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Help would be much appreciated!

In a poker game five cards are dealt at random from an ordinary deck of 52 playing cards. Find the probability of: “Two pairs”, where the number of face value of the pairs are distinct, and the remaining card has a different number or face value than either of the two pairs.

My solution: $((4/52)*(3/51)*(4/50)*(3/49)*(44/48))*13*nCr(???)$ Not sure if my approach is right.

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  • $\begingroup$ Well, the pattern is $AABBC$ where $A,B,C$ are distinct. Warning: there is a symmetry here in that swapping $A,B$ doesn't change the hand. You have to take that into account. $\endgroup$ – lulu Sep 14 '17 at 14:47
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The probability to have "two pairs" in a hand of $5$ cards should be $$\frac{\overbrace{\binom{13}{2}}^{\mbox{values of the pairs}}\cdot \overbrace{\binom{4}{2}^2}^{\mbox{suits of the pairs}}\cdot \overbrace{(52-2\cdot 4)}^{\mbox{remaining card}}}{\binom{52}{5}}= \frac{198}{4165}\approx 0.047539$$

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Not sure if this is correct, but this would be my approach:

First I "choose" the value of the two pairs, that can be done in 13 nCr 2 ways. Then i "choose" which of the four cards of each chosen value I will have, that can be done in 4 nCr 2 * 4 nCr 2 ways. Then the last card can be "chosen" in 44 nCr 1 ways. That leaves us with:

13 nCr 2 * 4 nCr 2 * 4 nCr 2 * 44 nCr 1 possible hands with exactly two distinct pairs. If you divide that with the total hands, 52 nCr 5, you should get the probability you are looking for.

I couldn't really understand your approach, but there are often multiple ways to solve problems like this.

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If you wanted to know the probability that the first two cards form a pair, the second two form a different pair, and the fifth card is different from both, that would be $\frac 3{51}\times\frac{48}{50}\times\frac{3}{49}\times \frac{44}{48}$, since the second card has to be one of 3 that pairs with the first, the third card has to not be one of the remaining 2, the fourth has to be one of the 3 that pair with the third, and the fifth card has to not be one of the 4 that match either of the pairs.

However, this is only one pattern. How many patterns are there? The unpaired card can be any of 5 options. Once you've decided this, there are 3 ways to pair off the other four cards - the first of those can go with any of the 3 others, and that determines everything. So you need to multiply the probability above by $15$.

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