Explore BrainMass

Explore BrainMass

    Die throw problem

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    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.

    © BrainMass Inc. brainmass.com March 5, 2021, 12:28 am ad1c9bdddf


    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.