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Coscott comments on Open Thread for February 11 - 17 - Less Wrong Discussion
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Ebyy n friragrra fvqrq qvr naq n svsgl guerr fvqrq qvr (fvqrf ner ynoryrq mreb gb A zvahf bar). Zhygvcyl gur svsgl-guerr fvqrq qvr erfhyg ol friragrra naq nqq gur inyhrf.
Gur erfhyg jvyy or va mreb gb bar gubhfnaq gjb. Va gur rirag bs rvgure bs gurfr rkgerzr erfhygf, ergel.
Rkcrpgrq ahzore bs qvpr ebyyf vf gjb gvzrf bar gubhfnaq guerr qvivqrq ol bar gubhfnaq bar, be gjb cbvag mreb mreb sbhe qvpr ebyyf.
You can do better :)
Yeah, I realized that a few minutes after I posted, but didn't get a chance to retract it... Gimme a couple minutes.
Vf vg gur fnzr vqrn ohg jvgu avar avargl frira gjvpr, naq hfvat zbq 1001? Gung frrzf njshyyl fznyy, ohg V qba'g frr n tbbq cebbs. Vqrnyyl, gur cebqhpg bs gjb cevzrf jbhyq or bar zber guna n zhygvcyr bs 1001, naq gung'f gur bayl jnl V pna frr gb unir n fubeg cebbs. Guvf qbrfa'g qb gung.
I am glad someone is thinking about it enough to fully appreciate the solution. You are suggesting taking advantage of 709*977=692693. You can do better.
You can do better than missing one part in 692693? You can't do it in one roll (not even a chance of one roll) since the dice aren't large enough to ever uniquely identify one result... is there SOME way to get it exactly? No... then it would be a multiple of 1001.
I am presently stumped. I'll think on it a bit more.
ETA: OK, instead of having ONE left over, you leave TWO over. Assuming the new pair is around the same size that nearly doubles your trouble rate, but in the event of trouble, it gives you one bit of information on the outcome. So, you can roll a single 503 sided die instead of retrying the outer procedure?
Depending on the pair of primes that produce the two-left-over, that might be better. 709 is pretty large, though.
The best you can do leaving 2 over is 709*953=675677, coincidentally using the same first die. You can do better.