88 people stand in a circle, each having a hat with a number from 0 to 87 written on it. Everyone can see the numbers on other people's hats but can not see his own number. They simultaneously write a number on a piece of paper and give it to the judge. If at least one of them wrote a number that is on his own hat then everyone wins, otherwise everyone loses. What strategy should they use to guarantee victory?
(Numbers on the hats do not have to be all different. People can not exchange any information during the procedure but can agree on some strategy beforehand.)
88 was a bit too large for me, so i decided to start easy, and see if a simple logical plan could be deduced (without reviewing the solution provided).
One hat is very easy. Just pick the number.
Two hats is slightly more difficult, but not much. Gurer ner gjb tebhcf bs ahzoref, bar jurer gurl ner gur fnzr, bar jrer gurl ner qvssrerag (rnpu tebhc unf gjb zrzoref). Gurersber, gurl fubhyq ynory gurzfryirf N naq O. N fubhyq pubbfr gur fnzr nf O, naq O fubhyq pubbfr gur bccbfvgr bs N. Gung jvyy pbire obgu pnfrf.
I'm working on three, on and off. I'm guessing the strategy should be somewhat similar to two, but it is obviously different being each person has to deal with multiple others. Four and on, i would think, should then be similar.