AA vs AA

[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

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[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. [img]/images/graemlins/tongue.gif[/img]

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. [img]/images/graemlins/smile.gif[/img]

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