>
>I believe that it is possible to keep the 9 first (or last)
>prisonners alive no matter what, and the 10th would have
>50%, because he can't recieve any data from behind.
>
>Please help!

I think I see a way to do this. The key is to pass as much information as possible forward, and the problem is that in your solution, only half of the prisoners pass information forward. The hard part is that the prisoners seem to have a choice: pass information forward OR save themselves. But it is possible to do both! The method makes use of "white hat parity," i.e., the oddness or evenness of the number of white hats.

Let prisoner number 10 say "white" if he sees an odd number of white hats, otherwise he says "black". Then he has a 1/2 chance of going free. Then prisoner 9 counts the white hats he sees in front of him, and if he agrees with prisoner 10 about the parity, he knows his must be black, so he says "black," otherwise he says "white." But this not only saves prisoner 9, it also tells prisoner 8 what to expect: Since 8 heard what 9 and 10 said, he knows what parity 10 saw, and also knows whether 9 saw the same. Therefore he knows what parity to expect, and just as before, if he counts the white hats he sees and comes up with the same parity as 9 did, he knows his hat is black, otherwise it is white. This clearly works for all the prisoners.

A simple way to state this rule is: Every prisoner starts by predicting even parity, and switches this predicted parity every time he hears the word "white." He then counts the white hats he can see to find his observed parity. If it agrees with the prediction, he says "black," otherwise he says "white."

There is of course no way to guarantee saving prisoner 10, but all others can be guaranteed saved by this method. Thanks for an interesting puzzle.

--Stuart Anderson