Odds of this?

[franknagaijr]

Actually, when you allow 'bleedover' in your calculations, where you have a card that could fall into either group, it gets really messy really quickly and so should be avoided.

However, there is a reasonable way to backdoor the answer as follows:

There are 48 unknown cards and C(48,4) combinations.

Non-Ace combinations = c(45,4)
One ace combinations = c(45,3) * c(3,1)
two ace combinations = c(45,2) * c(3,2)
three ace combinations = c(45,1) * c(3,3)

Add these all up and confirm you get c(48,4).

I just did, and I stand by my original answer, which is 0.15495, multiplied by 8 players to get 0.123959.

[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I originally made a mistake!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?

[Buzz]

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

[/ QUOTE ]Usually there is. The rigorous way to go from 1 to 8 is usually very tedious. I used a successive approximation method a few years ago. Been a while since I have thought about it.

Bryan Alspach, the Canadian mathematician, did something similar with "semi deals" a few years back, and his method is explained, or used to be explained on his web site. (However, I don't even know if he still has a web site).

Hopefully 0.015 is small enough so that there is not too excessive deviation by just multiplying by 8 and then rounding down, to get a rough first approximation, which is probably good enough for our purposes.

Sometimes you can figure out a clever way to set up a simulation, or several simulations to get what is needed.

<font color="green">This time, multiplying by 8 works just fine.</font>

Buzz

[bigpooch]

Huh? The multiplication by eight for nine-handed gives the
right answer since not more than one opponent can hold AAxy
given that another does. So for somebody to hold AAxy among
the other eight hands is 8 x 2970/194580 = 8 x 33/2162 =
4 x 33/1081 = 132/1081 or approximately 0.122109158.

Also, there is a small detail not mentioned: somebody could
hold AAAx which for a specific opponent happens with a
probability of 1/4324. Then, the chances of an opponent
(can't be more than one obviously) holding AAxy or AAAx is
134/1081 or about 0.123959297.

[RobNottsUk]

For Hold'em I did a long and tedious calculation to figure out the liklihood of Axs, when you hold Kxs and there is a 3-flush on the river against 9 opponents who always play Axs no matter what.

I actually worked out the right answer, that agrees with a mathematicians Hold'em Odds book. The working is based on algorithmns which aren't too easy to understand, so I'm not sure how easy they'd be to extend to Omaha.

The reason why I think you can't multiply the probability like that is that as soon as 1 player hold Annn, then it becomes very hard for a single opponent to hold the case AAnn.

Some proponents of the '2 &amp; 4' rule in Hold'em; make an error that you can count outs twice going from flop to river. In actual fact you have to sum the probabilities by doing (1 - p) where p is probability of missing on both streets. p = ((#unseen mis cards) / (#unseen cards flop))* (unseen miss cards turn) / (unseen cards turn))).

If you multiply the probability of throwing a 6 in 2 dice throws you get 1/3. Actually the right answer is (1 - 5/6 * 5/6) = 1 - 25/36; and therefore 11/36 which is less than 12/36.

Such small differences are a significant edge built!

[bigpooch]

It's not a matter that it's very hard once one opponent
holds AAxy, it's IMPOSSIBLE for another opponent to hold
a pair of aces (given you hold Axyz) if one opponent already
holds AAxy. The inclusion/exclusion calculation is right;
just that the "adjustment terms" are zero.

[RobNottsUk]

Yes, so once you've one, you can stop looking.
But, when you've found an Annn hand you have to adjust the probability and go from there.

It can't be rigorous, to make the assumption.

If you figure out all the possible Omaha hands that could be dealt with the unseen cards to 8 opponents, then you can look for AAnn, and indeed there will be many reflections, because each opponent will have same terms, given the type of hand a previous opponent has been dealt.

An nnnn hand will increase the chances of a subsequent AAnn hand for each remaining opponent.

In the Axs given K-hi flush Hold'em example, noticing all the 'reflections' in the terms was what made the calculation feasible, otherwise I would still be attempting to do it 2 years later, rather than having a worked out solution.

[Buzz]

[ QUOTE ]
Huh? The multiplication by eight for nine-handed gives the
right answer

[/ QUOTE ]Big Pooch - You're right. I was originally wrong. Sorry.

[ QUOTE ]
Also, there is a small detail not mentioned: somebody could
hold AAAx

[/ QUOTE ]If you'll read my second post above, you'll see, "Here's the chart, with trip aces included. AAAX....1*45 = 45"

Buzz

[bigpooch]

Okay, I overlooked your post about AAAx.

But multiplication by eight is correct since there is NO
CHANCE that more than one opponent can hold AA or AAA given
that one given hand is Axyz.

By the inclusion-exclusion principle, if A[j] denotes the
jth opponent has AA, then with A being the union of the
A[j] for j=1 to 8,

Pr(A) = sum(j=1 to 8)Pr(A[j]) - sum(j&lt;k)Pr(A[j] and A[k])
+sum(j&lt;k&lt;l)Pr(A[j] and A[k] and A[l]) - ...

The events (A[j] and A[k]) are empty when j&lt;k, so the
probabilities of the intersection of events are all ZERO.

Thus, we are only left with the first term. Since the
P(A[j]) are identical, the LHS is just 8 P(A[1]).