Two persons each divide $1000 into three different amounts and compare highest with highest, middle with...
decision/divide.1000 New
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Problem
Two persons each divide $1000 into three different amounts and compare highest with highest, middle with middle and smallest with smallest; whoever is larger in two of the three comparisons wins. What division is best against an opponent who divides randomly? Two persons each divide $1000 into three DIFFERENT amounts. Then they compare their highest, medium and smallest amounts. The person who has 2 of the 3 times the highest amount, gets $1000. The question is what division is most profitable, assuming that your opponent divides randomly? The following division emerged from my computer simulations: 401/400/199. But there should be a better one. Can someone come up with a better solution? -- Marco Meijers
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nh <noelh@hooked.net> wrote: >I'd go with 1 + 499 + 500 >figuring I'd always lose the minimum, but win the medium and >highest, winning the 1000. I'll take 2 + 3 + 995 and beat you. :-) Actually, I analyzed all 83,333 different partitions of 1000, and found that 231 + 384 + 385 wins over 63833 of them, while 1 + 499 + 500 wins over only 21,082. I'm still trying to decide if each of the 83,333 different partitions is equally likely in a random partitioning, or if some of them are more likely than others. I counted them all equal in the above. -- Dave Dodson <dodson@convex.com> [ Editor: 83,333 = round(1000^2/12) is the number of partitions of 1000 into three positive parts with repeats allowed; with three different parts, as the problem requires, there are round(997^2/12) = 82,834. ]