Dumb Noob Question (Trips in the Hole)

[kolotoure]

AAA3 is a favorite over any non-AA isn't it?

[bbartlog]

No, AAA3 is a dog to A2TJds (even if the AAA3 is suited, so long as the AAA3 is not suited in one of the same suits as the A2TJds...). Very tiny dog tho.

[T50_Omaha8]

bbartlog:
[ QUOTE ]
BTW, your trips calculation (5%) has to be off.

I don't believe so. Odds of a particular unpaired card in your hand getting matched = (1 - ((45/48)*(44/47)*(43/46)*(42/45)*(41/44))) = 28.7%. Odds of one of the four remaining cards on the board being one of the remaining two cards of that rank = (1 - ((45/47)*(44/46)*(43/45)*(42/44))) = 16.5%.
28.7% * 16.5% = 4.7%.


[/ QUOTE ]
This is an incorrect computation. You made a mistake in calculating the conditional probability of the second nine occuring.

It is rather difficult to use conditional probability to look at the probability of multiple events occuring on a five card board. For example, say the fifth and final card is the only nine on the board, then your first condition:

"Odds of a particular unpaired card in your hand getting matched = (1 - ((45/48)*(44/47)*(43/46)*(42/45)*(41/44))) = 28.7%."

Is true, so you would move on to the next condition to calcualte the probability of having trips. But the chances of having trip nines on such a board is of course zero.

It is much easier to use combinatorics to answer the question:
You have a four card hand, of which one card is a nine. The are three remaining nines in the deck and 45 remaining non-nines. The sample space of all possible boards is (48c5).

Our calculation is then
(3c2)*(45c3)/(48c5)
= (3)*(45*44*43/3*2)/(48*47*46*45*44/5*4*3*2)
= .02486123 approximately, or nearly 2.5%.

This ignores the (45c2)/(48c5) = .0005782 = .058% chance you get quad nines.

[ QUOTE ]
You can't just use (2.5% * 4) as your chance of getting trips with an unpaired hand by the river, though.

[/ QUOTE ]
True, the conditional probability is more complicated than that. But I'm guessing it works in the other direction than you assumed it does. My intuition tells me the conditional probablity of getting, say, trip threes given you don't get trip nines is higher than the probability of getting trip nines, since you KNOW that fewer than two nines are going to be on the board, which I believe carries positive information about getting trip threes.

I don't feel like doing the math to prove this, however.

[franknagaijr]

Do you have a link to the 101 level of this math (XcY)/(ZcQ)? I've been grinding stuff out on spreadsheets, but I'm clearly missing some elements.

Thanks,

[Split Suit]

http://en.wikipedia.org/wiki/Poker_p...ty_%28Omaha%29

this shud get u started with Omaha math...

[bbartlog]

Aha! Thanks for correcting that, I was wondering whether I had gone wrong - guess I didn't think about it enough. That explains things.

[T50_Omaha8]

[ QUOTE ]
Do you have a link to the 101 level of this math (XcY)/(ZcQ)? I've been grinding stuff out on spreadsheets, but I'm clearly missing some elements.

Thanks,

[/ QUOTE ]
I actually don't know of any. I learned how to do this stuff in probability class (which was absurdly difficult, way beyond counting and such) so I'm not sure of a good web link. That wikipedia omaha article doesn't say much about how actual combinatoric arguments work, and if you look up general math concepts on wikipedia it's very hard to understand them. Sorry.

If you have any direct questions I'll answer them as well as I can though.