You roll N (N > 1) number of M (M > 1) sided dice at once.
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?
Answer/Solution
1/(M^(N-1))
Steps/Work
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: C.
= 1/(M^(N-1)).
Answer: C.