AA vs AA

[Angel]

This is my first post and if I've chosen the incorrect forum within this site, I apologize in advance.

Twice recently, in as many days, I found myself on the river splitting the pot with Aces against an opponent who also held Aces. While it's not something that comes up often, twice in two days piqued my curiousity and for my own edification I decided to calculate the odds of this event happening once. Prior to actually picking up pen and paper I happened upon an old article by Brian Alspach (Poker Digest Vol. 5 No. 2) which stated that the probability is 1/136. He said that the math for the above calculation was in his previous article. I've got two problems. First, I don't have that issue (presumably Vol. 5 No. 1) and secondly, my attempts to arrive at the 1/136 number through my own efforts has failed. Would anyone care to assist me in this?

Thanks,
Angel

[Wake up CALL]

To begin with you need to know the number of players who are dealt in the hand. After that I am not sure anyone can reliably tell you the probability that you will both see the river and showdown, much less split the pot.

[Angel]

You're correct and I was remiss in some of the details of my question. The only number I am looking for is the probability of both of us being dealt the same pocket pair during the same hand (in this case - AA). Alspach's 1/136 was based on a ten player game - which is fine. I assume that if I have a bit of help on how to arrive at the answer I'm looking for - I can substitute the number of players into the equation easily enough. As far as the likelihood of us both seeing the river and splitting the pot - that does complicate the matter but it's not something I'm trying to determine.

Thanks for helping me to better present my question.

Angel

[The Dude]

Well, let's try the calculation this way. When I hold a pocket pair, what are the odds that somebody else holds the same pocket pair? We'll assume 10 handed.

You have two cards, so there are only 50 left. That means there are 2450 possible starting hands left (50*49), and nine chances at it. Although this is an imperfect calculation, that would indicate that the odds are roughly 1/272. Am I wrong?

Alspach must be saying 1/136 of the time you hold a pair. Since you will only be dealt a pair 1/17, he can't be saying that in 1/8 of those hands you will be against an identicle hand.

I am curious to see his calculation, because I don't see how it's possible to happen that often.

You will find better success with this post in the Probability forum.

[Angel]

Thanks for your input, I'll take this over to the probability section - thanks for the suggestion. In the meantime, you wrote:

" You have two cards, so there are only 50 left. That means there are
2450 possible starting hands left (50*49), and nine chances at it.
Although this is an imperfect calculation, that would indicate that the
odds are roughly 1/272. Am I wrong?"

I wouldn't say so and am glad to see you did it the same way as I did - even if we're both wrong /images/graemlins/smile.gif

"Alspach must be saying 1/136 of the time you hold a pair. Since you
will only be dealt a pair 1/17, he can't be saying that in 1/8 of those
hands you will be against an identicle hand."

Here's what he said, "You are playing in a 10-handed hold'em game and find that you have been dealt pocket 8s. When thinking about possible hands for other players, it is natural to wonder about the possibility of other players having been dealt pocket pairs. In my previous article, I determined the probability of another player also holding pocket 8s. What we learned is that the probability of this occuring is about 1/136." (Obviously I exchanged AA for 88 in my question - but it is equally obvious that the math doesn't change by my doing so.)

" I am curious to see his calculation, because I don't see how it's
possible to happen that often."

If I find the answer on my own - I'll post it.

Peace,
Angel

[memphis]

I think they came up with it this way.

If there are 9 other players and 2 Aces already gone, then the possibility of one of them receiving an ace on their first card is (2/50)*9. When one of the does get an ace on their first card, then the odds of them getting the 2nd ace is 1/49. Therefore the odds of one of the 9 getting the other 2 aces is (2*9)/(49*50)= approx. 1/136

[Kurn, son of Mogh]

How will knowing the answer to this help your game? In fact, I'll go so far as to say that during the play of any given hand, knowing the exact probability that my opponent holds essentially the same hand as I hold could pretty much be the most irrelevant piece of information on the planet.

[M.B.E.]

Angel never said that this information would be useful for making better playing decisions. It's simply a matter of curiosity.

[JTG51]

Can't people just ask questions out of curiosity? What's wrong with just wanting to know something that isn't very important?

Besides, knowing how to figure out the answer could very well be helpful in other ways.

[Copernicus]

The correct calculation is:

One player has (2/50)*(1/49) chance of duplicating your pair = .000816327

His chance of NOT duplicating your pair is 1 minus that, or .999183674

The chance of 9 players all NOT duplicating the pair is that to the 9th power or .992677006.

The chance of 1 of the 9 players duplicating the pair is 1 minus that or .007322994. 1/136=.007352941. Close enough for government work.

[Nottom]

[ QUOTE ]
You have two cards, so there are only 50 left. That means there are 2450 possible starting hands left (50*49), and nine chances at it. Although this is an imperfect calculation, that would indicate that the odds are roughly 1/272. Am I wrong?


[/ QUOTE ]

You are double counting here, basically you are counting AhAd and AdAh as 2 different hands. The correct number of hands is thus 50*49/2=1225 and the odds of someone in a full game holding the matching pocket pair with you is 1 in 136.

This number is misleading though since it is only saying that when you are fortunate enough to get dealt AA that 1 in 136 of the time you will be up against AA not that 2 people will get AA in the same hand every 136 hands.

[OffTilt]

[ QUOTE ]
How will knowing the answer to this help your game? In fact, I'll go so far as to say that during the play of any given hand, knowing the exact probability that my opponent holds essentially the same hand as I hold could pretty much be the most irrelevant piece of information on the planet.

[/ QUOTE ]

Cmon dude, how did making that post help your game? I'd go so far as to say taking the time to criticize someone's curiosity may be the most pointless thing to do on the planet.

OffTilt

------------------------------------------
Start winning hundreds at Texas Holdem with Power Holdem+
http://www.online-pokerguide.com

[Kurn, son of Mogh]

Wow. One grouchy post in 1600 and they hit you over the head with a baseball bat.

Problem is, when you hit me in the head with a bat, all you do is break a perfectly good bat. /images/graemlins/tongue.gif

I stand chastised.

[RockLobster]

Gee, I laughed when I read your first post, it even had a smiley in the subject line... I assumed you were being 100% sarcastic. I still think you were. Either that or the Romulans are gathering forces along the neutral zone, and you're preparing for battle...

[Rushmore]

It really DID seem a bit much, especially given the facts that the poster was a newbie AND that he used the word "curiosity."

Go easy, dude. Be a Pooh-Bah, not a Poo-Bah.

[JTG51]

Be a Pooh-Bah, not a Poo-Bah.

LOL. Maybe I'm just in a strange mood, but I can't help but to laugh at that.

[Angel]

I'm afraid I missed the continuation of this post until just now due to the title change -

To answer your question of how this knowledge could help my game, I don't know that it ever will. Better however to be overprepared than under prepared and it certainly won't hurt my game to know. The greatest reason for me asking however was that I sincerely dislike making math errors - and I abhor making poker math errors, this problem I fiqured incorrectly may have no value - but the same error in another application may prove to be costly. I have enough ego to be embarressed that I had to ask and to never forget the math. /images/graemlins/smile.gif

Peace,
Angel

[Cyrus]

"How will knowing the answer to this help your game?"

Isn't it obvious how? By getting to know that the answer will not help his game.

[BruceZ]

I just ran across this thread. Actually memphis had it exactly right, and your method is an approximation. If you hold AA, the probability of a partcular player holding AA with you is 1/C(50,2) = 1/1225, and the probability that one of 9 players holds AA with you is exactly 9/1225. The reason this is exact is that no more than 1 other player can hold AA with you, so you can sum these 9 probabilities since the event of one player holding AA is mutually exclusive of other players holding AA. This would not be exact in other cases where more than 2 players could hold a better or equal hand. The reason your method is approximate, though very close, is because the event of one player holding AA is not independent of another player holding AA, so raising this probability to the 9th power is approximate.

[ropey]

I'm seeing this a little different...I'm getting 1/6016.

There are two events that have to happen...first, EXACTLY 2 aces need to be dealt in the first 10 cards (for 10 players). Then of the remaining cards, the same two players must recieve the 3rd and 4th ace in the deck.

-ropey

[Nottom]

You making it way to complicated and thinking about it the wrong way. You are only concerned with how often 1 out of the other 9 people will have AA when you do, which accounts for 2 of the aces immediately (they are in your hand). The other responces in this thread already explain the math and reasoning behind the calculation, which is correct.

[ropey]

I certainly see why its 1/136 for another player to have pocket aces when you do...

However, the odds of any two players recieving pocket aces on the same hand is 1/6016.

-ropey