You roll N (N > 1) number of M (M > 1) sided dice at once. What is the probability they will all land on the same number?	Answers
A) N/(M^N) B) M/(N^M) C) 1/(M^(N-1)) D) 1/(N^(M-1)) E) 1/(M-1)^N
The correct rolls are any M-ples: 11...1, 22...2, ..., MM...M. This is makes M possibilities. The N dice can roll M*M*...*M (M^N) number of ways. Thus the probability of landing on the same number is M/(M^N)   = 1/(M^(N-1)).   Answer: 1/(M^(N-1)).
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You roll N (N > 1) number of M (M > 1) sided dice at once. What is the probability they will all land on the same number?