the odds of flopping a set

[gaming_mouse]

[ QUOTE ]


i think it should be (48 choose 2)/(50 choose 4) for the turn, which is half of your number.

[/ QUOTE ]

Yep. You're right.

gm

[Dave H.]

Hmmm...I'm always trying to make these more difficult than they are, I guess. The flop was easy. And the turn turned out to be OK too.

I can see now, based on your explanation, that the river is just C(48,3)/C(50,5)...CORRECT?

But when I was thinking about it (quads by the river) prior to seeing your explanation, I was saying to myself:

I could get one of my paired cards on the flop, miss on the turn, and get it on the river. Or I could get none of my paired cards on the flop, but get one on the turn and the river, etc. Then I started the "Are they mutually exclusive" stuff, and then I got twisted up in my shorts...This probably just comes with practice, but if you have any "mental process" that you can come up with, it would be appreciated. I realize that's a pretty nebulous request, but I guess it can't hurt to ask.

I suppose that, after seeing your formulas, what I should have said to myself (about the river) was:

a. How many 5 card combinations are there in the unseen 50 cards?

b. Two of the 5 cards must be my pocket denomination. So that leaves 48 cards to be chosen 3 ways.

[gaming_mouse]

Dave,

Yes, your river answer is correct.

As far as mental process, I think you're getting it. There's no magic formula except for seeing lots of problems solved -- after that, you'll start to see recurring patterns.

With poker calculations, I would say there are easily less than 10 "types" of problems, all requiring only basic combinatorics. Just keep messing around with problems, and they'll get easier and easier. I can already tell you know alot more than you did after your first post.

gm