Random Math / Theory Question

Assume a situation where my opponent must hold either a set or a specific pair. Assume that the odds of him being dealt the specific cards for the pair or the specific cards for the set are equal. Under these circumstances is it much more likely for my opponent to have a set than a pair if his actions are equally indicitive of either?

Edit: See my third post, it makes much more sense [img]/images/graemlins/frown.gif[/img]

[Spicymoose]

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Assume a situation where my opponent must hold either a set or a specific pair. Assume that the odds of him being dealt the specific cards for the pair or the specific cards for the set are equal. Under these circumstances is it much more likely for my opponent to have a set than a pair if his actions are equally indicitive of either?

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You have to count the combos of different hands your opponent could have. Then, if you want to, weigh down some of the possibilities if you don't think he always plays like that.

The entire point of the question is that it's made under the assumption that all permutations and other outstanding factors aside from the difference in odds to flop either hand type are equal.

[Spicymoose]

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The entire point of the question is that it's made under the assumption that all permutations and other outstanding factors are equal.

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I think your post is too general, and it depends. Please give an example that uses your assumptions, and we can figure out what is most likely.

"Does how likely my opponent is to have flopped a certain type of hand affect how likely he is to have that hand when he represents either it or another hand equally, all other factors withstanding?"

Edit: Adding an example I gave to a friend over MSN:

Like, if somone stuck a gun to your head and said "your opponent must have either 77 for a set or AK for a pair, he's equally likely to be dealt either 77 or AK (not the case in reality, but assume)" you'd choose AK and expect to be correct by a large margin?

[Guy McSucker]

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"Does how likely my opponent is to have flopped a certain hand affect how likely he is to have that hand when he represents either it or another hand equally, all other factors withstanding?"

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I don't think this makes sense, although I kind of see what you're getting at.

The point is that we can see the flop, and we can see his betting. The unknown information is his hole cards. If the flop is K 7 3 and you tell us that it's equally likely he has a pair of kings or a set of 7s, then guess what? Under those assumptions, it's equally likely he has a pair of kings or a set of 7s.

This is just because you've given us the distribution of his hole cards as being 50% 77 and 50% Kx.

The reason there's a skewing of likelihood between the set and the pair in real life is precisely because there are fewer 77 combinations that Kx combinations.

Guy.

[StellarWind]

Here is an example of how to do this:

Board = KQ52
Your hand = AQ

Our estimate of the probability that he would play his hand the way he actually played if he holds:

55 = 10%
KK = 100%
QQ = 100%
AK = 50%
Everything else: 0%

55 = 3 possible hands * 10% = 30
KK = 3 possible hands * 100% = 300
QQ = 1 possible hand * 100% = 100
AK = 9 possible hands * 50% = 450

There is only one possible QQ hand because you can already see two queens.

There are nine possible AK hands because there are three unseen aces and three unseen kings.

P(55) = 30 / (30+300+100+450) = 30/880
P(KK) = 300 / (30+300+100+450) = 300/880
P(QQ) = 100 / (30+300+100+450) = 100/880
P(AK) = 450 / (30+300+100+450) = 450/880

[Spicymoose]

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Here is an example of how to do this:

Board = KQ52
Your hand = AQ

Our estimate of the probability that he would play his hand the way he actually played if he holds:

55 = 10%
KK = 100%
QQ = 100%
AK = 50%
Everything else: 0%

55 = 3 possible hands * 10% = 30
KK = 3 possible hands * 100% = 300
QQ = 1 possible hand * 100% = 100
AK = 9 possible hands * 50% = 450

There is only one possible QQ hand because you can already see two queens.

There are nine possible AK hands because there are three unseen aces and three unseen kings.

P(55) = 30 / (30+300+100+450) = 30/880
P(KK) = 300 / (30+300+100+450) = 300/880
P(QQ) = 100 / (30+300+100+450) = 100/880
P(AK) = 450 / (30+300+100+450) = 450/880

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For anyone using PokerStove, this process can be coppied to a degree. If you think there is a 50% chance of AK, you can 4 out of the 9 AK hands from a players range.

Make sure you are careful when doing this though. For example, if there are 16 combos of AK, and you feel there is a 50% chance he is playing AK like this, you can remove any 8 combos of AK as long as there is a rainbow board. If suits matter though, make sure you don't take out any hands that are on flush draws, as the equity could change drastically. Also, always make sure to count how many total combos there are before you start removing. Although PokerStove will allow 16 combos of AK, if there is a K on board, Pokerstove will ignore the 4 combos of AK with the given king, so removing A [img]/images/graemlins/diamond.gif[/img]K [img]/images/graemlins/heart.gif[/img] when the K [img]/images/graemlins/heart.gif[/img] is on the flop, won't actually change anything.

Thanks Stellar. So if I summarized this thread by saying, "The likelyhood of what an opponent may hold post-flop is independent of the likelyhood of him making that particular hand pre-flop," and add, "The cards seen (flop, hand, etc) affect the permutations of our opponents' possible holdings," I'd basically be correct, right?