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.

I think it might be simpler if you think from the viewpoint of what is the prob of player NOT getting at least one ace, then subtract that from 1.

It was some time ago, so I wouldn't take this as rock solid since my memory may well be fuzzy. With that word of caution, I have seen this question before, and I'm thinking the answer was somewhere around 50%.

I always heard if you are dealt an ace, you are 70% probable to be the only one with an ace....

It also depends on how many players are getting cards as well...
that 70 % thing might be true for 2 or 3 players.

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.

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.

I remember some cool member having a sig about this.

http://www.cardschat.com/members/four-dogs/ (http://www.cardschat.com/members/four-dogs/)

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 ([48-2N]/[51-2N])}

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%

Click on the ABSENCE OF AN ACE Hyperlink in my signature