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I have been working on the problem of probability of poker hands,

I have been able to calculate the probability of each hand except one pair and high card hand.

Here is what I have

P1 P2 X1 X2 X3(here P1 P2 are the same)

13C1 * 4C2 (total counts of one pair)

Now total counts of 3 cards distinct from the pair cards

(12C1 * 4) * (11C1 * 4) * (10C1 * 4) / ??

obviously there is repetition in the second term, How do I remove it?

According to wiki ?? valus is 6, I am not able to understand How to arrive at that value

In case of Triple hand, I did this

(13C1 * 4C3) * (12C1 * 11C1)/2 * 4C1 * 4C1

because in this case each terms is counted twice in case of Two distinct cards.

Can someone please help me out here?

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In problems of this kind one usually asks for counting hands that have a pair and not anything better. Likewise for a high-card hand, which amounts to not having any pair or better, i.e. also no flush and no straight. –  hardmath Apr 6 '13 at 18:53
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2 Answers

up vote 2 down vote accepted

Another way to look at it: There are $\binom{13}{1}$ ways to pick which type of card ($2$, $Q$, etc.) your pair will be, and $\binom{12}{3}$ ways to choose the types of your other cards (so that you will have a pair, instead of a full house, three-of-a-kind, etc.). There are $\binom{4}{2}$ ways to choose the suits of your paired cards and $\binom{4}{1}$ ways to choose the suit of any one of your $3$ non-paired cards. Hence, the answer is $$\binom{13}{1}\cdot\binom{12}{3}\cdot\binom{4}{2}\cdot\binom{4}{1}^3.$$

Note: $\binom{n}{r}$ is an alternate notation for $_nC_r$.


As for high-card hand, it's probably simpler not to calculate it directly. You say you've calculated all other hand types, so just subtract them all from the total number of possible hand types--that is, from $\binom{52}{5}$.

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Oh ok, How did I not think of it :(. Many thanks –  Dude Apr 6 '13 at 20:54
    
Any time. If it makes you feel better, I used to do it the way you did it, before someone else pointed out the obvious (in hindsight) to me. –  Cameron Buie Apr 6 '13 at 20:55
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Actually I liked wiki's calculation for the High Card, (13C5 - 10)*(4^5-4) –  Dude Apr 6 '13 at 20:59
    
Oh, that is nice! –  Cameron Buie Apr 8 '13 at 3:27
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Your calculations are right. And the value you represented with "??" is 6 indeed. It is 6 because you have $3\cdot 2\cdot 1$ ways of ordering the extra cards (and in poker, order of cards clearly does not matter).

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