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math exercise (no prize for this :)

Posted By: Joe Russell
Date: Tuesday, 10 May 2011, at 12:41 a.m.

In Response To: math exercise (no prize for this :) (Chuck Bower)

The first question seems easy: The chances that all four picks will be in different quadrants are 12/15 * 8/14 * 4/13 =.1407. Now you need 2 wins by each player and two additional wins by one player. Sounds like .1407 * .5^10= 1 in 7280. The chances of not getting a correct pick is 7279/7280. (7279/7280)^5046 = .5 You would need 5047 different entries to have more than a .5 chance of hitting. Guessing that 1% of the entries are duplicates, you would need 5098 to get 5047 different entries.

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