Set to Set on unpaired board

[sfwusc]

Any know this?

[gaming_mouse]

What is your question? We can't read minds. Does "set to set on unpaired board" mean "What is the chance that two people both have a set on an unpaired board?" How many players total are there? Do we assume that any player dealt a pocket pair will play it and see it to the river? You need to be specific with your questions.

gm

[sfwusc]

Sorry, this is what happens when you ask a question while 4 tabling [img]/images/graemlins/smile.gif[/img]

What are the odds of two sets on an unpaired board?

SFWUSC

[gaming_mouse]

Still not enough information.

Also, I would guess that what you really want to know is: Given that you have a set on an unpaired board, what is the chance that at least one other player does too?

Note that this is quite different from the question you are asking -- the chance (before any cards are dealt) that we are about to deal a round in which 2 or more players eventually make a set.

In addition, to do the calculation I think you want done (the "given that you have a set" one), we have to assume a random distribution of hands for our opponents. If you are encountering heavy betting on the flop and turn, this assumption crumbles and it becomes much more likely that you are up against a set. Even worse, it is impossible to quantify exactly how much weight to give the information of opponent betting.

The other question (the "before any cards are dealt" one) is easier to answer using pure calculation -- or I should say, it makes more sense to use pure math to answer that question.

So again, you need to be very specific about what you want to know. I'll be happy to answer your question -- you just first need to figure out exactly what the question is.

gm

[slickpoppa]

[ QUOTE ]
I'll be happy to answer your question -- you just first need to figure out exactly what the question is.


[/ QUOTE ]
You want anything else, your majesty?

[gaming_mouse]

[ QUOTE ]
[ QUOTE ]
I'll be happy to answer your question -- you just first need to figure out exactly what the question is.


[/ QUOTE ]
You want anything else, your majesty?

[/ QUOTE ]

Dude, slickpoppa: Chill out. His question is vague. I'm telling him I can't answer it until he clarifies. I didn't mean to sound like I had attitude.

Cheers,
gm

[slickpoppa]

i was kidding man

[gaming_mouse]

In that case, it's kind of funny. Hard to read tone sometimes...

I should have known, though, when it reminded me of the Simpson's episode where Homer tries a new bar and asks for a clean glass. If you were quoting that, I apologize doubly.

gm

[boondockst]

fine, i'll ask it LoL

I hold 33 and it's 3-handed.....flop is 3 Q 9 rainbow...the guy in fact had 99....what are the odds?

[gaming_mouse]

[ QUOTE ]
fine, i'll ask it LoL

I hold 33 and it's 3-handed.....flop is 3 Q 9 rainbow...the guy in fact had 99....what are the odds?

[/ QUOTE ]

The following is a slight cheat, but will give you a very accurate approximation to the odds, and the method can be extended to more players.

There are 47 cards left. Your opponents could have any one of (47 choose 2) = 1081 possible hands. 3 of those hands are pocket nines, 3 of them are pocket queens. I'll assume there was no preflop raise, so that we can fairly rule out queens. In that case the chance that one of your 2 opponents holds nines is:

1 - (1078/1081)^2 = .0055, about 1 in 181.

gm