Two pair
 

I keep losing to two pair on the river . Why? Why don't I catch 2 pair? What are the odds of my opponent hitting 2 pair by the river?

Re: Two pair
 

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I keep losing to two pair on the river . Why? Why don't I catch 2 pair? What are the odds of my opponent hitting 2 pair by the river?

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maybe if you figured out how to use the search functionality to find one of the hundreds of threads that already discussed this ad nauseum you wouldn't lose to two pair as often. god has a weird sense of humor that way.

but while we're on an unrelated subject, who is the braintrust who thought it was a good idea to have CTRL-W close a browser window without checking to see if you really meant to abandon the contents of the form on the page?

Re: Two pair
 

I will take a shot at this. You have to make some assumptions that you did not give. First I will assume that you do not have a pair, and second I will assume the board does not pair.

The probability that you end up with two pair exactly hitting both of your cards is as follows.

=3*3*combin(11,3)*Power(4,3)/combin(50,5) = 4.48 %.

The second part of your question is much more difficult, you did not give enough information. Did you hit any portion of the board ie. do you have one pair. Do you want to include times that your opponent has a pair to start with. All of these will effect the probabilities of your opponents having two pair.

Assuming your opponent does not have a pair, the board does not pair, and you have not hit any portion of the board the following is the probability I got. Hopefully, someone will check my math.

One opponent>

Total hand combo's = combin(45,2) = 990
Total combo's giving a pair = combin(5,2)*9 = 90
Likely hood of one opponent pairing = 90/990 = 9.1%

This puts your opponent on a random hand at the river, obviously this is unlikely.

Re: Two pair
 

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=3*3*combin(11,3)*Power(4,3)/combin(50,5) = 4.48 %.


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....what does Power(4,3) refer to?? and what does it mean? is the numerator raised to the 4/3 power??

this is what i came up with, what did i do incorrectly?

3*3*C(44,3) / C(50,5) = 5.6%

Re: Two pair
 

Your answer would be valid but what you did was allow some of the 44 remaining cards to pair. What I did is count the number of boards that do not have a pair. I did this by choosing 3 out of the remaining 11 ranks of cards and choosing 1 out of the four cards of that rank. Power(4,3) is an excell expression of 4 raised to the third. Another way of looking at the second portion is =3*3*((11*4*10*4*9*4)/3!) That is one of 11 ranks time one of four cards * one of ten remaining ranks times one of four cards etc. I then need to devide by three factorial since I double counted several of the hands.

I hope this helps.

Cobra