What's Wrong with my math ?? Odds of Ace appearing

[BillFranklin]

I'm new to probabilies and trying to learn how to calculate them to figure odds in various poker situations. I'm not too confident in my math so i tried to figure out the calculations for " commonly known poker situations"

For example: According the Super System the odds of no player including yourself being delt an Ace (10 handed)is
13.28% I can't figure how to come up with this number. I came up with 20.08 %
here is my math:
1326 starting hands 196 hands with Aces
(1-196/1326)*(1-196/1325).... *(1-196/1317)= .2008
.2008 *100 = 20.08

Any help is appreciated. Also does anyone know any good books on poker odds or just probabilites in general. I recently bought Petriv's odds book. I've read some of it, haven't finished yet but this doesn't seem to be what i'm looking for. (odds for calculating common starting hands, i'm looking for more complex situations)

[bobbyi]

[ QUOTE ]

1326 starting hands 196 hands with Aces
(1-196/1326)*(1-196/1325)

[/ QUOTE ]
The second term here shouldn't be 1-196/1325. You are essentially saying that out of the 1326 original combinations, only one of them is now used up. But that's not true. The two cards dealt to the first player make it such that lots of the other combinations are now impossible (any combination that contains either of those two cards). Note that some of the dead combinations contain an ace and some don't.

[mosdef]

here's a conceptually easier approach:

in order to have all ten hands not contain an ace, you need to deal 20 consecutive non-aces off the top of the deck

the prob that the 1st is not ace = 48/52
2nd is 47/51
3rd is 46/50

and so on.

so prob(no aces) = (48*47*....*30*29)/(52*51*...*33)

most of the middle terms cancel:

prob = (32*31*30*29)/(52*51*50*49) = 13.28%

[chadash]

well I'll give a simpler way to do it with combinations.
we want to draw 20 cards from 48 (i.e. non-aces). so the odds are:
48c20*4c0 / 52c20 = 13.3%
hope this is helpful