**Probability of Another player being dealt an ace.**

Hi, I've thought about this problem before, but got no where, and this is really bothering me that I can't figure it out.

The problem is, given N players, and given that I am dealt an ace and a non-ace card, what is the probability that another player holds 1 ace? Let's not worry about 1 or more Aces just yet.

I know about the chart that shows you the probability that your ace is dominated:

http://en.wikipedia.org/wiki/Poker_probability_%28Texas_hold_%27em%29
But I can't figure out the simpler question, of whether an opponent has an ace.

Let's simplify it to just 1 opponent. Given 3 remaining aces, and the fact that we've seen 2 out of 52 cards, the P his first card is an ace is 3/50. (I'm almost 100% sure this part is right). The P that his second card is an ace , given that his first card is not an ace is 3/50. So I'm adding 3/50+3/50 to get 6/50 for 1 opponent.

Then if there were 9 opponents, I'd have 3/50*2*9=54/50=1.08, which is > 1!

I know that can't be right! A probability of >1 is nonsensical, P's can only be 0<=P<=1. Plus we know that just because you are 9 handed and have an ace, it doesn't mean someone else ALWAYS gets dealt an ace.

What am I doing wrong here? This is driving me nuts. My first revision was to say, OK, Player1Card1ProbAce=3/50, then Player1Card2ProbAce=3/49, given that his first card was not an ace. But if you start adding up those numbers, you get an even bigger number than 1.08 for 9 players.

I don't think you multiply the 3/50's by each other either. Then the number you get is too small.