Mind if I ask where the other 13/16 went? Since rand() % 10 can only output 0-10 there are only 10 possibilities, all of which are covered in your two ranges.
Either people frequently have problems with statistics as well, or random numbers are really complicated ;)
Well, I'm totally not an expert in "randomness assisted by computer", I just know it's something, like cryptography, where you should be careful before implementing your own solution.
And I am bad in maths, and statistics/probas are so tricky (at least for me), you are right.
However, a code demonstration (in Perl, sorry, for perf/convenience):
# We fake a bad `rand()' function by only taking
# values between 0 and 15.
$cnt{rand(15) % 10} += 1 for (0..1_000_000);
print $_, "\t", $cnt{$_}, "\n" for (sort keys %cnt);
What you said is correct; the only problem is that with the numbers you quoted, the total probability of rolling a number in the range [0,10) is 2/16 + 1/16 = 3/16 < 1 :) The numbers you meant to quote, as borne out by your data above, are a 10/16 = 5/8 chance to get a number in the range [0,5), and a 6/16 = 3/8 chance to get a number in the range [6,10). (Equivalently, a 12/16 = 3/4 chance to get a number in the range [0,5] and a 4/16 = 1/4 chance to get a number in the range [6,9], but since that doesn't partition the [0,10) range evenly, it's less instructive.)
