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Atleast 2 are dice need to have same number, which means 2,3,4,5 and 6 can have same.

So we can compute the prob of all have different and subtract that from 1 which will give us the answer.

For the first dice, the prob is 1 since any value is fine.
For the second, the prob has to be other than first. So 5/6
For third, the prob has to be other than first and second. So 4/6.
For fourth, 3/6
For fifth. 2/6
For sixth, only one remains so 1/6.

So the probability is (1).(5/6).(4/6).(3/6).(2/6).(1/6) = 5/324.
So the answer is 1-5/324

why do not we start with 6/6 and end with 2/6, instead we start with 5/6 and end with 1/6 ?

for example:

to find number of ways 6 objects can be organized in group of 6 (order matters) 6*5*4*3*2*1

to find number of ways 6 objects can be organized in group of 4 (order matters) 6*5*4*3

from here i thought we should start with 6/6 and end with 2/6...?

Please explain y not, thank you.

I think what defenestrate is trying to get at is the fact that 6/6 is just 1, i.e. doesn't matter what number we get in the first roll. What we are concerned with is the fact that this number not show up again, hence starting with 5/6 (the prob of 2nd roll) and so on and so forth. In the case of 6 rolls, the last roll has a prob of 1/6 [5/6*4/6*3/6*2/6*1/6] = 5!/6^5 which is the same as 6!/6^6 if you put 6/6 as the first roll instead of 1.

In the case of 5 rolls, the last roll has a prob of 2/6 [5/6*4/6*3/6*2/6]= 5!/6^4 which is the same as 6!/6^5 if you put 6/6 as the first roll instead of 1.

Additionally, for the sake of completeness, the prob of 5 dice is [1- (120/1296)] = 49/54 not 8/9.

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