Chance of other player having an ace when...

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Whats the chance of another playing holding an ace when I hold one ace in the hole and one ace comes on the flop? (10 handed game)

[tech]

I get 62.4%.

I calculated 1 - P(none of the other 18 player cards are an A). This works out to:

1 - 45/47 * 44/46 * ... * 28/30 = 1 - 37.6% = 62.4%

Someone please check my math. It is WAY past my bedtime. /images/graemlins/cool.gif

[BruceZ]

[ QUOTE ]
I get 62.4%.

I calculated 1 - P(none of the other 18 player cards are an A). This works out to:

1 - 45/47 * 44/46 * ... * 28/30 = 1 - 37.6% = 62.4%


[/ QUOTE ]
This is the correct answer for the case of 9 opponents seeing the flop. I got the same answer by inclusion-exclusion for this case. The important thing is how many players see the flop, not how many players there are originally, since the opponents can't hold an ace if they've already folded it.