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Two pair with unpaired hold cards and board not paired Two pair with unpaired hold cards and board not paired

04-10-2014 , 02:13 PM
Quote:
Originally Posted by hauturi
I guess you noticed that OP also changed his mind about what he was asking mid-thread?
No, he actually didn't. You simply missed a stipulation in the OP.

Quote:
Originally Posted by apkrnewb
Having a hard time getting this.
If i have 2 non paired cards and the board doesn't pair,
how often in 100 tries will i have 2 pair on the river?
that said, i do think that the question "how often will i pair both of my hole cards" is not the same as "How often will i have 2 pair by the river", though all the answers itt seem to answer the first question.

Last edited by Turyia; 04-10-2014 at 02:21 PM.
Two pair with unpaired hold cards and board not paired Quote
04-11-2014 , 02:59 PM
Quote:
Originally Posted by Turyia
that said, i do think that the question "how often will i pair both of my hole cards" is not the same as "How often will i have 2 pair by the river", though all the answers itt seem to answer the first question.
It depends if he means he will have 2 pair, or that he will have 2 pair.
Two pair with unpaired hold cards and board not paired Quote
04-11-2014 , 03:51 PM
Quote:
Originally Posted by David Sklansky
I can't stand it anymore.

Say the board is A2345.

There are ten different two pair.

Each two pair has nine holdem hand two card combos. So there are 90 qualifying combos.

That 90 is out of the number of two card hands that can be made from the remaining 47 non board cards. 47x46/2 which is 1081.

Except that the OP specified non pair hands. There are six combos each of sixes through kings and three each aces through fives. That is 63. Leaving 1018.
I didn't see this post before when I asked what was simpler than mine. That's the problem with linear forum mode, it's easy to miss stuff. Yours is simpler.

Yours:

9*C(5,2) / [C(47,2) - 5*C(3,2) - 8*C(4,2)]

Mine had powers of 4, but if I cheat by canceling 4^3 I had:

9*C(11,3) / [9*C(11,3) + 6*C(11,4)*4 + C(11,5)*4^2].

Of course yours as the advantage of being able to do it in your head, assuming you can do C(47,2) in your head. But wait, I can cheat some more and cancel C(11,3) to get just

9/(9 + 6*8 + 8*7/5*4) = 9/101.8 = 90/1018.

And I could probably have done those cancellations while driving. Assuming of course I didn't blow the factor of 2 as I did even when typing it out.

Last edited by BruceZ; 04-11-2014 at 04:25 PM.
Two pair with unpaired hold cards and board not paired Quote
04-11-2014 , 04:19 PM
Mine is simpler not because it is shorter but rather because it is simpler to understand.
Two pair with unpaired hold cards and board not paired Quote
04-11-2014 , 04:48 PM
Even a ******ed 8-year old could understand either solution. Conceptually they're both completely straightforward; however, yours requires one to pretend that the board was dealt first and then the hand which is less natural. There are actually people for whom that would be harder to understand. Mine follows the standard pattern of Bayes' theorem. My denominator counts the 3 cases for boards that match 0,1, or 2 cards. You have to consider 3 cases too, for total hands, hands with a pair of A-5, and hands with a pair of 6-K. Taking total hands and subtracting pairs is a little simpler than computing 0,1, and 2 matches, but that pattern of taking C(11,k)*4^k comes up so often that it's completely second nature, though granted it wouldn't be to many people, like the same ones who don't understand that you can reverse the order of the board and the hand.

My original response was prompted by statmanhal who was explicitly computing the probability that the board had 5 different ranks. Dividing 2-pair hole cards by total possible hole cards as you did is definitely better.

Last edited by BruceZ; 04-11-2014 at 08:26 PM.
Two pair with unpaired hold cards and board not paired Quote
04-16-2014 , 03:29 AM
Hello,

I am new at this so please bear with me. Can someone please dissect this equation below for me in human terms?

3*3*C(11,3)*4^3 /
[3*3*C(11,3)*4^3 + 2*3*C(11,4)*4^4 + C(11,5)*4^5]


i know that the 3*3 is from C(3,1)*C(3,1) to fill the first two cards but I don't understand the C(11,3)*4^3 expression. Why are we choosing 11 out of 3 available cards and multiplying that by 4^3...?

Also it would be nice if someone can explain the mathematical expression in the denominator for the sample space...


Thanks!
Two pair with unpaired hold cards and board not paired Quote
04-16-2014 , 04:16 AM
We are choosing 3 ranks out of 11 for the 3 board cards that aren't paired, and then there are 4 suits for each of those ranks, so 4^3. The first term in the denominator is the same as the numerator for a 2-pair.* The second term is for matching only 1 hole card which we can do 2 ways times 3 suits each, and then there are C(11,4) ways to choose the 4 ranks of the other 4 board cards, and 4^4 ways to choose their suits. The third term is for matching 0 hole cards, so there are C(11,5) ways to choose those ranks, and 4^5 ways to choose those suits.

David's method might be easier to understand.


*Hey, I'm only counting 3 things, and David's counting 4. His things are a little easier to count because the C(47,2) doesn't require considering suit, but the others do, with 3^2 in one, while the others just involve C(4,2) and C(3,2). Though his certainly seems subjectively simpler, I'm finding it hard to quantify why because the difference isn't as great as it first appears. I think it's mostly because the subtraction of the pairs can be regarded as something of an afterthought, with the rest being really simple. But you still have to do the pairs for the exact answer, and there are 2 types. They are simple, but so is what I did. Plus you have to think about the problem backwards, i.e.,they announce the keno numbers, and then we draw a random keno ticket. David likes simple solutions, and I think it's important to be able to think that way. It's also important to understand general principles like Bayes' theorem, and if you think about the problem in the forward direction, the Bayes' theorem solution is obvious. And if you wanted the probability of 1 pair or 0 pair, with mine you only have to move a different term from the denominator to the numerator. If you hit on that first, you can stop thinking. You generally stop running once you've already caught the bus. But it's always best to have a second solution as a check for anything this complex. Had I done David's solution as a check, I would have found my missing factor of 2. But David's solution is so conceptually simple that I might not even feel a need to check it. So maybe that proves it's simpler.

Last edited by BruceZ; 04-16-2014 at 02:20 PM.
Two pair with unpaired hold cards and board not paired Quote
04-19-2014 , 04:31 AM
i get it! Thanks!
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