*In Hold’em, once all five cards are on the board, the nuts will always be three queens or higher. If you’ve figured out what you think is the best possible hand and it isn’t three queens or better, then look again because you missed something.*

This morning, Mr. Multi sent me an email with his proof of this statement. With his permission to repost that email, here's today's guest post from MM on the subject:

Here's my proof of this factoid, which I believe to be true. We have a full board and a two card hand.

First, it should be apparent that the worst nut hand has to be at least a set. That's because you can hold a pocket pair and hit your set on the board. It's why set mining is so powerful--a single card on the board can give you a much better hand. And it's why we hope that the board pairs, because we understand how weak it may be.

But which set is the lowest nut hand?

Let's find a board that is as weak as possible. We need to eliminate the potential flushes, quads, full houses, and straights. Flushes are easy, there are no three cards of the same suit on the board. For quads and boats, there can be no pairs on the board. That leaves straights.

To eliminate straights, the board needs to avoid two gaps that could be filled by a hand. The 2 and 3 are the lowest cards on this board, but a following 4, 5, or 6 would allow straight possibilities (e.g., a 532xx board could be filled with an A4 or 64 hand). The next cards then must be 7 and 8 so that there are three ranks between the board cards. Similarly, the 9, T, and J would allow straight possibilities, making the next lowest card a Q. That means a Q8732 (rainbow) board cannot have a straight, a quad, or a boat possibility, leaving only a set of queens as the nuts.

Note that the Q8732 is the worst board for the trip queens, not necessarily the only board. It turns out there are 16 such boards, ranging from QJ762 down to our original Q8732. In every case the board contains a queen, a seven, and a two--each five ranks apart from the next to prevent straights. If you've ever heard the maxim that all straights must have a ten or a five, then this is the converse: there cannot be a straight if the board contains Q72. Or K72, K82, K83, those three cards each separated by at least five ranks. That's a total of 64 board where a set of kings or queens is the nut hand.

Amusingly, an ace on the board assures that the worst nut hand is at least a straight. Since the ace plays both high and low, there's always a straight possibility.

Having determined the ranges of sets as nut hands, should we concerned about remembering and using this. In short, no. We've found 64 boards where a king or queen set is the worst nut hand. There are 2,595,960 possible boards. Don't count on a set being the best hand; you're not losing much equity.

'Nuff said.

All in for now...

-Bug

Sorry, "there cannot be a straight if the board contains Q72. Or K72, K82, K83" is not correct (e.g, QJT72 and K9853).

ReplyDeleteThe best you can do is turn the statement around: "The only boards where no straight is possible contain Q72, K72, K82, or K83."