odds of getting two cards 10s or higher in the next 5 hands

[Zaebos]

...including all pairs. How much do the probabilities of getting this kind of hand go down per hand (as in 4 hands left, 3 hands left....). And how do you figure this sort of thing out?

[BruceZ]

[ QUOTE ]
...including all pairs. How much do the probabilities of getting this kind of hand go down per hand (as in 4 hands left, 3 hands left....). And how do you figure this sort of thing out?

[/ QUOTE ]

The probability of getting 2 cards 10 or higher is C(20,2)/C(52,2) = 20*19/(52*51) = 14.3%. The probability of getting it in N hands is 1 minus the probability not getting it for N hands. This is

1 - [1 - C(20,2)/C(52,2)]^N.

N=5: 53.8%
N=4: 46.1%
N=3: 37.1%
N=2: 26.6%
N=1: 14.3%

[Zaebos]

that was quick thanks
Is there a website or a book where information on how to calculate things like this(and other calculations) are explained