have each prisoner make a mark on the wall of the living room the first time they go into the room. When the 100th guy goes in, he can count 99 marks (hopefully he can count) and will know that he is the last one.
I can't think of any real solution to this riddle...
well, using statistics, he can only be sure in the 100x99+1st day, which is the 9901 day.
Because chances that it won't be him that day (don't try to understand how I reached that, I used some on/off lightbulb logic) are only the chances that 99 prisoners were selected in 9901 days, making it 99100*9901), or rounding to 99:99000, being 0.001:1, or 0.1:100, 0.1%, giving him 99.9% certainty.
So if this riddle is based on statistics, it is illogical.
So I hope there is some catch here I missed.
I'd be interested to see a working solution for it.
1. Theoretically they could never all reach the living room as the random pick could always hit the same guy.
2. The first point aside it was possible that they can see the light in the living room from their cells. Then they could count the changes if everyone would toggle the light on his first visit.
But that would be a boring solution.
Originally posted by Hitro I'd be interested to see a working solution for it.
1. Theoretically they could never all reach the living room as the random pick could always hit the same guy.
2. The first point aside it was possible that they can see the light in the living room from their cells. Then they could count the changes if everyone would toggle the light on his first visit.
But that would be a boring solution.
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100*9901), or rounding to 99:99000, being 0.001:1, or 0.1:100, 0.1%, giving him 99.9% certainty.
