Posted Wed May 04, 2005 9:45 am GMT by NickFlynn
This is probably really stupid, but I'll risk the embarrassment to get it clarified.
Here's the specific situation that got me thinking about this question, although I think it applies to other types of pot odds questions as well. You are heads up on the river against one opponent, who bets half the pot. The pot is laying you 3:1 odds on a call, so the general concept is that if you think you might be ahead 1 out of 3 times, you should call because of the positive EV.
You will lose twice, costing you $200 and win $300 once for a net of $100 (assuming you started with a $200 pot).
Ok, here's the part that throws me off a little bit - the actual breakeven point is winning once out of every FOUR times, correct? Because you will lose $300 on three misses, and get $300 on the time you were right, making it an even EV play, correct?
So, then, is it correct to assume if you are getting pot odds of X:1, then you have a positive EV play if you think you can beat (X+1):1, even by a little bit? I realize this is kind of theoretical, and you really never be in a situation where you know the odds this precisely.
I realize this is a little 2+2ish, but I'm posting it here so that I might get a reasonable answer. :D
Thanks,
- Nick
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Posted Wed May 04, 2005 11:35 am GMT by suitedaces84
Odds vs Probablility
Odds
-bad:good
Probability
-good in total
total = bad + good
so if my odds are X:Y the probability will be Y in (X+Y)
so 3:1 is the same as 1 in 4
It sounds like you have a perfect understanding of the concepts, but just didn't understand the terminology completely.
Posted Wed May 04, 2005 12:35 pm GMT by NickFlynn
Thanks.
- Nick
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