How to efficiently calculate probabilities of not getting two pair?

Question

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?

Answer 1

The total number of ways of dealing four cards where order matters is $52\cdot 51\cdot 50\cdot 49$.

Let us instead count the number of ways that at least one of the players gets a pair. To do this, we will approach via inclusion-exclusion.

For the first person to get a pair and for us to not care what the second person gets (a pair or not), pick the first card. The second card must match the first. The remaining two cards can be anything. There are then $52\cdot 3\cdot 50\cdot 49$ number of ways that the first person can get a pair ignoring whether or not the second person also gets a pair.

Similarly, there will be $52\cdot 3\cdot 50\cdot 49$ number of ways that the second person can get a pair ignoring whether or not the first person also gets a pair by symmetry.

Now, the way that both players get a pair is either both players' numbers are different or the same. $52\cdot 3\cdot 48\cdot 3$ number of ways for the first case and $52\cdot 3\cdot 2\cdot 1$ for the second case.

Let $A$ be the event that the first player gets a pair, $B$ be the event that the second player gets a pair. The number of ways that at least one player gets a pair is $|A\cup B|=|A|+|B|-|A\cap B| = 2\cdot 52\cdot 3\cdot 50\cdot 49 - 52\cdot 3\cdot 48\cdot 3-52\cdot 3\cdot 2\cdot 1 = 741624$

Dividing by the number of ways of dealing, we have the probability that at least one gets a pair is $\frac{741624}{52\cdot 51\cdot 50\cdot 49}=\frac{2377}{20825}\approx 0.114$

The probability then that neither player gets a pair is $1-\frac{741624}{52\cdot 51\cdot 50\cdot 49}\approx 0.886$

Answer 2

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 $88.6\%.$