Optimal Boards: Part III

Brian Alspach

Poker Digest, Vol. 5, No. 12, May 31 - June 13, 2002

In Parts I and II of this series, we determined the number of boards making pocket aces the nuts. Recall that we are using the term ``absolute nuts'' to mean the player has a winning hand that no other player may even tie; ``relative nuts'' to mean the player cannot lose, but other players may tie; ``the board plays'' to mean all players left in the hand tie; and ``the hand is vulnerable'' to mean the board produces a situation for which there is a potential for another player to win.

We now are going to consider player's hands that are not pairs. Our goal is to find the beginning hold'em hand having the best chance of being the nuts after the board is dealt. As is frequently the case, my initial thoughts about the problem were overly simplistic. There are some subtle features and surprising aspects not immediately evident. In fact, this problem is turning out to be a lot of fun.

For the rest of this article, ranks $x$ and $y$ are distinct. Let me now pose a question for you to ponder. Which hand will have more boards making it the nuts: $x$-$y$ offsuit or suited $x$-$y$?

I expect most or all of you are going to answer suited $x$-$y$ based on good intuition. Let me even tell you that you are correct, but now I am going to ask you to prove it.

This may produce some sputtering, and I realize saying something fifty times or repeating it louder may work for a politician, but that is not a proof. So let's go ahead and prove the statement that suited $x$-$y$ has more boards making it the nuts than $x$-$y$ offsuit.

We don't want to have to count the numbers of boards for all possible choices of $x$ and $y$. Instead, we are going to introduce a clever, but common, way of producing a proof. How, you ask?

For each board that makes $x$-$y$ offsuit not vulnerable, we find a corresponding board making suited $x$-$y$ not vulnerable. We then show the corresponding boards for suited $x$-$y$ have no duplicates. This means, so far, we have just as many boards for suited $x$-$y$. We finish the proof by finding one more board for suited $x$-$y$ so that we know there are more of them.

For specificity, let's suppose the two hands are the $x$ of hearts with the $y$ of diamonds, where $x$ is a higher rank than $y$, and the $x$-$y$ of hearts. First consider boards making $x$-$y$ offsuit not vulnerable, where the best hand allowed by the board is quads, a full house or a straight.

If such a board does not contain the $y$ of hearts, then use the same board for the $x$-$y$ of hearts. Think about it a little and you will see that $x$-$y$ of hearts has exactly the same relationship to this board that $x$ of hearts, $y$ of diamonds has.

If the board $B$ for $x$-$y$ offsuit contains the $y$ of hearts, then form a board for $x$-$y$ of hearts by replacing every diamond in $B$ with a heart of the same rank, and replacing every heart in $B$ with a diamond of the same rank.

Again, a little reflection will lead you to conclude this new board has the same relationship with $x$-$y$ of hearts that $B$ has with $x$ of hearts, $y$ of diamonds.

The only other kinds of boards making either of the two hands in question not vulnerable, are boards that give at least one of the hands a flush. It seems apparent that $x$-$y$ of hearts should have more boards making a nut flush than $x$-$y$ offsuit simply because four suited cards are needed for the latter, while only three hearts are needed for the former. In spite of this being apparent, we need to be more careful here than what we did earlier.

The $x$ of hearts and $y$ of diamonds may end up with either a heart flush or a diamond flush. If the board produces a nut heart flush and the $y$ of hearts is not in the board, then use the same board for the $x$-$y$ of hearts. Clearly, this board gives us a nut heart flush (it may improve to a straight flush or the board may be a royal flush and play) for the $x$-$y$ of hearts.

If the board produces a nut heart flush and the $y$ of hearts is in the board, then replace the $y$ of hearts in the board with the $y$ of diamonds. This new board clearly maintains the nut flush characteristics for the $x$-$y$ of hearts.

The boards we have produced so far do not overlap because none of the boards in the first collection have the $y$ of diamonds, whereas, those in the second collection all have the $y$ of diamonds.

Now consider a board $B$ with four diamonds and the $y$ of hearts so that the player has a nut diamond flush. The corresponding board we use for the $x$-$y$ of hearts is obtained by making a devious change to $B$. First, suppose the nut hand is a straight flush. This means the four diamonds on board together with the $y$ of diamonds in the player's hand produce the best possible straight flush. In this case, we change the four diamonds to the corresponding hearts, and change the $y$ of hearts to the $y$ of spades.

Under the preceding alterations, it is clear the player with $x$-$y$ of hearts now has the nut straight flush. The problem is that we must make certain this board does not duplicate one already appearing. It has no $y$ of diamonds in it, so it does not duplicate any of those with the $y$ of diamonds. On the other hand, those without the $y$ of diamonds were obtained by not altering the board at all. Thus, if it duplicates one of those, then the four hearts that make a nut straight flush with the $y$ of hearts are in the board for the $x$-$y$ offsuit hand. So some other player could have the $y$ of hearts, that is, the $x$-$y$ offsuit hand is vulnerable. So the board was never there in the first place.

A board with four diamonds, the $y$ of hearts, and making a nut flush for the $y$ of diamonds that is not a straight flush is altered in a different way. We change the three diamonds of largest ranks to the corresponding hearts, change the smallest diamond to a spade, and change the $y$ of hearts to the $y$ of spades. Clearly, this board makes the $x$-$y$ of hearts the nut flush. These boards are distinct because they are the first collection with only three hearts.

The boards with four or five diamonds and no $y$ of hearts are altered in the same way as those we just did. When the $y$ of diamonds produces the nut straight flush, we convert all diamonds to hearts and leave the others unaltered. When the $y$ of diamonds produces the nut flush that is not a straight flush, we convert the three diamonds of biggest ranks to the corresponding hearts, the remaining diamonds to spades of the same rank, and any other cards unaltered.

It is easy to see all the boards make $x$-$y$ of hearts the nuts. None of the new boards duplicate any of the preceding boards using arguments similar to what we did above.

The above scheme has established a correspondence between all the boards for which $x$-$y$ offsuit is a nut flush and some of the boards for which $x$-$y$ of hearts is a nut flush. All we need is one extra board for the latter hand. We obtain it by choosing a set of three hearts which gives us a nut flush together with $x$-$y$ of hearts and adding two clubs that do not pair the board. This board clearly does not duplicate any of the others.

We now have proved there are more boards making suited $x$-$y$ the nuts than there are boards for $x$-$y$ offsuit.

In Part I, I wrote that there would be three parts, but the discussion has expanded as hidden facets have emerged. In the next part, we shall consider the ranks of the suited cards.

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