Explore BrainMass

Die throw problem

We repeatedly throw a die, stopping when the value of the throw exceeds the value of the first throw. Compute the expectation value of the number of throws.


Solution Preview

It is convenient to consider infinite sequences of die throws and then consider the subsequence of the subsequent entries starting from the first entry that satisfies the rules. So, if the sequence is 3, 2, 2, 1, 5, 2, 6,...., then the subsequence is 3, 2, 2, 1, 5, because the throw of 5 is the first one that is equal or larger than the first throw. The probability that under the rules you would have n die throws is thus the same as the probability that a subsequence has length n. The probability that the subsequence of a sequence starting with r has a length of n is:

p(r,n) = 1/6 * [(r - 1)/6]^(n-2) * (7- r)/6

The first factor of 1/6 is the probability that the first die throw yields r, the second factor is the probability of having n-2 die throws that are below r, the ...

Solution Summary

We solve this die throw problem from first principles. The value of the number of throws are given.