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I'm trying to determine the game 5 card poker when order is matter all my trails ended with fail except the one pair. The way I did one pair assuming it is $\{k,k,x,y,z\}$

$^{13}P_1 \times ^4P_2 \times ^{12}P_3 \times (^4P_3)^3$

Taking 1 permutation out of 13 cards and then multiply it with how many suit permutation for the first 2. in this case we will have the possible hands of the first two cards with order. Then by multiply it with the second part of taking 3 different cards from the rest (12 cards) including the 4 different suits I got the right number. 13,178,880

Now by applying the same rule to Two-pair $\{kk,dd,x\}$ using two different ways.

1- $^{13}P_1 \times ^4P_2 \times ^{12}P_1 \times ^4P_2 \times ^{11}P_1 \times ^4P_1 = 988,416$

2- $^{13}P_2 \times(^4P_2)^2 \times ^{11}P_1 \times ^4P_1 = 988,416$

the result is wrong. Considering the permutation between the three group which means multiply the result by $^3P_3$ = 6 is equal to 5,930,496 is still wrong the correct result based on the book is 14,826,240.

My trials stopped just here waiting to find a way and the two find the answer for the rest of the game.

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  • $\begingroup$ Sorry, the book says there are more ways to get two pair then there are to get one pair? That seems wrong. $\endgroup$ – lulu Mar 28 '16 at 13:27
  • $\begingroup$ As a (minor) variant on your method, I would count your particular two pair $AABBC$ as $52\times 3\times 48 \times 3 \times 44=988,416$. (no condition on first card, second can then only be one of three, third can be anything of different rank, fourth is one of three, fifth must have a new rank). Same answer you got. $\endgroup$ – lulu Mar 28 '16 at 13:31
  • $\begingroup$ the question is talking about order so it might change the answer if we consider AABBC, BBAAC, BBCA, ...... $\endgroup$ – Feras Mar 28 '16 at 13:42
  • $\begingroup$ There are $\frac {5!}{2!\;2!}=30$ ways to permute $AABBC$ . Of course switching $A$ and $B$ does not give us a new configuration, so we need only consider $15$ of these. Thus your final answer is $15*988416=14,826,240$ (as the book reports). I am still very surprised that this is greater than the number of single pair hands...but I haven't checked that calculation. $\endgroup$ – lulu Mar 28 '16 at 13:49
  • $\begingroup$ Oh, Ok. I misread your calculation. For single pair hands: Starting with the pattern $AABCD$ we have $52\times 3\times 48 \times 44 \times 40=13,178,880$ But that's just one pattern. I thought you were saying that was the final answer. There are $\binom 52 = 10$ possible patterns (according to where the $AA$ goes) so the total number is ten times that. That makes more sense. $\endgroup$ – lulu Mar 28 '16 at 13:53
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The calculation that got you 988416 is correct. However, there should be $\displaystyle \frac{5!}{2! \cdot 2! \cdot 2!}$ possible permutations for $\{kk,dd,x\}$.

$5!$ is the number of permutations for 5 cards, and the three $2!$'s in the denominator represents the numbers of permutations of the 2 $k$'s, the 2 $d$'s, as well as $kk$ and $dd$.

$$ 988416 \times \frac{5!}{2! \cdot 2! \cdot 2!} = 14826240 $$

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  • $\begingroup$ Now I got it, you applied the order partitions rule on the result to get the possible permutation from the main possible results. but the last factorial $2!$ shouldn't be $1!$ $\endgroup$ – Feras Mar 29 '16 at 8:36

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