Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Problem Poker dice is played by simultaneously rolling 5 dice. How many ways can we form "1 pair", "2 pairs"?

For one pair, I got the answer right away. First I consider there are 5 spots for 5 dices. Then I pick 2 places out of 5, which means there are 3 places left, so we have to choose 3 out of 3 which is 1 way. Hence, I have: $${{5}\choose{2}} \cdot 6 {{3}\choose{3}} \cdot 5 \cdot 4 \cdot 3 = 3600 $$ However, I couldn't figure out why I got two pairs wrong. First, I pick 2 places for the first pair, then its rank. Next, 2 places for the second pair, and its rank. Since there is only 1 place left, I pick the rank for the last dice. $${{5}\choose{2}} \cdot 6 {{3}\choose{2}} \cdot 5 \cdot 4 \cdot 3 = 3600$$ why the correct answer is 1800, which means I need to divide by a factor of 2. I guess that might be the order of two pairs can be switched, but I wonder is there a better way to count it? I'm so confused :(! Any idea?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

You’ve correctly identified the mistake: you’ve counted each hand twice, because the pairs can be chosen in either order. For instance, you’ve counted the hand $11223$ once with $22$ as the first pair and $33$ as the second pair, and once again with $33$ as the first pair and $22$ as the second pair.

Here’s a way to count that avoids that problem. First pick the two denominations that will be pairs; this can be done in $\binom62$ ways. Then pick the remaining denomination; this can be done in $4$ ways. Now choose which of the $5$ dice will show the singleton; this can be done in $5$ ways. Finally, choose which $2$ of the remaining $4$ dice will show the smaller of the two pairs; this can be done in $\binom42$ ways. The total is then $\binom62\cdot4\cdot5\cdot\binom42=1800$ ways. The key to avoiding the double counting is to choose the positions for a specific pair. Once you know where the smaller pair and the singleton are, you automatically know where the larger pair is: there’s nothing to choose.

share|improve this answer
    
Thanks a lot. I saw my problem now. In fact, I was confused this problem with counting poker hand. –  Chan Sep 14 '11 at 9:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.