3 ms·

Because there's no limit to the amount of time they can play. Not just if P2 is never at the switch, but imagine even if the "counter" player is never at the sw

Because there's no limit to the amount of time they can play. Not just if P2 is never at the switch, but imagine even if the "counter" player is never at the switch too. Statistically, when talking about an infinite run-time, there's no such thing as "never". Even if it takes a million years, they'll eventually all get turns in the switch room. Or, if you want to look at it more realistically, the run-time is limited by their lifespans I suppose. But given that they're prisoners who all presumably have life sentences, then they'll just play until they die, since the alternative is to simply be in prison until then anyway.

Udo 13y ago
I knew why I always hated these games, because there are some rules which apparently aren't a big deal, while others seem to matter a lot. In this case, the problem asked for a precise solution, which was given in the resolution example, with no thought at all to that nagging little probability problem. So in order to "solve" this, I would have to know that everybody is basically supposed to disregard certain aspects of the problem. That's gotcha-type stuff that I always get wrong.
- chrismcb 13y ago
  What aspect is being disregarded? Previously you asked " Prisoner 2 is never at the switch at all before the Counter gets to his magic number" Except that if Prisoner 2 is never at the switch, the counter can never reach the magic number. Assume there are 3 prisoners (1,2,C), the magic number is 4. But if the sequence is 1,C,1,C,1,C.... The count will never be greater than 3 (2 is the switch starts as off) The warden said everyone will eventually visit the room. The one thing that was assumed (but not spelled out explicitly) is the game goes on forever. Which means at some point 2 visits the room a bunch of times, and then C will visit again, and will meet the magic number. Nothing to do with probability. And there is no guessing, 44 is a precise number.

Udo 13y ago

I knew why I always hated these games, because there are some rules which apparently aren't a big deal, while others seem to matter a lot. In this case, the problem asked for a precise solution, which was given in the resolution example, with no thought at all to that nagging little probability problem. So in order to "solve" this, I would have to know that everybody is basically supposed to disregard certain aspects of the problem. That's gotcha-type stuff that I always get wrong.

chrismcb 13y ago
What aspect is being disregarded? Previously you asked " Prisoner 2 is never at the switch at all before the Counter gets to his magic number" Except that if Prisoner 2 is never at the switch, the counter can never reach the magic number. Assume there are 3 prisoners (1,2,C), the magic number is 4. But if the sequence is 1,C,1,C,1,C.... The count will never be greater than 3 (2 is the switch starts as off) The warden said everyone will eventually visit the room. The one thing that was assumed (but not spelled out explicitly) is the game goes on forever. Which means at some point 2 visits the room a bunch of times, and then C will visit again, and will meet the magic number. Nothing to do with probability. And there is no guessing, 44 is a precise number.