Some probability questions for you math masters

[WarDekar]

Given 3 players that all see the showdown, what's the probability of:

Winning with 3-of-a-kind given there is only 1 of the 3 on the flop
Having the winning hand with at least 3 of a kind
The best hand in a hand is 2-pair
Winning with 2-pair given an opponent has a 4-flush on the flop

Any help would be appreciated!

[AaronBrown]

Your questions are not precise enough to answer easily. For example, the chance of winning with a pair in hand matching one card on the board depends on what the board is. If your three is the lowest card on the board the chances of winning are obviously less than if it is the highest. If the board makes a flush, your trips are worthless.

A bigger problem is the probability will be affected by what the other players stay in on. If you deal three hands and everyone stays until showdown their hands are likely to be worse than if you dealt ten hands to tight players and everyone stayed in with heavy betting.

[WarDekar]

As I said in my post, assume it's 3 random hands (although in some of the questions, your hand may not be random, but the hands are random unless stated otherwise) and they all see the showdown.

[Abbaddabba]

Is there any particular reason you'd want to know those figures?

My guess is that they'd be pretty hard to work out just because of all the conditions that need to be satisfied.

For the first, you need to figure out how often a set will occur... which is a function of how often a pocket pair is dealt. That's pretty simple in itself.

You then need to discount the times that the set fills up, the times that you make quads and the times you win with a flush despite having a set (ie: 4 to a flush on the unpaired board and you hold the winning flush despite also holding a set).

You also need to figure what percentage of the time other random hands will complete a flush, straight, or anything better as a percentage of the times one of three hands makes a set. This will be a percentage of a percentage of the time that it happens. Ie:


[the sum of the probabilities of one of three players making a set and only a set]

minus

[(the sum of the probabilities of one of three players making a set and only a set) * (the sum of the probabilities of one of three players making better than a set)]


That's the basic algorithm you'd need to follow. If you have pokerstove and a lot of time on your hands, you can probably get the answer yourself.

[AaronBrown]

[ QUOTE ]
As I said in my post, assume it's 3 random hands (although in some of the questions, your hand may not be random, but the hands are random unless stated otherwise) and they all see the showdown.

[/ QUOTE ]
I don't think you did say that, but it makes it easier to answer.

In the first question, your best chance is obviously if there are no straight and flush prossibilities, no pairs on the board and your trip is the highest card on the board. Then you can't lose.

For every card on the board that's higher that your trips, the chances that a single player has a pair of them in hand is 1 in 330. If there's a higher pair on the board, the probability that a single player has a match is 43 in 330. If there's a lower pair on the board, the probablity that a single player has quads is 1 in 990. With any single pair on the board, higher or lower than your card, the chance of a full house without a pocket pair is 14 in 990. You'll have to count the straight and flush outs yourself, they depend on the exact structure of the board.

It's not quite correct to add up all the ways to lose and multiply by 2 for two players. It overstates the risk of losing, but not by much if the answer is low. Figuring this one exactly means counting up all the cases and figuring which ones overlap.

Pokerstove will do the calculation for you if you give a specific board and your hand. So you could get exact answers for a few different cases, would that help?