The Game

There are two players, player AA with data set AA and elements aia_i and player BB with data set BB and elements bib_i. The arithmetic mean of each data set is 10 and 20, respectively.

A "game" is played in which each aia_i is compared to its corresponding bib_i with the same base. If the element from set AA is less than its corresponding element from set BB, then player AA is awarded 1 point and visa versa with player BB.

What is the probability that player BB has more points than player AA

You may manipulate the following or exclude them if they are not necessary. This is to either a) make this problem easier, or b) to make the problem possible.

A) The range of the data in both sets. (Make this 0,0,\infty if possible)

B) The number of elements in each set. Notes:

     ~~~~~ • The number of elements in sets AA and BB must be the same

     ~~~~~ • The number of elements in set AA determines the number of matches.

     ~~~~~ • I don't think the number of elements matters, but it might

Note by Trevor Arashiro
4 years, 6 months ago

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So many possibilities here, depending on the particular distributions. A "game theory" question could go like this: Suppose player AA must choose NN (positive integral) scores with the only stipulation that they have a mean of μA\mu_{A}, and player BB must similarly choose NN (positive integral) scores that have a mean of μB.\mu_{B}. The scores are then ordered by each players and compared, in order, to those of the other player with the tally system you have indicated. What is the best strategy for the two players to achieve the maximum respective tally? Given that the two play optimally, what is the likely final tally as a function of N,μAN, \mu_{A} and μB\mu_{B}? I know that this isn't quite what you had in mind, but I thought I'd just throw it out there, anyway. :)

Brian Charlesworth - 4 years, 6 months ago

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Yes, this is a much more feasible task. That turns this from more of a computer science problem into a combinatorics/NT problem.

Trevor Arashiro - 4 years, 6 months ago

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Assuming that the number of elements in each set is the same and knowing the fact that the arithmetic mean from B is two times as the arithmetic mean from A, I think the probabilitiy is 1/2

Catalina Villegas - 4 years, 2 months ago

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