Prisoners and Boxes

Originally posted by uzless
There must be a formula. The math guys here should be able to figure it out. PBE6 knows this stuff better than i do. Fabian seems good also.

Me? I have not the faintest idea.
Perhaps I could count something out if I had the time. As a prisoner I would have the time and hence perhaps solve the problem. But for here and now, I just enjoy the discussion.

Originally posted by XanthosNZ
They either succeed or they don't, it's 50/50.

ITS NOT 50/50,IF THE 1ST PRISONER DOSE NOT GET IT THE 2ND WILL

Originally posted by XanthosNZ
They either succeed or they don't, it's 50/50.

That's how I figure the odds when I play poker...I figure I've got a 50% chance of winning each hand. I'll either win the hand or lose the hand.