08-26-2005, 11:50 AM
Yesterday aces_dad and I were discussing a hand and we disagreed about the best course of action. The hand is not really relevant, but something that really struck me was that I seemed to be using different calculations than he and another poster were using to figure out which course of action was correct. Here is the scenario, simplified for the sake of discussion:
On the river, there are 3 big bets in the pot, it is heads up, and you are 100% sure you are behind. There is no rake. If you bet, there is some chance that your opponent will fold. If he calls, he will win at showdown, and if he raises, you will fold, because you hold garbage. What percent chance must there be that your opponent will fold to make betting correct?
I say the breakeven point is 25%. If you bet, there will be 4 big bets in the pot, so you need a 25% chance of winning to make risking one bet correct. aces_dad disagrees. His argument is as follows:
[ QUOTE ]
Pot odds are determined by the size of the pot before we add our bet, not after. You can't add your bet before doing the calculations. So in this example, assuming we're behind and the pot is exactly 3BB, then we're trying to win a 3BB bet with 1 bet, or 3:1. Hence we need to win this bet 1/3 times to make it exactly break even getting 3:1.
Mathmatically the Expected value of such a bet would look like:
EV = (pot size) * (win %) = 3 * ( x )
For EV = 0, the required x = 1/3.
Our exact breakeven point here is 1/3; we need to make this 1 bet 3 times, to get the 3 back we invested.
If we only win 1/4, we need to make this bet 4 times, to win 3. Hence if it only works 1/4 we actually lose .25BB every time we make this play.
I'm having trouble describing it but basically, you don't really 'win' the last bet you put in, you just get it back, so it can't be considered part of the pot when you make your calculations.
[/ QUOTE ]
This argument is intuitively very appealing. It is precisely how I thought about the issue before I read SSHE about a month ago. But when I was reading SSHE I noticed that my calculations of the necessary odds to call vs. theirs were always off by one bet. I eventually realized that the discrepancy is caused by the one bet I would be putting in by calling. That is, you do 'win' the one bet you put in the pot if you end up winning the hand. I thought of a long, contrived analogy to demonstrate this point, but I'm pretty sure this post is long enough already. So suffice it to say that I think a dollar in the pot is a dollar in the pot, no matter how it got there.
I'd be much obliged if my poker superiors would care to comment on this matter.
On the river, there are 3 big bets in the pot, it is heads up, and you are 100% sure you are behind. There is no rake. If you bet, there is some chance that your opponent will fold. If he calls, he will win at showdown, and if he raises, you will fold, because you hold garbage. What percent chance must there be that your opponent will fold to make betting correct?
I say the breakeven point is 25%. If you bet, there will be 4 big bets in the pot, so you need a 25% chance of winning to make risking one bet correct. aces_dad disagrees. His argument is as follows:
[ QUOTE ]
Pot odds are determined by the size of the pot before we add our bet, not after. You can't add your bet before doing the calculations. So in this example, assuming we're behind and the pot is exactly 3BB, then we're trying to win a 3BB bet with 1 bet, or 3:1. Hence we need to win this bet 1/3 times to make it exactly break even getting 3:1.
Mathmatically the Expected value of such a bet would look like:
EV = (pot size) * (win %) = 3 * ( x )
For EV = 0, the required x = 1/3.
Our exact breakeven point here is 1/3; we need to make this 1 bet 3 times, to get the 3 back we invested.
If we only win 1/4, we need to make this bet 4 times, to win 3. Hence if it only works 1/4 we actually lose .25BB every time we make this play.
I'm having trouble describing it but basically, you don't really 'win' the last bet you put in, you just get it back, so it can't be considered part of the pot when you make your calculations.
[/ QUOTE ]
This argument is intuitively very appealing. It is precisely how I thought about the issue before I read SSHE about a month ago. But when I was reading SSHE I noticed that my calculations of the necessary odds to call vs. theirs were always off by one bet. I eventually realized that the discrepancy is caused by the one bet I would be putting in by calling. That is, you do 'win' the one bet you put in the pot if you end up winning the hand. I thought of a long, contrived analogy to demonstrate this point, but I'm pretty sure this post is long enough already. So suffice it to say that I think a dollar in the pot is a dollar in the pot, no matter how it got there.
I'd be much obliged if my poker superiors would care to comment on this matter.