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View Full Version : "Expected Value" in HOH


bigmatt25
07-29-2005, 02:29 PM
As I'm rereading HOH Vol. 1, I'm having trouble understanding "expected value." In talking about figuring out your implied odds, he says, "To be more precise in your calculation multiply the size of your intended bet by the probability he will call it. The resulting 'expected value' of his call should then be added to the pot to get your implied odd." Can anyone give me an example of this? I'm just not sure how the math would work out.

AZnuts
07-29-2005, 08:07 PM
I'm not sure exactly what you mean without a specific example from the book, but how about this . . .

First, if you were faced with simply calling a river bet, that's straight forward: If the current pot is $50, then a bet of $25, you have pot odds of $75/$25, 3-1 on your call. If you repeated the same situation over-and-over, the odds would still be the same.


Now, if you're doing the betting in the same situation, the so-called pot odds are different because you're not sure if it will be called. Assuming you have the winner, your expectation will be between not getting called (just the $50 pot) and getting called ($75 net, $50 pot + $25 your opponent puts in). To figure the expected value over many occurences, you need to multiply that bet amount by how often it will get called.

In this case, if you figure an opponent to call it 50% of the time, your net expected value for that move is $50 + ($25)x(.5) = $62.5.

Depending on what part of the book you are talking about, there may be a lot more to it. You don't know if you have the best hand (unless it's the nuts), so the real calculation has to also include the probability that you've won.

If you have a page number from the book, we can do better.