If you choose three distinct integers between 1 and 63, what is more probable: that their sum will be less than 95, or greater than 95?
The integers could be described as being between 64-63 and 64-1, (those are subtraction signs), or between 64-1 and 64-63. The total would be 64*3 minus a random triplet whose members are between 1 and 63. That's 192 minus the original distribution, so the probability that the sum is less than 95 is the same as the probability is greater than 192-95, which is 97. The probability that the number is more than 95 is higher than that it is over 97, and thus higher than that it is over 95.
You could also just recognize this as being the same as tossing 3 63-sided dice and therefore have a symmetrical distribution, and, as the mean for the total is 3*62/2 = 96, it's more probable that the sum be greater than 95.
Posted by Charlie
on 2006-12-01 08:49:39