Re: math problem
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So what we do is take the 3 opponents from the loser finalist and pair them off with each other . So we have 3 possible choices to determine the second heaviest from that side of the bracket . Now we take this coin and pair it off with the 3 losers on the winners side of the bracket . Now we have to compare the heaviest of the 6 weighings with the loser finalist and we get 22 in total . Best case scenario is that it can be done in 21 attempts but not always !
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There are 4 possible 2nd place coins, and up to 5 possible 3rd place coins.
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10-12-2007, 12:37 PM
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