Optimal Boards: Part IV

Brian Alspach

Poker Digest, Vol. 5, No. 13, June 14 - 27, 2002

We are ready to determine the two-card hand having the most boards for which the hand makes the nuts in hold'em. Might a dark horse come through? We soon shall see.

In my previous article was a proof that suited $x$-$y$ is better than offsuit $x$-$y$, and prior to that we determined the numbers for A-A. This brings us to suited aces. A hand of the form suited A-x makes the absolute nuts only by making quads, straight flush or nut flush. Let's take a quick look at how we determine the appropriate numbers which we then list in a table. We'll assume the suit is hearts for simplifying the exposition.

There are 1,081 boards of the form A-A-A-y-z. The player with A-x has the absolute nuts unless the board allows a straight flush. There are exactly 36 y-z combinations allowing a straight flush. Thus, there are 1,045 boards giving the player with quad aces the absolute nuts.

If the board is x-x-x-y-z, then the player with A-x has the absolute nuts unless y-z is a pair of larger rank than x, or the board allows a straight flush. These numbers vary with x and are easily determined.

The straight flush numbers below are not difficult to determine and we move to nut flushes, where things get more interesting. The bulk of the nut flushes for A-x of hearts arise from having three hearts on board. There can be no pair on board as the flush would not be the absolute nuts. Furthermore -- and here is the source of the surprise about to be sprung -- the three hearts on board cannot allow a straight flush.

The number of choices for the three hearts on board is $C(11,3) = 165$. Combinations such as 2-3-4, 2-3-5, etc. are forbidden since they allow straight flushes. So for A-K of hearts, there are 46 combinations of three hearts allowing straight flushes. For A-Q of hearts, there are 40 combinations allowing straight flushes. Similarly, for A-J of hearts there are 34 combinations; for A-10 of hearts there are 28; and for A-9 of hearts there are 22. Aha! Here is the first appearance of A-9 of hearts -- our dark horse.

This does not continue beyond A-9 of hearts because for A-8 of hearts, any choice of three ranks from 9,10,J,Q,K allows a straight flush. Thus, there are 26 combinations for A-8 of hearts. The striking point, though, is that A-9 of hearts has the maximum number of absolute nut flushes with three hearts on board. This pattern continues for four and five hearts on board.

ABSOLUTE NUTS


Suited ace A-K A-Q A-J A-10 A-9
straight flush 1,081 1,126 1,171 1,216 226
quads 2,099 2,084 2.069 2,054 2,048
nut flush 52,016 55,242 58,520 61,801 65,021
totals 55,196 58,452 61,760 65,071 67,295

So surprise, surprise! We have discovered that A-9 suited has more absolute nut boards than suited A-K, A-Q, A-J or A-10. The above numbers are much bigger than the numbers for A-A so that pairs are no competition. We saw above that A-8 of hearts begins to lose ground to A-9 of hearts with three hearts on board. This continues for boards with four and five hearts so that suited A-9 appears to be the winner.

If one checks suited kings, you will find the numbers diminishing there as well. The absolute nut flushes take a big drop because there must be an ace of the suit on board. The restriction gets worse for suited queens because now there must be both an ace and a king of the suit on board.

The last task to complete is to look at the numbers for relative nut boards (other players may tie) and boards that play themselves and all players win. We give only the numbers here.

RELATIVE NUTS


Suited ace A-K A-Q A-J A-10 A-9
quads 473 473 473 473 473
full houses 7,560 6,354 5,268 4,266 3,222
straights 24,000 24,000 24,000 24,000 4,650
totals 32,033 30,827 29,741 28,739 8,345

The board plays numbers are essentially the same for all the hands we are checking. There are three royal flushes, there are 33 quads with an ace kicker on board, and there are 324 ace high straights with the sole exception of A-9 suited which has 450.

You will note that A-9 suited doesn't do nearly so well for relative nut boards. This suggests we come up with some kind of weighted value. We do so as follows. For any board which yields the relative nuts for a hand, we divide by the number of potential players who can tie, and if the board plays, we multiply by 1/10.

Performing these operations gives us the following weighted values for the suited ace hands (rounded to the nearest integer): A-K (65,130), A-Q (67,783), A-9 (70,206), A-J (70,548), and A-10 (73,358).


Home | Publications | Preprints | MITACS | Poker Digest | Poker Computations | Feedback
website by the Centre for Systems Science
last updated 5 December 2002