you're getting upset because you're making a terminology error, you say "digit" but mean "digits". A digit is either a finger, or the numbers 0-9, singular.
The pattern is clear and ubiquitous -- if you actually know something, and if you're foolish enough to post a clear exposition, the morons who never post anything coherent will downvote your posts with a probability approaching certainty.
The modulo operator is the obvious solution to the original question. And if there were't an atmosphere of pervasive anti-intellectualism at HN, this thread would be much shorted than it is.
0 % 7 = 0
1 % 7 = 1
2 % 7 = 2
.
.
.
56 % 7 = 0
57 % 7 = 1
58 % 7 = 2
59 % 7 = 3
0 appears nine times.
1 appears nine times.
2 appears nine times.
3 appears nine times.
4 appears *eight* times.
5 appears *eight* times.
6 appears *eight* times.
You're arguing against something I never said. Large numbers reduce the bias, they don't eliminate it. Notice that earlier I considered the objection of a hypothetical clever child by using the minute digits as well. All these steps only minimize the bias in favor of small numbers, that cannot be eliminated. Even a 64-bit random number possesses this property. And the classic remedy for this bias is to apply the modulo operator to get the desired range.
You've failed to acknowledge the trade-offs of using your solution versus the OP's solution (or the solution above), and you're calling people who are trying to point out the trade-off "morons", that's why you're getting downvoted.
Would that solution be "multiply by the number of children and divide by ten"? Let's see:
n n * 7 / 10
---------------
0 0
1 0
2 1
3 2
4 2
5 3
6 4
7 4
8 5
9 6
> ... you're calling people who are trying to point out the trade-off "morons", that's why you're getting downvoted.
No, I am being downvoted because I'm right. Being right is simply rude, but being right about something trivially proven is beyond the pale.
No; his/her solution is: If 0-9 couldn't be divided evenly among the kids present, leftover digits would result in a re-roll, exactly as I said above.
> No, I am being downvoted because I'm right
Are you? This is math we're talking about, right and wrong are very precisely defined. For the sake of discussion, can you repeat the question you claim to have the right/correct solution to, and your answer?
This makes it less efficient - in this case, 18/60 chance of having to redraw, versus 4/60. Taking the modulus of the largest range you can leaves less residue (what lutusp is alluding to, without realizing others are describing a simple way of getting an exact uniform distribution).
If you want to make it more efficient (in terms of entropy used), you need to save the 1-of-prime (eg bits, trits, etc) that have been successfully chosen (uniformly). For example, with 6 children and an 8 sided die (3 bits), a roll of 6 or 7 would narrow the choices to only children 0-2 or 3-5 respectively. Your subsequent rolls could then be done with a 4 sided die.