There are 100 prisoners, numbered 1-100. The prisoners' numbers are written onto 100 cards, put randomly into 100 envelopes numbered 1-100. They can't communicate, but are invited, one by one, to open 50 envelopes. If any prisoner fails to find his number, they all get killed.
-
Show this thread
-
At best, there's a 50% chance any one prisoner can find his number. It would seem like there's a probability of (½)¹⁰⁰ or 0.00000000000000000000000000008% that every prisoner could find his own number, if each were to check 50 envelopes at random.
2 replies 0 retweets 1 likeShow this thread -
They could certainly do even worse than this. If they all agreed in advance to open envelopes 1-50, then fifty of them would be guaranteed never to find their number and they'd all die. But could they come up with a strategy in advance which would improve their survival chances?
3 replies 0 retweets 0 likesShow this thread -
What makes this my favourite combinatorics problem is that there *is* a strategy which increases their likelihood of survival from 0.00000000000000000000000000008% to about 30%. So, what is the strategy, and why does it work?
12 replies 0 retweets 3 likesShow this thread -
Replying to @propensive
Coud the 2-3 first guys that will open the envelope sort them from 1 to 100? (or by pile of 10)
1 reply 0 retweets 0 likes
No, that would constitute communication, so it's forbidden. :)
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.