you randomly choose t=1000 distinctive numbers between 1 to n=2010, i.e., 1000 #s from the set {1, 2, ..., 2010}.is the probability that their sum is divisible by p=5 greater than, equal to, or less than 1/5?(WARNING: MC does not work well here, the answer is within an extremely tiny neighborhood of 1/5)the original problem has n=2000. anyone has a solution for the general case (t,n,p)?

My first answer on this forum.I think it's greater than 1/5.If you take how many numbers you pick between 1 and 5, you have the following case:* if you take 1,2,3 or 4 numbers, their sum is 0,1,2,3 or 4 (mod 5) with equal probabilities* if you take 0 or 5 numbers, their sum is 0 (mod 5) and you can have a recurrence with (995,2005,5) or (1000,2005,5)As 2010 is divided by 5, you will have some cases and your probability is greater than 1/5.For 2011, the probability is lower than 1/5.I'm a bit lazy to calculate exactly but for exemple (5,10,5) is 52/252>0.2Hope it will help

welcome, Vanubis. yes, your answer is correct.for anyone interested in the exact prob (and a more detailed treatment), i was pointed to this paper by Wagon and Wilf.