Re: Approximating Multiple Opponents
Dear Gamey Mouse
Your are correct -- at least I got same answer as you did of 35%. I used a brute force counting scheme (numerical enumeration). Three players: A, B, & C. A beats B 70% of the time; & A beats C 30% of the time. 100 permutations covers all possible cases.
Evidently you solved it the easy way. I did it using brute force on an excel spreadsheet -- what a waste if time my way was....
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