Two persons each divide $1000 into three different amounts; whoever has the larger amount in two of the three comparisons wins.
decision/three.amounts New
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Problem
Two persons each divide $1000 into three different amounts; whoever has the larger amount in two of the three comparisons wins. What division is best against an opponent who divides at random? 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? -- Marco Meijers [ Editor: the title sentence above is a summary; the original question follows it. ]
Show solution
A theoretical solution to this problem can be found in: http://www.mathsoft.com/puzzle/soln11/soln11.html It turns out that the solution is indeed 231 + 384 + 385, provided that the other person always makes a random choice, of course. -- J. H. Meyer <wwjm@wwg3.uovs.ac.za> [ Editor: the analysis at the MathSoft page is not in the collected material; "indeed" refers to earlier postings in the thread that are also not collected. ]