size of the players' stacks, and a zillion other factors. But it makes no sense at all to make a blanket rule against checking the flop just because there is an ace sitting out there and one fears a possible straight being made on the turn. That always has to be a consideration (except for when you have one of the four peculiar flops I listed previously), but there is no good reason to make it the dominant, determining factor above all others.

Sorry, Doyle. You're right about a whole lot of things, and you've probably forgotten more about poker than I'll ever know, but on this one specific point you're making a distinction where there is no meaningful difference (i.e., flop with an ace versus flop without an ace). I look forward to your reply in the comments section, since I'm certain you're a daily reader. (Ha!)

Incidentally, I figured out something else along the way, which I will probably never have occasion to put into a blog post again, so I'm throwing it in here as a free added bonus. I wondered how many full boards of five unpaired community cards there were for which no straight was possible. I couldn't find any systematic, mathematical way to determine this, so I had to work it out the long, hard way, with a bunch of cards laid out on the table and pen and paper in hand. If I counted correctly, there are 31 such boards with an ace (e.g., A-2-6-9-J), and 50 without an ace (e.g., 3-5-8-J-K), for a total of 81. There are 1287 possible unpaired 5-card boards (treating all suits as the same, as I have everywhere in this post).*** So of all the full, unpaired Texas Hold'em (or Omaha, for that matter) boards, 1206/1287, or 93.7%, have a straight possible, given the right hole cards. You always wondered that, didn't you? So now you know.



*Showing my work, as endless school teachers always told me I had to do: If the second card is a K, that leaves 11 different possible ranks for the third card (2 through Q). If the second card is a Q, there are 10 different possible ranks for the third card (2 through J). And so on, down to the case in which the second card is a 3, leaving only a 2 to make an unpaired flop. By this way of counting, the second card can't be a 2, because all of those possibilities were accounted for with a higher-ranked second card. (We're not concerned with the order the cards are in here; I just arbitrarily did the counting assuming the cards came in descending order of rank.) Finally, 11+10+...+1 = 66.

**Please, please don't make me show my work on this one! I'll just shorten it up to say that there are 55 unpaired K-high flops, 45 Q-high, 36 J-high, 28 10-high, 21 9-high, 15 8-high, 10 7-high, 6 6-high, 3 5-high, and 1 4-high. The sum of those is 220. Again, the order of the cards does not matter, nor do their suits.

*** I'll approach this one somewhat differently, because I think it's easier when the numbers get this high. Again ignoring suits, the number of unpaired boards is 13 x 12 x 11 x 10 x 9 = 154,440. But that counts every different order, so it vastly overcounts what we're interested in here. Each unique set of five cards (e.g., K-J-9-5-2) can have 120 different orders, so since we don't care about the order, the number of different unpaired boards for current purposes is 154,440/120 = 1287. I felt a bit unsure that I had this right, so I actually double-checked it by adding up the possibilities the same way as the previous calculations, and got the same number, so I'm confident about it. I.e., there are 495 unpaired A-high boards, 330 K-high boards, etc., down to 5 7-high boards and 1 6-high board, adding up to 1287.