[andreareina]

Take a 3-bit counter:

    0->A
    1->B
    2->C
    3->A
    4->B
    5->C
    6->A
    7->B

A and B get hit three times while C only twice, so it will see 66% utilization compared to A and B

EDITED s/once/twice/ thanks CyberDildonics

[CyberDildonics]

while C only once

You listed C twice

[woodruffw]

That's a typo. You can still see they're correct about the ratio (two "C"s for every three "A"s and "B"s).

[manigandham]

There's no ratio. It's even across all of them, as long as the integer keeps incrementing. One more number (9) instead of stopping at 8 and there would be an even spread.

[ahmedtd]

Not when the counter overflows back to 0. If it's a 3 bit counter, 0 is A again, not C.

[manigandham]

The comment says a 32-bit signed int. Where is the 3-bit assumption coming from?

[iforgotpassword]

The comment starts with that assumption for the sake of a concise example. do you expect them to write the whole sequence out using a 32bit counter? :-D

[andreareina]

Edited, thanks

[manigandham]

That doesn't change anything... it's still round robin. You just stopped at an arbitrary number of 8 integers instead of 9.

[lofatdairy]

It's not arbitrary, GP stated it was a 3 bit counter. In GGP (or something, not sure how far down in the thread we are), they were referring to a 32 bit int counter until overflow. If you bin each number from 0 to 2^32 - 1 by mod 3, you don't get 3 bins of equal sizes, 1 bin always comes out smaller.

[manigandham]

Smaller by a single number at most, it's effectively insignificant. Not sure where the 3-bit counter assumption came from as the original post said 32bit signed int.