Ax vs. AA = 441:1?

[Ocho]

In a B&M tournament tonight, I was given AQs against AA and within 30 mintues, given AKo against AA. Am I correct that the odds of someone else having AA when I am holding an Ace are 441:1?


Needless to say, I didn't get to the last table. But I did finish in the money.

Ocho

[Ocho]

No. I don't think this is right. But what are the correct odds and how'd ya get them? thanks.

[daryn]

ok.. so you're holding an ace and a non ace.. there's 50 cards left in the deck, and 3 of them are aces. the chances of another player being dealt AA are just (3/50)(2/49) = 407.3-1 against.

[uuDevil]

Daryn, I think your number is for any one opponent while the poster probably wants to know what the odds are he is up against anyone having AA.

If you have 9 opponents, P(any opponent holding AA)= (3/50)(2/49)*9=.02204, or about 44.4:1. so originally posted number isn't far off, except for the decimal pt. /images/graemlins/wink.gif

[irchans]

From twodimes:

http://twodimes.net/h/?z=65286
pokenum -h as ad - ah qc
Holdem Hi: 1712304 enumerated boards
cards win %win lose %lose tie %tie EV
As Ad 1567203 91.53 123893 7.24 21208 1.24 0.921
Qc Ah 123893 7.24 1567203 91.53 21208 1.24 0.079


That's about 12:1.

[LondonBroil]

Thanks for letting us know what his chances of winning are, but that's not what he was asking.

[tpir90036]

a rough approximation that is close enough:

P(one of your opponents having AA) = (3c2)/(50c2) = 3/1225

since there is not a chance of more than one person having AA we don't need to do any fancy inclusion-exclusion stuff so just multiply by the # of opponents:

P(someone else having AA when i have an Ace) = (3/1225)*n where n = # of opponents.

if n=9: (3/1225)*9 = .02204 = 44.3:1

[Ocho]

ack! approx. off by a decimal place becuase i forgot to multiply by number of opponents. Maybe i didn't deserve any money at all with that kind of sloppy math!!

Thanks,
Ocho

[BruceZ]

Correct, you've been reading, but why do you say it's a rough approximation? The fact that only 1 player can have AA makes this solution exact.

[tpir90036]

oops! i forgot to go back and delete my first sentence after i wrote the sentence about not needing to worry about double counting since only one person can have AA. ha ha ha...slowly but surely...

-tpir