Hello,Mathematicians! How\'s My thinking On this one??
 

Let's assume that I "read" my opponent as having one of the following hands: i(i=1,N). Assume that the probability of winning with each of his hand respectively is Wi.

Then the probability of his winning with one of the "N" hands is (1/N)(SUM(Wi))
Any flaws in my thinking?? [img]/forums/images/icons/shocked.gif[/img]
Just wondering,
SittingBull

Re: Hello,Mathematicians! How\'s My thinking On this one??
 

No.

Assuming that each of his N hands are equally likely. Other wise it would be a a weighted average.

Re: Hello,AC! Is there enough info to...
 

arrive at the correct solution? If so, state the formula;if not,what additional info would u need? Elaborate a little on the idea of "weighted average" .I'm trying to determine whether I'm a 'money favorite' against one of "N" possible heads-up hands when each one of the "N" hands has a certain probability of beating my hand on "S" Str. in a 7-stud high game given the pot odds as P to 1 on "S" Str.
Happy pokering, [img]/forums/images/icons/laugh.gif[/img]
SittingBull

Re: Hello,AC! Is there enough info to...
 

Your question was: Any flaws in my thinking?
My answer was: No

Weighted avg is multiplying the probability that a hand will win by the prob that the person holds that hand. Which is of course an extension to your example.

For each percent of the time a particular hand multiply by the chance that he actually holds that particular hand. Divide by num of hand possibilities (ensure that the sum of all the probabilities of holding each particular hand is 100%).

If Pi is the probability of holding a hand that has a Wi chance of winning then the over all chance of winning is

Sum(pi * Wi)/N
assuming that sum(Pi)=1.0.

Sorry,AC! Thought u stated my thinking was flawed! n/m
 

Re: Hello,Mathematicians! How\'s My thinking On this one??
 

As sitting bull implied, each of the N hands is probably not equally likely, so you may want to adjust by using a weighted average.

For example, suppose you know that your opponent has either AA, AK or KK. There are 6 different ways of being dealt AA, six different ways of being dealt KK, but 16 ways of being dealt AK (for a total of 28 different hands he could have). So instead of just dividing your sum by 1/n your probability would be (6/28)*P(KK) + (6/28)*P(AA) + (16/28)*P(AK).

Note: if you consider each of the ways of getting AA's, KK's, etc. as being a different hand, then this would break back down to your formula, since having any one particular hand (ranks and suits specified) would be just as likely as any other (assuming no knowledge of other cards out).

Re: Hello,AC! Is there enough info to...
 

An example would be something like You hold AJc and flop Flop AcJsTs ... based on your read of you opponent, you put him on either AA (1 way), KK (6 ways), QQ(6 ways) or AK (8 ways). Each holding has its own chance of winning, but his chances of holding each hand are not the same so you would need to adjust it for the likelyhood of the hand (The AKs should be considered independantly of the other AK hands for completeness).

I'll leave the actual math for the experts, but hopefully that makes some sence.