Further solutions were submitted by Joe Thornton.  Further solutions were submitted by Philippe Fondanaiche, Nancy Schwarzkopf, Lou Cairoli, Ron Welch, Bill Webb, Stuart Gale.

The product of the resulting differences will be odd only if each of the differences is odd. (Note that the integer zero is even.)

If you are looking at the integers from 1 to 9, there are five odd integers and four even integers, so at least one of the factors will be the difference of two odd integers, which is even.  In this case the resulting product will always be even -- a suckers bet.

If you have the integers from 1 to 10, then an odd product is possible -- consider the rearrangement 10,9,8,7,6,5,4,3,2,1.  For this to happen, however, the odd integers in the rearrangement must sit in the one of the five odd positions, while the even integers must sit in one of the five even positions.  The probability of this happening is (5! 5!) / 10!, a little less than .4% of the time.

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