From outside Bradley University, correct solutions were submitted by Will Troxel, Monty Gray, Burkart Venzke, Cyril Terakopiantz, Olus Kayan, and by Philippe Fondanaiche, who also solved the more general problem.

There are 500,000 even integers between 1 and one million (inclusive), and, of course, the same number of odd integers.  The probability the first number chosen is odd is .5 with the same being true for chosing an even number.  The probability the second number chosen has the same parity as the first is 499,999/999,999, while the probability it is of opposite parity is 500,000/999,999.  For the sum of two integers to be even, they must have the same parity.  Therefore the probability is greater that the sum is odd.

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Page last updated 3 April 2000.

ã2000 Alberto L. Delgado