Odds of this?

[franknagaijr]

I think BigPooch was skipping my contributions, but I'll try be thick-skinned about it. I agree with Big Pooch, that the number will be exactly times 8, as opposed to a slightly smaller number.

I think I'm reinventing the wheel here, but here's a methodology I used to confirm what seemed intuitive:

1) I calced the percentage chance of the first hand having
a) No Aces 0.7657
b) One Ace 0.2187
c) Two Aces 0.1526
d) 3 aces 0.0002

2) I calced the percentage chance of same, if the first hand had zero or one aces. (I didn't bother worrying about the numbers if first hand had two or more aces, obviously)

3) I mutiplied the chance of the corresponding second hand having two or more aces times the distributions of the first hand as follows:

a) If no aces - 0.014107
b) If one ace - 0.01387604
c) If two aces - 0.152636 (100% times chance of first hand having two aces)
d) If 3 aces - 0.00231267

4) I added these chances together to come up with
0.30989824

5) I multiplied the chance of the first hand having two aces
0.30989824

6) I used the excel x=x test to confirm these were the same number.

So I expect if I extrapolate this technique 8 times, it will still come up with the answer of the shorter method, multiplying times 8.

[Buzz]

Big Pooch and Frank - I'm sorry I got started on this. I'm not a mathematician. Perhaps the reason is my math teachers were interested in proofs. I'll agree that proofs are important. But proofs never were my cup of tea. I hope I have not offended any mathematicians. I like mathematicians. I just don't want to prove anything.

And yet here I am, feeling that I need to prove something to you. And it's something I don't know how to prove.

But maybe I can make it clear to you that what you are proposing doesn't work. I'll give it a shot.

Suppose you were dealt T[img]/images/graemlins/spade.gif[/img],2[img]/images/graemlins/spade.gif[/img],3[img]/images/graemlins/heart.gif[/img],4[img]/images/graemlins/heart.gif[/img] in the unraised big blind. And also suppose the flop was
Q[img]/images/graemlins/spade.gif[/img],J[img]/images/graemlins/spade.gif[/img],9[img]/images/graemlins/spade.gif[/img].

You might be wondering if any of your eight opponents had been dealt the
A[img]/images/graemlins/spade.gif[/img] or K[img]/images/graemlins/spade.gif[/img].

Let's not worry about any of the other 43 missing cards yet.

But let's calculate the probability the small blind was dealt either of these cards. P=1-C(43,4)/C(45,4) = 0.1717
Agreed?

Now let's calculate the probability any one of your eight opponents was dealt either of these cards.
P=1-C(43,32)/C(45,32) = 0.9212
Agreed?

I realize I haven't rigorously proved anything.

But perhaps you can see that you don't get the correct probability any one of your eight opponents has the
A[img]/images/graemlins/spade.gif[/img] or K[img]/images/graemlins/spade.gif[/img] by multiplying the individual probability (0.1717) by 8.

Edit: I see, however, that multiplication by eight is correct in the particular case where Hero has AXXX and wants to know the probability an opponent (one of eight) holds AAXX. Thanks, Big Pooch.

Buzz

[bigpooch]

In this case, of course multiplication by eight won't be
right, BECAUSE it is POSSIBLE that two or more of your
opponents could have the holding in question (namely, at
least one of the As/Ks); obviously, one opponent could hold
the As and another could hold the Ks.

The reason now that multiplying by the number of opponents
does not work is that the "higher" terms in the summation
for the inclusion-exclusion equation are nonzero and a
"negative adjustment" is needed for the overcounting of
the cases when more than one opponent (exactly when two
opponents) has a key card (one of As, Ks).

On the other hand, the probability that one of your eight
opponents has both the As and Ks in his Omaha hand is
EXACTLY eight times that for the probability that a specific
opponent has both the As and Ks; that's because not more
than one opponent could have such a holding.

If it is not possible that more than one opponent can have
such a holding, the "higher terms" are zero in the
inclusion-exclusion; similarly, if at most two opponents
can have the holding in question, only the first two
summations on the RHS need to be computed.

[Buzz]

[ QUOTE ]
1) I calced the percentage chance of the first hand having
a) No Aces 0.7657
b) One Ace 0.2188
c) Two Aces 0.01526
d) 3 aces 0.00023

[/ QUOTE ]fixed your post

Hi Frank - That's from
a) No Aces 148995/194580
b) One Ace 42570/194580
c) Two Aces 2970/194580
d) 3 aces 45/194580.

Buzz

[Buzz]

Big Pooch - Sounds like you know what you're talking about more than I do. And sounds like you're very sure of yourself. I'll do some more thinking about this and then get back to this post and do some editing of my posts in this thread if I need to.

Thanks.

Buzz

[franknagaijr]

Buzz - I don't think that the situation you describe is analogous to the one we're dealing with. In your follow-up, it is fair to use the 32 cards in a group. However, in the original question, we must think of discrete (is that the correct math word) sets of four cards, and although each set of fours probabilities for having aces changes based on whether prior sets have aces or not, in the aggregate, the numbers do not change.

For fun, I took the number sets I had so far, and checked against a 3rd set of four cards, and I came up with 3 times 1.5495 as follows.

Distribution of possibilities after 2 dealt hands:
No Aces dealt: 0.57123
1 Ace dealt in either hand: 0.36077
1 ace dealt in each hand: 0.037002
2 or more aces dealt in one hand 0.3039898

Now if you take those numbers and calc the subsequent probabilities for the third hand, it looks like this

Chance of ANY hand having 2 aces or more based on prior hands
No aces in first two hands: 0.01272
1 ace only in first two hands: 0.002775
1 ace in each of first two: 0.00 (no chance)
2 or more aces in on of the hands: 0.30309898

Add those up to get 0.046485, which is exactly 3 times the base number.

[Buzz]

Hi Frank - I try to stick with what I know works, while avoiding what I'm not sure about.

I didn't think that multiplying by eight would give quite the right answer, but I can see that it does.

As long as the individual probability is very low, I'm going to use it as a rough approximation in a game anyhow, so for me, for the most part, it's moot whether it's precise or not.

But Big Pooch is talking the talk and seems very sure of himself. Must be respected.

I keep being busy primarily with something else, but I'll definitely get back to this, and maybe be able to incorporate one more concept into my ever growing grasp of probability as it applies to Omaha-8. (I think some of you are sort of in the same boat as me).

Buzz

[Buzz]

Let’s give Hero K[img]/images/graemlins/spade.gif[/img],K[img]/images/graemlins/diamond.gif[/img],J[img]/images/graemlins/club.gif[/img],T[img]/images/graemlins/club.gif[/img], and let’s make the board on the river:
A[img]/images/graemlins/club.gif[/img], K[img]/images/graemlins/heart.gif[/img],Q[img]/images/graemlins/diamond.gif[/img],J[img]/images/graemlins/diamond.gif[/img],J[img]/images/graemlins/spade.gif[/img].

An individual opponent needs two aces in his hand to win, one of the following hands:
AhAsAdX 40
AhAsXY 780
AhAdXY 780
AsAdXY 780

2380 is total

2380/123410 = 0.01928531
0.01928531*8 = 0.15428248

Now we simulate, and If *8 works, Hero should win about 84.6%

Indeed, as simulated, (100,000 runs) Hero wins 84.6%

Good enough for me.

I'll go back and make the necessary corrections in my previous posts.

Thanks for making me do it right, Big Pooch.

And Frank, you were right all along.

Buzz

[franknagaijr]

[ QUOTE ]

And Frank, you were right all along.

Buzz

[/ QUOTE ]

Somewhere in the world, a three-legged blind pig has just found an acorn.

[RobNottsUk]

The problem with the argument that the approximation is exact, is shouldn't you be using conditional probability?

In the calculation at this step :

"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

You can say, that I have AA, now how many nn hands are there.

But you're taking out the Ace in the deck, to use c(45,2), because it is true that only one player can be dealt AA.
But with other opponents an Annn hand does not preclude an AAnn holding; so the probability of the 2nd opponent holding AAnn is conditional on the 1st opponents holding.

When you're dealing out 8 hands to opponents, you simply cannot de-Ace the deck, it is not a valid simplification.

You must use the c(46,4) and then the next player has c(42,4) possibilities. The following hand probabilities are conditional on the previous dealt cards.