Being dealt A2xx in Omaha/8

[donkeyradish]

How do you go about calculating this probability?

ie. Chance of being dealt both at least one ace and at least one deuce in 4 cards.

My real-life results are 6.5% but I can't figure out how to calculate the true probability.

And how would you extend this for example, to calculate the odds of being dealt A23x? It sounds like it should be easy but I'm not sure where to begin.

[gaming_mouse]

8*(50 chooose 2)/(50 choose 4) = 0.0425531915, just over 4%

[MickeyHoldem]

[ QUOTE ]
8*(50 chooose 2)/(50 choose 4) = 0.0425531915, just over 4%

[/ QUOTE ]
This doesn't seem right...

Exactly 1 ace and exactly 1 duece = 4*4*(44choose2)/(52choose4) = 0.055909133

[gaming_mouse]

[ QUOTE ]
[ QUOTE ]
8*(50 chooose 2)/(50 choose 4) = 0.0425531915, just over 4%

[/ QUOTE ]
This doesn't seem right...

Exactly 1 ace and exactly 1 duece = 4*4*(44choose2)/(52choose4) = 0.055909133

[/ QUOTE ]

typo. change the denominator to 52 choose 2

EDIT: Fukc. I mean, 52 choose 4

[AngusThermopyle]

ie. Chance of being dealt both at least one ace and at least one deuce in 4 cards.

Odds of being dealt at least one ace:
1 - oddsNoAce =
1 - (( 48*47*46*45 )/(52*51*50*49)) =
0.281263274540585

Given the fact that you have an Ace, odds of having at least one deuce in the remaining 3 cards =
1 - oddsNoDeuce =
1 - (( 47*46*45)/(51*50*49) ) =
0.221368547418968

Odds at least one Ace, at least one Deuce, multiply to get:
0.0622628425273517 about 6.23% or roughly
1 in 16.

"Ballpark computation"
Ace, one in 13, four tries = 4/13
Deuce, on in 13, three tries = 3/13
Both: 12/169 = roughly .071
I know it is not proper probability, but gives a quick approximation which is usually a bit on the high side.

Extended to A23x,
Odds of a 3 in the remaining 2 cards =
1 - (46*45)/(50*49)
and multiply that result by the A2 result.

[gaming_mouse]

Angus,

Whoa! There is no need for such a long answer. The answer is:

4*4*(50 chooose 2)/(52 choose 4)= 0.07239819

Mickey, btw, the 8 in my first post was wrong too. Somehow I got 4*4 =8 when I was doing the math in my head.

gm

[AngusThermopyle]

mouse,

Whoa! Since you do not explain why/where/how you came up with your formula (and since you already admit to one mistake), I will believe my answer.

Unless you can find where I made a mistake.

[bohemian]

Check out this thread:
http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=otherp&Number=1420747& Forum=,All_Forums,&Words=&Searchpage=0&Limit=25&Ma in=1420747&Search=true&where=&Name=18572&daterange =&newerval=&newertype=&olderval=&oldertype=&bodypr ev=#Post1420747

[AngusThermopyle]

Fine, but what is your answer to this question?

[gonepunting]

gaming_mouse's answer is slightly out as it double counts hands with AA2 or A22.
I make it
A2xx 4*4*(44*43/2)=15136
A22x or AA2x 6*4*44=1056 times 2 = 2112
All aces or twos (8 choose 4) -2 = 28

Total 17276
Divded by (52 choose 4) = 6.38%

PS My very first post what do you think?? /images/graemlins/ooo.gif

[AngusThermopyle]

All aces or twos (8 choose 4) -2 = 28

Since AAA2 and A222 each can be made 16 ways, for 32 ways, and AA22 can be made 36 ways, for a total of 68 ways to make all Aces or Twos, with at least one of each....

I don't think much of your defense of mouse's computations.

[pzhon]

There are 52C4 Omaha hands, each equally likely.

/images/graemlins/diamond.gif A2

The hands involving A2 are A2XX, A22X, AA2X, A222, AA22, and AAA2, where X can't be an ace or deuce.

A2XX: 4 * 4 * 44C2 =
A22X: 4 * 4C2 * 44
AA2X: 4C2 * 4 * 44
A222: 4 * 4
AA22: 4C2 * 4C2
AAA2: 4 * 4

That is 17,316 out of 270,725 hands, so 6.396% of the time you have A2.

gonepunting was right except for an arithmetic error.

/images/graemlins/diamond.gif A23

The hands involving A23 are A23x, A233, A223, and AA23.

A23x: 4 * 4 * 4 * 40
A233,A223,AA23: 4 * 4 * 4C2

That totals 2,848 hands out of 270,725, so 1.052% of the time you have A23.

[AngusThermopyle]

Good news, bad news.
Good news, I can follow your logic and computations (I'm not sure if there is enough explination/comments for the original poster to follow unless he understands the '4C2' notation and how to compute it), and using them I come up with the same answer.

Bad news, I cannot find the error in my method. But will work on it.

[gaming_mouse]

[ QUOTE ]
gaming_mouse's answer is slightly out as it double counts hands with AA2 or A22.

PS My very first post what do you think?? /images/graemlins/ooo.gif

[/ QUOTE ]

You are right. I've really botched this thread up good /images/graemlins/smile.gif.

Nice first post,
gm

[pzhon]

[ QUOTE ]
Bad news, I cannot find the error in my method.

[/ QUOTE ]
You conditioned on the first card being an ace rather than that there was at least one ace. As an example of the difference, the expected number of aces given that the first card is an ace is 20/17 = 1.176. The expected number of aces given that there is at least one ace is 16660/15229 = 1.09397.

If you used your method to compute the probability of getting dealt A2 in Hold'em, you would multiply the probability of getting at least one ace, 33/221, by the probability that your second card is a deuce given that your first card is an ace, 4/51, to get 44/3757 instead of the correct 16/1326.

Without breaking it up into cases or using the probability of getting dealt A2, I don't see a way to read off the probability of getting a 2 given that you are dealt at least one ace. I think you either need to use cases (say, the number of aces you are dealt) or use a different method.

[gaming_mouse]

[ QUOTE ]
mouse,

Whoa! Since you do not explain why/where/how you came up with your formula (and since you already admit to one mistake), I will believe my answer.

Unless you can find where I made a mistake.

[/ QUOTE ]

Angus,

Didn't mean to offend. And it looks like you have now been vindicated /images/graemlins/smile.gif.

Nonetheless, I think it was still worth pointing out that your approach is not as simple as it could be. Solving these arguments using counting methods is almost always easier than the kind of conditioning you were attempting. I made a simple double-counting error, but the basic method is still much cleaner. pzhon showed how to do it properly.

gm

[donkeyradish]

Ah, good explanation, It makes the problem much easier to split it up like that.

[Beavis68]

there are 52*51*50*49/24 starting hands = 270725
There are 4*4*50*49/2 A2xx hands = 19600

Total probablility 19600/270725=7.2%

[Beavis68]

[ QUOTE ]
Angus,

Whoa! There is no need for such a long answer. The answer is:

4*4*(50 chooose 2)/(52 choose 4)= 0.07239819



[/ QUOTE ]

that is exactly right, I have no idea what this formula is though.

[gaming_mouse]

[ QUOTE ]
that is exactly right, I have no idea what this formula is though.

[/ QUOTE ]

actually it was a little off.

the reasoning of 4 aces, times 4 twos, then select 2 cards from the remaining 50, doesn't quite work. That's because you will double counting hands in which more than 1 ace or more than 1 2 occur.

For an exact answer, you need to break it down into non-overlapping cases like pzhon did.