Odds of this?

[macawboy]

About a couple of months ago i asked the odds of being dealt AAxx and thanks to all that contributed to the answer (about 1 in 40).
Suppose i am in a full ring game(9 players)under the gun i am dealt Axxx can any one tell me the odds of comeing up against AAxx?My first thought was about 1 in 5.
40 divided by the 8 players left but because i hold one of the Aces surely this can,t be right?
Any formulas to work this one out?

[franknagaijr]

You have Axxx, so there are 48 unknown cards in the deck. There are 3 aces in play. For any given player, there are
c(48,4) combinations available to them.
There are 1 *45 ways to get three aces.

There are 3 ways to get two aces * c(45,2) ways the other two cards can come.

Multiply this times the number of hands being dealt to figure out the odds.

I came up with a 1.55% chance of any given player getting this, multiplied by 8 for about a 12.4% chance of somebody getting two aces.

[macawboy]

franknagaijr,
Thanks for takeing the time to work this one out much appreciated.
12.4% thats about 1 in 8 of running into AAxx.

[Buzz]

Hi Macaw Boy - Frank Nagai Jr. is correct.

Buzz

[brian64]

If n is the number of other players, then I think it should be c(n,1)*c(3,2)*c(46,2)/c(48,4)= n*1.6% assuming that you're including AAAx.

c(n,1)= choose the player who gets AA
c(3,2)= choose the Aces he gets
c(46,2)= choose the other 2 cards he gets from the 46 remaining after taking his 2 aces from the deck
c(48,4)= total # of possible hands from 48 cards

Buzz- would really like to see your reasoning if you think it is different

[RobNottsUk]

I've worked this out before, but frustratingly the forum it's on was terminated. Whilst I haven't looked this up and check, I think I remember the basics, so quickly...

The easiest way is to realise as mentioned that there's 3 ways of being dealt AA. Now you simply multiply 3x Number of Hold'em hand combinations possible from unseen cards. I think Buzz gets the divide by 2, because 46x45x44! / (44! x 2) = 46x45/2 is the formula for working 2 card combo's out. Number ways of combining 2 things.

On the flop, you can reduce the number of Hold'em hands possible, as you've seen more cards; an Axx flop, makes AAxx less likely, but if the betting pattern strongly suggests AAxx and not other hands fitting with it, then you should use Bayesian probability, and not statistics of card dealing to inform your decision.

[franknagaijr]

Assuming Buzz is correct (easy assumption) I think my mistake above was to not divide the (c(3,2) * c(45,2) ) number by 2, which means that I was dealing with a permutation that I needed to turn into a combination.

However, the coffee hasn't sunk in, so I'll just take your word for it.

<font color="green">Sorry, Frank. I was wrong this time. - Buzz</font>

[Buzz]

Hi Brian - If Hero was dealt AceBCD, then BCD are known and there are 48 unknown cards.

Then if one opponent, who we'll call Villain, has a pair of aces (or trip aces):

If Villain has trip aces, there's only one way he can have them and 45 ways he can have one of the other 45 cards.

Of if Villain has a pair of aces and two other cards, there are C(45,2) ways he can have the other two unknown cards (2 cards out of 45 cards).

Trip aces doesn't amount to much, and you can argue that Villain wouldn't play a hand with trips anyhow. Or if you include it, fine. No big deal one way or the other. (You get about 6% either way).

But let's include it, just to show that you have to make a chart for Villain.

Here's the chart, with trip aces included.<ul type="square">AAAX....1*45 = 45
AAYZ....C(3,2)*C45,2) = 2970[/list]then combining, 2970+45=3015

And then 3015/C(48,4) = 3015/194580 = 0.0155.

That's a small enough number that we can roughly use the following approximation: if the probability any one individual opponent has a pair of aces is 0.0155, then the probability any one of eight opponents has a pair of aces is eight times as much. Usually you can't simply multiply the probability an individual has something by the number of opponents, but in the particular case (and some others) you can.

I originally made a mistake! Sorry!

Buzz

[RobNottsUk]

After last post, I thought about this some more, and remembered what I'd done.

Where 1 player has Annn and wants to know probability of an opponent having AAnn where n is a non-Ace card.

c(3,2) AA pairs. 3! / 2! = 3
c(45,2) Hold'em hands without an Ace. = 45! / 43! * (45-43)!
= 45x44 / 2 = 990

Therefore 3 x 990 = 2,970 AAnn hands.

That's a proportion of all the 4 card opponent possible hands, which may or may not include Aces.

c(48,4) = 48! / (48-4)! * (4!) = 48!/(44! * 4!)

48x47x46x45 / 24 = 4,669,920 / 24 = 194,580

2970 / 194,580 = 0.0152636

Using small probability approximation, and multiplying by 8, on a 9 seat table; my calcuation is 12.2%.

So my working after a few mistakes along the way (and not enough coffee), comes out with the Buzz answer. Hopefully those unfamiliar with combinations and the formula can follow that better with more working shown.


Now, obviously there is an issue on the fallacious multiplication of the probability to go from 1 to 8 opponents.

Can we figure out the number of combinations of dealing 8 Omaha hands, and then seeing how frequently 1 of them has AAnn?

Are the numbers just gonna be too big to handle?