Question 601511
The probability of getting doubles is p = 6/36 = 1/6 = 0.1667


Let's compute the probability of getting doubles x = 0 and x = 1 time. To find this, you would use the binomial probability distribution formula.



P(X = x) = (n ncr x)*(p)^(x)*(1-p)^(n-x)

P(X = 0) = (8 ncr 0)*(0.1667)^(0)*(1-0.1667)^(8-0)

P(X = 0) = (8 ncr 0)*(0.1667)^(0)*(0.8333)^(8-0)

P(X = 0) = (1)*(0.1667)^(0)*(0.8333)^8

P(X = 0) = (1)*(1)*(0.232494)

P(X = 0) = 0.232494


P(X = x) = (n ncr x)*(p)^(x)*(1-p)^(n-x)

P(X = 1) = (8 ncr 1)*(0.1667)^(1)*(1-0.1667)^(8-1)

P(X = 1) = (8 ncr 1)*(0.1667)^(1)*(0.8333)^(8-1)

P(X = 1) = (8)*(0.1667)^(1)*(0.8333)^7

P(X = 1) = (8)*(0.1667)*(0.2790035137489)

P(X = 1) = 0.372079



Now add these final results:


0.232494 + 0.372079


0.604573



So P(X <= 1)  =   0.604573




Now subtract this from 1 to find P(X > 1)


P(X > 1) =  1 - P(X <= 1)


P(X > 1) =  1 - 0.604573


P(X > 1) =  0.395427



So the probability that you will get "doubles" more than once is roughly 0.395427