my math must be flawed, cuz this doesnt sound right
 

The chances someone gets dealt aces preflop at a 10 player table (only one person gets them):
2*10=20-->number of cards dealt out preflop
52-20=32-->number of cards not dealt
C(32,2)=561-->total number of 2 card hands that weren't dealt
[C(10,1)*C(4,2)*C(28,0)]/561=0.10695
0.10695*100=10.695% one person gets dealt aces preflop at a 10 handed table.
Now this seems too high; what did I do wrong?

Re: my math must be flawed, cuz this doesnt sound right
 

it should be 10 * C(4,2) / c(52,2) = minus a miniscule factor accounting for multiple rockets... 4.5 %

Re: my math must be flawed, cuz this doesnt sound right
 

yeah i jsut figured it out to be the same thing, idiot me didnt know what iw as doing for a second. thanks though.

Re: my math must be flawed, cuz this doesnt sound right
 

Or you can just say 10/221 which gets you 4.5 % also and is close enough for gummint work.

Re: my math must be flawed, cuz this doesnt sound right
 

If my A.M. inclusion/exclusion principle is correct, to meet your condition that only one person has A-A:

C(10,1) * C(4,2)/C(52,2) - C(10,2) * C(4,4)/C(52,4) = 4.5083% as opposed to 4.5249%

Re: my math must be flawed, cuz this doesnt sound right
 

Can you explain why it would be 4.5083% opposed to 4.5249%

There are 1326 possible hand combinations with AA making up 6 of them.

6/1326 = .0045249 * ten seats = 4.5249%

Re: my math must be flawed, cuz this doesnt sound right
 

Because of the "(only one person gets them)" clause. If more than one person gets them, it's counted twice.

Re: my math must be flawed, cuz this doesnt sound right
 

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Or you can just say 10/221 which gets you 4.5 % also and is close enough for gummint work.

[/ QUOTE ]

I have always seen it come up trying to figure possibility of AA being out 10 handed when you have KK.

Similar roughhand calc 9*6/1225 = 4.4% ~22:1 against

Re: my math must be flawed, cuz this doesnt sound right
 

So you are saying that with all the different permutations of the 1326 hands dealt to 10 people, whenever AA comes up twice within the 10 hands you count it TWICE towards the % of AA coming up?

That does not make mathematical sense. Once the condition (AA) is met you do not look for more satisfied conditions within the range (that set of 10 hands). You are inflating the results.

Re: my math must be flawed, cuz this doesnt sound right
 

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That does not make mathematical sense

[/ QUOTE ]
I agree, that's why I fixed your calculation.

[ QUOTE ]
You are inflating the results

[/ QUOTE ]
No, I'm deflating them, from your inflated figure.

Edit: And actually, this doesn't even factor in the fact that you want only hands where one player has them, it just removes the double counting of the times when two people have A-A.