Probability of 3 dice having at least two same number

Probability that at least two have the same number = 1 - probability that all three numbers are different = 1 - (5/6)*(4/6) = 1 - 20/36 = 1- 5/9 = 4/9 = 0.444...

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There are 6³ = 216 possible sets of three numbers.  If you take Melody's 96 and divide by 216 you also get 0.444...

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Lets assume the die are rolled one at a time (it makes no difference)

The number of ways you can get 1,1, not 1 is 5

The not 1 can be 1st 2nd or 3rd.     That is 3 different ways so that makes   5*3=15

then you could have 1,1,1,                                 so that makes 15+1=16

This would be the same for all the other pair combinations and there are 6 numbers on a die so

16*6 =  96 possibilities.     

That is what i think :)

Probability that at least two have the same number = 1 - probability that all three numbers are different = 1 - (5/6)*(4/6) = 1 - 20/36 = 1- 5/9 = 4/9 = 0.444...

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Thanks Alan,

I guess i didn't read the question properly :(

Oh well at least i answered 'my' question correctly    

Thanks unknown people of the internet.