Examples of this this situation are listed in , but what are the actual odds of at least one person having an A?

My math says that it's about 64% in a full 6-max table. My calculations:

(3/47)*10=0.638

That is, three cards (the aces) of 47 (the unknown cards after the flop) have ten chances to appear (two cards for each opponent).

Is this the correct way to calculate these numbers?

So for a 5-player table:
(3/47)*8=0.51

And 4-player:
(3/47)*6=0.383


So how does this inform strategy? I raise PF and usually like to bet out on the flop. If I get called, I often slow down. If I get raised, I'll usually muc it. I'm not sure if this is optimal play, though.

Comments?

Your calculation is off. I don't know the right answer off the top of my head, but if if we do it your way for full ring, 3/47*18 > 1, which obviously isnt true. Have to go now, but will post later

First, it's exponents not multiplication... duh. So (3/47)^10 would be closer.

I've got it backwards... I think the easy way to do it is to calculate the probability that one will not hit:

44/47 * 43/46 * 42/45 * 41/44 * 40/43 * 39/42 * 38/41 * 37/40 * 36/39 * 35/38=

.94 * .93. * 93 * .93 * .93 * .93 * .93 * .93 * .92 * .92 = about 48%

basically .93^10

for 5-handed = about 56%

for 4-handed = about 65%

for 10-handed - about 24%

So it's a crooked coinflip at a 6-handed table -- and that's assuming that opponents play any A.

geez, i was realy tired when i posted that last night.