In Texas Hold'em poker, what odds should I take into consideration when I bet or call?

Answers

Answer #1 | 22/12 2013 10:02

>> "I have to consider the overall chances of my hand improving, or should I only call based on the chances that my hand improves until the next round."
If you think he'll check on the next round, or better yet if the current bet puts one of you all-in (so there can be no more betting later), then you use the overall chances.
But if you're expecting to face another bet on the next round, you only use the chances of improving by then. These are lower chances but it also means you have implied odds: if you hit you can get more money in, if you miss you can get away cheaply (though this is more realistic with a small pair than KK unless you're absolutely sure he has AA). So you would use the chance of flopping a set to determine the combined (pot+implied) odds you'd need, then decide if you had the right implied odds or not. If he has AA then your implied odds will tend to be high.
That answers your question but I also notice you made some math mistakes which I'll point out.
>> "If I consider the entire future board of 5 cards, then the chances of me hitting a K of two Ks left out of 5 tries out of 48 cards left is 5*2/48, which is 21%"
No, that's an over-estimate, it double-counts the ways you can make quads. You can start with that but then you'd have to subtract as follows:
5(2/48) - (5C2)(2C2 / 48C2) = 19.9468%
(That's an application of the inclusion-exclusion principle.)
But the simplest way is to find the chance of NOT hitting, and subtract from 100%.
1 - (46C5)/(48C5) = 19.9468%
Equivalently, strictly in terms of combinations: (48C5 - 46C5)/(48C5)
>>"I multiply that by the chances of him not getting and A"
No, not that simple, because the two events (you hitting a K and him not hitting an A) are not independent. If you were gonna multiply .199468 by something, it would have to be the conditional probability of him missing given that you hit. But that would be more work so no reason to do that.
So instead we can do it in a separate calculation as follows:
(2(44C4) + 44C3) / (48C5) = 16.63%
Or another way is to start with .199468 and subtract out the chance that you both hit.
Pr(both hit) = [(48C5) - 2(46C5) + (44C5)] / 48C5 = 3.3174%
.199468 - .033174 = 16.63%

Answer #1| 22/12 2013 10:02