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rljames
11-11-2004, 01:38 PM
I hold an Ace.
There is one Ace on the flop.
10 players (9 + me) in the game.

What is the probability that *at least* one of the other players holds an Ace at that point in the hand? Prior to the Turn.

Could you also provide the math to calculate for other number of players in the hand.

Thanx,
Bob

BruceZ
11-11-2004, 03:06 PM
[ QUOTE ]
I hold an Ace.
There is one Ace on the flop.
10 players (9 + me) in the game.

What is the probability that *at least* one of the other players holds an Ace at that point in the hand? Prior to the Turn.

[/ QUOTE ]

1 - C(45,18) / C(47,18) = 62.4%.

[ QUOTE ]
Could you also provide the math to calculate for other number of players in the hand.

[/ QUOTE ]

1 - C(45,2N) / C(47,2N)

where N is the number of opponents (not counting yourself). This is 1 minus the probability of no aces being dealt. C(45,2N) is the number of ways to select the 2N non-aces out of 45 non-aces. C(47,2N) is the total number of ways to select 2N cards.

<font class="small">Code:</font><hr /><pre># opp. P(at least 1 ace out)
1 8.4%
2 16.5%
3 24.1%
4 31.5%
5 38.4%
6 45.0%
7 51.2%
8 57.0%
9 62.4%</pre><hr />

JoeyT
11-11-2004, 03:10 PM
It's been a long time since I tackled a probability question, but I think this is how you want to do it:

It'll be easier to figure out what the probability is that nobody has an ace, and then subtract that from 1.

You have seen 5 cards, and 2 of them are aces. That means there are 47 cards left in the deck, 2 of which are aces and 45 of which are not. Let's label the other 9 players in the hand A-I.

The probability that player A does not have an Ace is (45/47 * 44/46). This is because the probability that his first card is not an ace is the 45/47, but then the probability that his second is not an ace raises to 44/46 since you now know another card is not an ace). Player B's chances of not having an ace are then (43/45 * 42/44)... etc.

So, you would come up with something like
(45*44*43*...*28)/(47*46*45*...*30), which reduces to (29*28)/(47*46) = 812/2162 = ~.376.

This would mean that the probability that somebody else has an ace is 1 - .376 or ~.624.

Hopefully this isn't way off. Soemone verify /images/graemlins/smile.gif

scandal
11-11-2004, 03:55 PM
Similar post about strength of kicker when holding Ax (http://forumserver.twoplustwo.com/showflat.php?Cat=&amp;Number=1198420&amp;page=2&amp;view=colla psed&amp;sb=5&amp;o=14&amp;fpart=1)

ZeeBee
11-19-2004, 09:26 PM
Suprised noone has posted anything on this so far...

I'm no expert so forgive me for bumbling through it....

In addition to being a pure probability question, this is also a conditional probability question.

The answers you've been given so far are the odds that someone is dealt an Ace. That's not the same as the odds of someone who is in the pot with you having an Ace. What you need to fully answer the question "what are the chances that if there are x players, one or more have an Ace" (or similar) is "what are the chances they have an Ace given they chose to enter the pot".

To calculate this you need to try to understand what cards they will play. If they only play AA for a raise and they've raised preflop, then you can be pretty sure they have an Ace regardless of what the odds of being dealt one say.

In more realistic situations you can try to estimate what sort of hands they will be playing with (based on observation of what cards they have shown, what their VPIP is or what a "typical player like them" is likely to be playing). You then factor in the cards you know about (e.g. the A you have and the one on the board) to figure the odds of them having an Ace.

In practice, this is complicated.

But one obvious observation is that players are likely to have an Ace (or a King) more than the pure probabilities of being dealt one would dictate simply because people (especially at low limits) like to play Aces &amp; Kings. The fact that they are in the pot increases the chance that they have an Ace or King (or any other good hand).

ZB