Tom K. answered 10/18/20
Knowledgeable and Friendly Math and Statistics Tutor
I analyzed the problem in the following manner.
Let X1, X2, XN be U[0,1], then let the percentage of Xi be Xi/∑Xi
All variables less than Xmax will be U[0, Xmax]
Then, to figure the probability that the sum of all Xi < Xmax be greater than Xmax, note that we can transform the variables, and let Xmax = 1, Xi ~ U[0, 1}, Xi not equal to Xmax
Then, the volume of ∑Xi <= 1 = 1/(n-1)!, and the volume over all the Xi is 1, so the probability that there will be a runoff is 1 - 1/(n-1)! or (n-1! - 1)/n-1!
Tom K.
The key is seeing that the problem is equivalent to finding the probability that the sum of n-1 U[0,1] random variable sum to greater than 1. How you solve that, or the complement, that the sum is less than or equal to 1, is up to you.10/19/20
Wilma J.
I understand that. I was wondering if there was a way to approach this from Bayes’ theorem or Bayesian statistics?10/19/20
Tom K.
You are using Bayes' theorem in that you are conditioning the distribution of the n-1 other variables on the value of the maximum; by knowing the max value, the distribution of all other variables is U[0, x max]. This is why, we can transform these to being U[0,1] with x max = 1 and just be calculating that the sum of the n-1 other variables is greater than 1 in order to have a runoff.10/20/20
Wilma J.
Thanks.10/21/20
Wilma J.
Are there other ways to approach this problem?10/19/20