Probability of being dealt a pair in a poker hand 
In poker, which is more likely to being dealt? A pair or two-pairs?
Solve using probability.

Attempt:
First, lets find the probability of a pair. The size of sample space is ${52 \choose 5}$. Now as for the ways to pick a pair, First we select the ranks, this can be done in ${13 \choose 1}$ ways and we want 2 suits from the 4 so this we do in ${4 \choose 2 }$ ways. Now, for the remaining three cards since one rank already been chosen we only have ${12 \choose 3}$ ways to pick the ranks of remaining three and each can be chosen in ${4 \choose 1 }$ ways (1 from each suit). Thus,
$$ P( pair) = \frac{ {13 \choose 1 }{4 \choose 2 }{12 \choose 3 } { 4 \choose 1 } }{ {52 \choose 5 }}$$
Now, for two pairs, like previous part, first pick rank ${13 \choose 1 }$ and ${4 \choose 2 }$ suits. Now, for the second pair, we now pick ${12 \choose 1}$ and ${4 \choose 2 }$ suits. Finally, the last card can be chosen in ${11 \choose 1 }4 $ ways. Thus,
$$ P( 2 \; pairs) = \frac{ {13 \choose 1 } {4 \choose 2 } { 12 \choose 1 } {4 \choose 2 } {11 \choose 1 } {4 \choose 1}  }{ { 52 \choose 5} } $$
Hence,
$$ \frac{ P(pair ) }{P(2 pairs) } = \frac{ {12 \choose 3 } }{ {12 \choose 1 } {4 \choose 2 }{ 11 \choose 1 } } \approx 20/3$$
So, getting a pair is 20/3 more likely than two pairs. IS this correct argument?
 A: You've made a couple of mistakes.  In computing the probability of a pair, you say that each of the remaining ranks can be chosen in $4$ ways.  This is correct, so you should multiply by $4^3$, not by $4$.  Also, at the end, I don't know how you computed the ratio, but that's not the answer I get.  As I said, the numerator is incorrect, so you have to recompute it anyway.
A: For one pair, the count should be
$$\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3$$
Explanation:


*

*Choose the rank for the pair:$\;\binom{13}{1}$ choices.

*Choose the two cards for that rank:$\;\binom{4}{2}$ choices.

*Choose the three non-paired ranks:$\;\binom{12}{3}$ choices.

*Choose the cards for each of those ranks:$\;\binom{4}{1}^3$ choices.


Thus, you forgot to apply the factor $\binom{4}{1}$ for each of the $3$ non-paired ranks.

For two pairs, the count should be
$$\binom{13}{2}\binom{4}{2}^2\binom{11}{1}\binom{4}{1}$$
Explanation:


*

*Choose the ranks for the two pairs:$\;\binom{13}{2}$ choices.

*Choose the two cards for each of those ranks:$\;\binom{4}{2}^2$ choices.

*Choose the non-paired rank:$\;\binom{11}{1}$ choices.

*Choose the card for that rank:$\;\binom{4}{1}$ choices.


But in your count
$$\binom{13}{1}\binom{4}{2}\binom{12}{1}\binom{4}{2}\binom{11}{1}\binom{4}{1}$$
the second pair could just as well have been the first, so you are counting each two-pair hand twice. Thus, your count is twice the correct count.

Using the correct counts, the ratio works out to ${\large{\frac{80}{9}}}\approx 8.9$.
