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Freeze-Out PricesDuring poker tournaments, it frequently occurs that there is much more betting on the outcome when it gets down to two or three players. It also happens that the players themselves may make some settlement so as not to be gambling for the full prize. (In other words, it is agreed that the eventual winner will pay the eventual loser something. This is known as a saver.) The problem is how to figure the chances of each of the players winning. Obviously with equal chips, the best poker player has the best chance. But, what if the chips aren’t equal? To show how this problem is figured, I will assume that the players are equally skillful. I will also assume that there are just two players, though the method will work for any number. Let’s say that two players are playing some type of basic poker game (it does not have to be poker). One player has twelve chips and the other has four chips and they are betting a chip at a time. They agree to play until one of the them is broke. What are the chances that the twelve-chip player will win the freeze-out? One way to think of this is to imagine that these two players play 1,000 such freeze-outs. How many of these freeze-outs should each player win? The key to this problem is the fact that it is fair game (such as flipping coins for one chip at a time). Because it is a fair game it necessarily means that the players must break even in the long run. Neither can show a long run profit if there is no edge. Now, if they play 1,000 freeze-outs, how is it that the players would break even chipwise? It can only happen if the poker player with the twelve chips busts the player with the four chips 750 times while he gets busted himself 250 times. This result breaks them even. Thus, we see that in a freeze-out of a fair game between two players with unequal chips, the chances of each player winning are exactly in proportion to the number of chips he has. If he has twice the chips of his opponent, he has twice the chance of winning. This generalizes to more than two players also as long as they play until one player gets all the money. Each player’s chance of winning pot is the same as that fraction of the total chips that he has in front of him. (In an actual poker game there is a slight deviation from this result because players with short stacks can go all in, but the effect is negligible.) Now that we see how to establish the odds in a freeze-out among two or more equal players, how do we alter them if the players are not equal? The answer, of course, is to adjust for the ability of the players by taking the fair game pot odds as the base and then adding something to the best player’s chances while subtracting from the others. For instance, with equal chips the best player may be sixty percent to win. Likewise, he may be even money with forty percent of the chips. When you do make the adjustment it is important to keep one thing in mid: The higher they are playing in proportion to the chips, the smaller the edge for the better player, and therefore the smaller the adjustment. After all, as they get more hands to play, the better it is for the best player. If however, they are playing $ 500-$ 1,000 and have only $ 10,000 between them, you’d better not stray far from the chips ratio if you are betting the favorite. @Copyright 2005-06 All Rights Reserved www.poker.tj |
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