I said "Each person in turn enters the room and opens the boxes one after another until either he finds the box with his name or he has opened 50 of the 100 boxes. "So it can be less than 50 also.....

QuoteOriginally posted by: animeshsaxenaQuoteOriginally posted by: cm27874A variant to the problem: what if people are allowed to peek into n-1 boxes? The probability of all people finding their rings can then be made at least ( (n-1)/n) )^n, which converges to 1/e.Again, one can do slightly better by considering permutations. If everybody peeks into all boxes but the one with his own name on it, the probability is the fraction of fixed-point free permutations (derangements), i.e. !n / n!. This expression converges to 1/e as well but stays above ( (n-1)/n) )^n (I guess...). For n = 4 it is 243/768 vs. 288/768.I think u got it!Person 1 enters the room...opens first box...finds 4 written on it...so he opens box 4...he finds 10...and he opens 10...it's a permutation cycle.Probability of permutation cycle...is all u need to know...which gives answer > 30%Sorry I made a mistake. Ur problem is very different. Peeking into n-1 boxes is totally different. U can increase the probability without it!You can only look @ 50 boxes...out of 100....(50 is max)...but can be less than 50..if u see the ring earlier! NO PEEKING NO CHEATING!

I can achieve zero for the probability that everyone finds their own ring: agree that everyone opens the first 50 boxes.

QuoteProbability of permutation cycle...is all u need to know...which gives answer > 30%Let me see... the probability of NOT everybody finding his ring is then equal to the fraction of permutations with a cycle of length > 50, which isNice!Just for completeness: in my variant of the problem, where n people are allowed to peek into n-1 of n boxes, the probability of everybody finding his ring can be made 1 - (n-1)! / n! = (n-1) / n.

QuoteOriginally posted by: outrunYou said "The 100 boxes are taken to a closed room, randomly shuffled and put down in a row.", so I bought 100 boxed and did that experiment.., and my sequence ended up have 50 cycles! ..ah, the catch is that you changed names into numbers to be able to do this..without the catch condition...it will become a simple formula application

previously under the radar: here (and recently), which points to Warshauer's paper proving the strategy is indeed optimal with the specified setting

QuoteOriginally posted by: wileyswpreviously under the radar: here (and recently), which points to Warshauer's paper proving the strategy is indeed optimal with the specified settingThanks....thats very interesting