Originally Posted by

**Sang-drax**
Solving this problem by expanding the cases isn't very good. It's easy to miss a few cases and when the problem gets more complex the method becomes really impractical.

The number of ways the three cards can be chosen: **50 C 3**

This is the same as I posted before.

The number of combinations *exactly one 5 and exactly one 10*: **3 * 3 * (50-6)**

(3 fives, 3 tens and the last card could be anything but one of those six cards)

Divide these two numbers and the resulting probability is **0.0202**

This is the probability to get two-pair (not full house or trips) with the two different cards you hold on your hand after the flop (three cards).

The chances are creater because you can get two pairs in more ways, for example 6, 6, 5 in the flop and 5, 10 in your hand, but I think 2.02% answers Brain's question.

Cool, my counting method would have worked, except I was tired and counted combinations that shouldn't have been counted.

Here was my working:

Code:

Total scenarios for the next three cards:
50 * 49 * 48 == 117,600
We want (5H or 5D or 5C) and (10H or 10D or 10S) and (one other)
Possible combinations they could come out successfully:
# 5 5 (47 * 3 * 2) = 282 // whoops, invalid combination
# 5 10 (44 * 3 * 3) = 396
# 10 5 (44 * 3 * 3) = 396
# 10 10 (47 * 3 * 2) = 282 // whoops, invalid combination
_____________________________
= 1356
* 3
= 4068
10 10 10 (3 * 2 * 1) = 6 // whoops, invalid combination
5 5 5 (3 * 2 * 1) = 6 // whoops, invalid combination
10 5 5 (3 * 3 * 2) = 18
5 10 10 (3 * 3 * 2) = 18
_____________________________
= 48
(4068 + 48) / 117,600 == 0.035 == 3.5%

If you take out the "whoops", you get:

Code:

((396 + 396) * 3) / 117,600 = 0.020204 == 2.02%

If you add the combinations where you get one 10 and two fives or one five and two tens, you get:

Code:

(((396 + 396) * 3) + (18 + 18)) / 117,600 = 0.020510 == 2.05%

So, the question being a little ambiguous, I think the answer is either 2.02% or 2.05%.