probability homework question

[pococurante]

OK, now that I'm working with the correct numbers...

There are 54,912 hands with three of a kind exactly (no quads or boat). Along with your trips, you have two cards of two different ranks, of which there are 12 possibilities. The problem can now be simplified down to "there are 12 options, I get 2 random picks, what are the odds I get the two I want".

Let's say you want #1 and #2. Reach into a hat and pull out a number... you have a 1/6 chance of picking a winner on your first pick. Then with your second pick, there are 11 remaining numbers and only 1 winner, so it's 1/11.

54912 x 1/6 x 1/11 = 832.

There are exactly 832 ways to be dealt trips with one higher card and one lower card in 5 card stud.

832/2598960 = 0.032% = you have a 1 in 3123.75 chance to be dealt that hand.

If AAA2K doesn't count, then it's 768 hands, 0.0029%, or 1 in 3384.

If both AAA2K and 222A3 don't count, then it's 704 hands, 0.0027%, or 1 in 3691.7.

cliffnotes: BruceZ is smart, and i r dumb