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Tue 16 Oct, 2012 10:10 am
Anyone could help me to learn how i can calculate the probability of winning/losing with following conditions in a long run:
A: a value with 10 possibilities (0,1,2,3,....,8,9)
B: a value with 10 possibilities (0,1,2,3,....,8,9)
if we have ONE random value of A and THREE random Value of B, what is the probability of having at least one of the three values of B to be tie(equal) with the value of A.
Thanks
@ablet,
Depends on whether or not you're sampling B with replacement.
With replacement:
The probability that all three samples don't match A is (9/10)^3. Therefore, the probability that at least one sample matches A is 1 - (9/10)^3.
Without replacement:
The probability that all three samples don't match A is C(9,3)/C(10,3). Therefore, the probability that at least one sample matches A is 1 - C(9,3)/C(10,3).
C(n,r) = n! / (r! * (n-r)!)
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