This is a discussion on Probability of Another player being dealt an ace. within the online poker forums, in the Learning Poker section; 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, 


#1




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 nonace 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. 
#6




Tink u got this totally screwed up.
Heads up for clarity and ease of thought. xy vs ab the possibility of any one of those unknown cards being an ace is 4/52. Once u see you do not have an ace, the possibility of villain holding an A is 4/50 = 8%, the possibility of villain holding AA is 8% x 3/49 ( a bit less than 8% so for simplicities sake lets say .08 x .08 = .0064 aka .64% which is rare indeed. Take that a step further and for 2 A's to even be in play pf should be about that same percentage. Rare indeed. 
#7




First, I'll show you how to compute the probability that any one opponent has at least one A, given that you have exactly one A:
Either his first card is an A (3/50 = 6%) or his first card is not an A and his second card is (47/50 x 3/49 = 94% x 6.12% = 5.76%). So the total probability that one opponent has at least one A is 6% + 5.76% = 11.76%. In order to check, we can also compute the probability as the complement of his having no A in either card (47/50 x 46/49 = 94% x 93.88% = 88.24%). The complement is 1  88.24 = 11.76%, which is the same answer as above. Yeah, we must have done it right. I need to go get ready for my birthday dinner, but I'll help with the extension to multiple players afterwards. 
#8





#9




I'm back... in order to extend the calculations to more than one opponent, it's easiest to continue using the complement approach, as follows:
To compute the probability of at least one opponent having at least one ace, you simply take the complement of all opponents having no aces. So, for two opponents we have 1  {(47/50) x (46/49) x (45/48) x (44/47)} = 22.55%. In general, for N other players we have 1  {(47/50) x (46/49) x ... x ([482N]/[512N])} This results in the following probabilities for varying values of N: N Prob that at least one player has at least one A 1 11.76% 2 22.55% 3 32.43% 4 41.43% 5 49.59% 6 56.96% 7 63.57% 8 69.47% 9 74.69% 