I Need Help With Calculating The Number of Possible Two-of-a-Kind Poker Hands

rayroshi

New member
Joined
Mar 14, 2011
Messages
11
I am doing something wrong, when trying to calculate the number of hands that can be drawn, when dealing two pairs in a poker game, but can't see what it is, so I would appreciate being shown where my mistake is and--most importantly--WHY I am wrong, because I just can't see anything wrong with the way that I am doing it. My erroneous answer is exactly double what it should be, so I must be counting things twice.

Here is what I have (incorrectly) tried:

a. For the first pair, I choose one rank/denomination, then two suites: (13C1)(4C2)
b. Then, for the second pair, I again choose one rank/denomination from the remaining 12, then two suites: (12C1)(4C2)
c. Then, for the fifth card, I choose one rank/denomination from any of the remaining ranks/denominations, then one suite: (11C1)(4C1)
d. Finally, turning the crank, I get [(13C1)(4C2)][(12C1)(4C2)][(11C1)(4C1)] = (13)(6)(12)(6)(11)(4) = 247,104.

But the right way to do it is to choose both of the pairs' ranks/denominations simultaneously, like this:

[(13C2)(4C2)(4C2)][(11C1)(4C1)] = (78)(6)(6)(11)(4), which = 123,552.

Why do I have to choose both ranks/denominations at the same time, to start? What is wrong with my way of doing it?

By the way, this method which I am trying works correctly when calculating the number of ways of dealing a full house, so why doesn't it work here?

Any help would be greatly appreciated.
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
3,326
I am doing something wrong, when trying to calculate the number of hands that can be drawn, when dealing two pairs in a poker game, but can't see what it is, so I would appreciate being shown where my mistake is and--most importantly--WHY I am wrong, because I just can't see anything wrong with the way that I am doing it. My erroneous answer is exactly double what it should be, so I must be counting things twice.

Here is what I have (incorrectly) tried:

a. For the first pair, I choose one rank/denomination, then two suites: (13C1)(4C2)
b. Then, for the second pair, I again choose one rank/denomination from the remaining 12, then two suites: (12C1)(4C2)
c. Then, for the fifth card, I choose one rank/denomination from any of the remaining ranks/denominations, then one suite: (11C1)(4C1)
d. Finally, turning the crank, I get [(13C1)(4C2)][(12C1)(4C2)][(11C1)(4C1)] = (13)(6)(12)(6)(11)(4) = 247,104.

But the right way to do it is to choose both of the pairs' ranks/denominations simultaneously, like this:

[(13C2)(4C2)(4C2)][(11C1)(4C1)] = (78)(6)(6)(11)(4), which = 123,552.

Why do I have to choose both ranks/denominations at the same time, to start? What is wrong with my way of doing it?

By the way, this method which I am trying works correctly when calculating the number of ways of dealing a full house, so why doesn't it work here?

Any help would be greatly appreciated.
1st you picked one denomination and then you picked a 2nd. Suppose you 1st picked 7 and then you picked K. Alternatively you could have picked the 1st denomination to be a K and the 2nd to be a 7. You see your method is double counting. However if you pick 2 different denomination from the 13 denominations available you have no double counting.

Note that using your method you did not have any trouble with picking the lone denomination card after picking the 2 denominations for the pairs (you had trouble with the way you picked the two pairs!). This is why you had no trouble with the full house. You picked one denomination for the 3 of a kind and one denomination for the 2 of a kind.
 
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