Odds of this?

macawboy · #1 10-08-2007, 11:11 AM

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 · #2 10-08-2007, 11:26 AM

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 · #3 10-08-2007, 12:49 PM

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 · #4 10-09-2007, 03:09 AM

Hi Macaw Boy - Frank Nagai Jr. is correct.

Buzz

brian64 · #5 10-09-2007, 07:09 AM

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 · #6 10-09-2007, 07:28 AM

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 · #7 10-09-2007, 08:01 AM

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>

RobNottsUk · #8 10-09-2007, 08:16 AM

The c(3,2) term is 3, from 6 permutations or 3 ways of holding AA when you hold an Ace. 3x2/2

Then your c(45,2) is the number of Hold'em hands. But if you take off 2 A's, and cards from 52, shouldn't that be 46?

52,51,50,49,48,47 leaves 46..1

Worked fair bit on Omaha card distributions at one point, and breaking it up into Hold'em hands, and possible 3 card flops etc, usually helped considerably.

franknagaijr · #9 10-09-2007, 09:09 AM

The 45 number was arrived at as follows:

OP had an Ace, plus 3 non-aces, which means there are 48 unknown cards including 3 aces, or 45 non-ace cards. There is one group of combos with 3 aces and one non-ace that I've accounted for, so any other combo we deal with is a mix of 2 of the 3 aces, and the other 45 unknown cards.

I was thinking more about it, and I don't see why my original number is wrong. Do I really have a permutation when I put two distinct combinations side by side?

To take the area where I am in doubt in micro form, assume I have 3 kings and 3 aces only to fill four slots, with precisely two aces and two kings. Is the correct number c(3,2) * c(3,2) , or is the correct number that same number divided by 2 to account for permutations?

Of the six cards in play, there are 15 combos - c(6,4), of which 3 are AAAK, 3 are KKKA, and the remainder must be 9, or c(3,2) * c(3,2).

Which means that I stand by my original number until Buzz can show me where I made the error.

RobNottsUk · #10 10-09-2007, 09:52 AM

It seemed simpler as the question is about a pair of Aces, is to declare them as 2 from 3, and then count all the rest as 46 unknowns...

But now I see, you rule out extra Ace's held in Axxx hand and AAAx in the AA hand.

So yes, if you exclude those then you are looking at 45 cards.

But I think it may be simpler, to allow the extra Aces and then subtract the illegal combos from the lot afterwards, as they are a subset.

Anyway I'm sure this is kind of more enlightening than just having Buzz write out a long explanation lol