WPT "stat" question

[PE101]

Hi,

I was watching a WPT show yesterday and and an interesting "stat" was mentioned at one of the breaks:

"You're at a 6 handed table, you are dealt pocket kings - what are the odd that someone at the table has pocket aces?

I don't remeber all the choices offered, but one of them was 220:1 (the one I thought was correct), but after the break, the answer was something like 30 or 40:1 (sorry, I don't remember the exact number).

I don't get it. I can see that it should be better than 220:1 since I know 2 cards that they DON'T have.

What am I missing? (Other than the details of the problem [img]/images/graemlins/wink.gif[/img])

Thanks

[LetYouDown]

For what it's worth...WPT has posted incorrect odds before, but I digress. 6 people have a 6/C(50,2) chance of getting A-A, or about 203-to-1. Factor in the fact that 5 people have this chance and you get an answer close to 5 * 6/C(50,2)...or roughly 40-to-1. It wasn't the question of what the odds were that any individual will have A-A, but that at least one of your opponents will have A-A. Heads up, the answer would be 203-to-1. At a 6 person table it's very close to 40-to-1. For an exact answer, you'd need to use the Inclusion/Exclusion principle to remove the times you're double counting when two players have A-A.

[PE101]

You're right. Thanks for de-confuseifing me.

[uuDevil]

The answer on the show (44:1) is wrong. It's about 39.9:1.

This came up after the last broadcast and was discussed here.

Sorry for the dunb question, but could you explain the following equations? This seems to be standard notation yet I don't follow.

5 * 6/C(50,2)

6/C(50,2)

[LetYouDown]

5 = The number of players remaining. 6 really is C(4,2)...it's the number of ways to choose 2 cards (A-A) from 4 cards. There are 6 ways you can have A-A.

C(50,2) is the number of ways to choose 2 cards from 50 cards. There are 50 cards remaining in the deck, so the other players have C(50,2) possible hands because two of them are gone (K-K).