Probability that someone else has an A

[Alexpoker]

Against X opponents, I try to evaluate the Probability that someone else has an A preflop (one or more), assuming I have an A.
=3C1*49/50C2 for one opponent
=(3C1*49/50C2)*x for x opponents

For x= 1 p=12%
for x=9 p=100% !!

What's wrong ?

thanks to refresh my math...

[Alexpoker]

In fact, I know how to calculate odds to improve my hands (preflop->river->turn) and odds to get pocket pair but I don't know to calculate odds when there's some opponents.

[World Jugador]

Let's assume X opponents. We have 2 cards, so 50 cards remain, 47 are non-aces.

So let's draw 2X cards (two for each of X opponents). Then the odds of NOT drawing an ace are:

47/50 * 46/49 * ... * (47-2X+1)/(50-2X+1) = P(47,2X) / P(50,2X)

The odds of drawing an ace is then 1 - P(47,2X)/P(50,2X). (If X is greater than 1, we can simply further, but this is all you need to know for a calculator to give you the answer.)

So X = 9 yields a probability of 75%

[World Jugador]

[ QUOTE ]
I know how to calculate odds to improve my hands (preflop->river->turn)

[/ QUOTE ]

A very strange game indeed if your sequence is preflop to river to turn.

[Cobra]

Assuming you have AX, there are 3 aces remaining and 47 non aces. Remember the probability that one or more of your opponents holds something is the same as 1- the probability that they do not hold it.

Probability no one else has an Ace.

9 opponents = (47c18)/(50c18) = .253

Probability that one or more aces are out = 1-.253 = .747

1 opponent = (47c2)/(50c2) = .882

Probability that he has an ace is 1 - .882 = .118

Cobra