TITLE: How to efficiently calculate probabilities of not getting two pair?
QUESTION [0 upvotes]: Suppose two guys are playing 2 card draw poker. I am trying to find the probability of neither players drawing a pair.
Let $C_1,C_2,C_3,C_4$ denote the cards drawn. 
We know that any card will do for the first card. We also can see 
$$\Bbb P(C_1\sim C_2) = \frac{3}{51}= \frac{1}{17}$$
For the first player's second card ($C_3$), I set up the problem as 
$$\Bbb P (C_ 3\sim C_ 1) = \Bbb P (C_3\sim C_1|C_2\sim C_1) \Bbb P (C_2 \sim C_1)+  \Bbb P (C_3\sim C_1|C_2\nsim C_1) \Bbb P (C_2 \nsim C_1) $$
$$\Bbb P (C_ 3\sim C_ 1) = \frac{3}{51} \frac{2}{50} + \frac{48}{51}\frac{3}{50}= \frac{1}{17}$$
I did something similar for Player 2 and got the same number. So the likelihood of each player not drawing a pair is $\frac{16}{17}$ and multiplying together gives us $\frac{256}{289}$
Assuming I did it correctly, this felt very clumsy and was very tedious to write out. Does anyone know a faster way I can do this?

REPLY [0 votes]: Alternatively, you can use a complementary counting method. Let us count the ways either A or B (or both) gets a pair. There are $13 * \dbinom{4}{2} * \dbinom{50}{2} = 95500$ ways that A can get a pair. This is the same for B, by symmetry, so we simply multiply that result by $2.$ The number of ways that A and B both get a pair is just $13 * \dbinom{4}{2} * (12 * \dbinom{4}{2} + 1) = 5694.$ The total count is $2 * 95500 - 5694 = 185406.$ The total number of ways to deal the cards is just $\dbinom{52}{2} * \dbinom{50}{2} = 1624350.$ The successful ways is then $1624350 - 185406 = 1438944$ (we subtract because this is the complementary count). Our answer is $\frac{1438944}{1624350} = \frac{18448}{20825},$ which is approximately $88.6\%.$
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