4 handed i'm on btn AA 3x raise, sb calls just me and him he has 44, flop is 442. i know flopping quads is about 4000 to 1, and me getting pocket aces with him hitting quads on flop is how much more? trying to add up all the ridiculous way im getting beat lately.

You do realize that when you play enough it's 100% that you will get beat in one ridiculous way or another. And you never win in a ridiculous way?

I have to save that continuously asking these hand probability questions is wrong on so many levels.

i don't actually see a question, but i thought it might be worthwhile if you examined the "flopping quads is about 4000 to 1".

i could be wrong, but given you have a pocketpair and the villian has a pocketpair he will flop quads significantly more often then 1 time in 4000 flops. i believe even if we consider his hand to be all possible hands while you hold a pocketpair the 1 in 4000 flops is an (overestimating) rounding upwards.

Also, it's not like he needed quads to beat you. A flop of 4xx would have sufficed. Villain was about 12% to flop a set or quads. Not the stuff of miracles. You never setmine against a big pair and spike the flop?

A sick beat is what I did to someone a few weeks ago: I flop top two vs villain's top set, we get it all-in on the flop, I runner-runner quads. Mathematically the worst possible suckout in poker.

46/17296 I think, or 1 in 376. (375 to 1).

Buzz

Thanks Guys. A lot of times i over estimate what it takes.

With respect, I think this is off, although not by much. I compute it to be 1 in 408.33. I'd love to know the "easy" way to compute this using combinations and permutations, but I brute-forced it. Assuming you hold a pp...

=(2/50)*((1/49)+(48/49)*(1/48)) + {this part is if the first flop card hits your pair}
(48/50)*((2/49)*(1/48)) {this part is if the first flop card misses}

Since you seemed to use combinatorics - how did you do it? Or if I missed something, please help! Thanks. Is the difference that you removed the other pp (aces)? My number is the general case of flopping quads with a pp, other cards unknown.

Answering my own question - yes, the above answer removes the AA (or another non-matching pp) from the deck before computing flop odds. My calc is the general case of a pp flopping quads, the 376 number is assuming another player holds a pp. Would still love the combinatorial equation though.

You basically used perms without knowing it. If you combine all your numerators and denominators, it becomes:

(2*1*48 + 2*48*1 + 48*2*1) / (50*49*48)

That denominator is the same as (50 P 3), the number of ways to pick 3 cards from 50 when order matters.

That numerator is 3*48*2!. Notice that you added 3 possibilities, namely the 3 possible placements of the blank card. Instead of adding we can just multiply by 3.

Order doesn't matter for this problem, which means if we're using permutations, we need the order to cancel out on the top and bottom. Our denominator counts the order of 3 cards so it therefore counts 3! arrangements. Our numerator counts the 2! arrangements of the quads cards, and when we multiply by 3 it counts the 3 placements of the blank card, so in total it counts 3*2! = 3! arrangements and we're happy.

Since order doesn't matter, you can use combinations. With combinations you don't have to cancel anything out or count arrangements of any kind. The combo method would look like this:

C(2,2) * C(48,1) / C(50,3) = 48 / C(50,3) = 48 / [(50 P 3) / 3!] = 3! / (50*49) = 3 / (25*49)

You see, C(50,3) is (50 P 3) with the order removed, and you remove the order by dividing out the 3! ways to arrange 3 objects. That's for if you're calculating it without the luxury of something like an nCr function on a calculator. On my TI calculator I'd have typed 48/(50 nCr 3). On a basic calculator I'd first reduce everything as shown above so I could just type type 3/25/49.

Depends on what problem you're solving. I was responding to ngFTW who had written Given you have a pocketpair and Villain has a pocketpair, and there are three cards on the flop, two of the cards on the flop would have to be the other two cards the same rank as Villain's pair. And then there are 46 remaining unseen cards that could be the third card.

And if we know four cards before the flop (our pocketpair and Villain's pocketpair), then there are 17296 possible different 3 card flops.

Thus the probability Villian with a pocketpair will flop quads is 46/17296, or 1 in 376.

It's not clear to me exactly what you're computing. But it's evidently not the same thing I was computing.

For sure somebody here can help you. However, I don't think I'm the best choice to help you (since I'm not a mathematician).

But the problem you want to solve has to be well defined.

Maybe I can get you started.

Do we know what pocket pair or could it be any pocket pair?

If we know it's a particular pocket pair, like aces, there are 6 possible 2-card combinations:
AA
AA
AA
AA
AA
AA
That's all there are. There is no other combination of two aces.
AA is the same two-card combination as AA (just a different permutation).

The "formula" for that would be C(4,2).
And C(4,2)=6

You can type "4 choose 2" into google and the solution for C(4,2) will be shown. C(4,2)=6.

That's probably enough for now.

Digest the above and then ask more questions. Someone here will be able to help you.

Buzz

I solved the general case of flopping quads - not the case of flopping quads removing a pair of another rank. I sorted it in my second response. Still - helpful to understand your approach as well, as I got to it through a longer process.

Thanks also to the post two up waling through the general solutions.

You eliminated the 2 cards that match hero's pair (AA). I don't see why you did that for the third card on the flop can be any of the 48 remaining cards.

"Given you have a pocketpair and Villain has a pocketpair, and there are three cards on the flop, two of the cards on the flop would have to be the other two cards the same rank as Villain's pair. And then there are 46 remaining unseen cards that could be the third card."

52- 2 (for Hero's pair)- 2 (for Villain's pair)- 2 (for the pair on the flop) = 46.

Buzz

OK. I misread and thought you were using 46 as the remaining deck before the flop.