Prob of no ace delt

icepick · #1 09-13-2006, 03:21 PM

Discussion I got into with a friend.

Given a table of 9 players, what is the probablity that none of them are delt an ace?

He claimed 1 in 13, which is just flat wrong. I guessed about 15% at the time. After figuring out, I believe it is 17.1%.

48/52*47/51*46/50*...*31/35 Right?

rjp · #2 09-13-2006, 04:01 PM

Looks good to me--17.13%.

BogusPomp · #3 09-13-2006, 04:01 PM

Yes, this is right.

one other method to get the answer would be:
(48c18)/(52c18) = 0.171302983

EasilyFound · #4 09-17-2006, 07:10 PM

[ QUOTE ]
Yes, this is right.

one other method to get the answer would be:
(48c18)/(52c18) = 0.171302983

[/ QUOTE ]

what does the "c" represent?

JimDrang · #5 09-18-2006, 07:02 PM

Its a probability term, 48c12 would be read as 48 choose 12.

http://www.newton.dep.anl.gov/newton...th/MATH123.HTM

MelchyBeau · #6 09-20-2006, 02:13 PM

xCy = x!/[(y!)(x-y)!]

in plain english xCy means you have x items and you want to choose y items, how many different ways can you choose y items, where order doesn't matter.

just a little sidebar xCy = xC(x-y)