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The solution assumes no prisoners die before they flip A twice. No fault tolerance : p
If no announcement is made when someone dies, however, you've got a problem.
The answer: http://www.cartalk.com/content/prison-switcharoo-0?answer
Doesn't this solution make the assumption that the guard will continue to pick prisoners to enter the switchroom indefinitely? If he picked P1, P2, P1, P2, Counter, Counter, would the solution still work?
Can the prisoners tell if the switch is in the on or off position?
I say mostly because the bit where one switch must be toggled and there are 2 switches makes it a bit different from 23C2. I don't think the difference is useful, though; the set of solutions is probably the same as the solution set for 23C2.
Earlier quoted context omitted.
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…
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…
Since there's a password article on the front page too... This problem is very clever, and there is a very clever way to solve this, assuming everyone plays nice. However, that solution assumes perfect collusion. These are prisoners we're talking about, so at least in some cases, we have to assume there may be bad actors involved. I mean this is a cute problem assuming the following: 1. All the prisoners have perfect…
Give current society and the normal constraints on these popular games the answer to all the above are incredibly obvious, the answers you ask can be deducted. (No prisoner can die, if they could it would have been hinted at)
It's like claiming you can't do cryptic crosswords because the questions are not clear or don't follow proper English.
Although I imagine you are just trying to be difficult :)
The answer includes a major (and I believe flawed) assumption, which is that the visits will be uniformly distributed. However, the puzzle states that "I may choose the same guy three times in a row".
The Counter could visit the switch room 44 times without flipping the switch if such visits were the first 44 chosen. After everyone visits (which would be 1,012 visits since "given enough time, everyone will eventually visit the switch room as many times as everyone else"), the Counter will be no closer to knowing the truth.
Of course, it is true that the greater the number of visits the greater the probability that the Counter's final visit will fall after everyone has visited, but there is no guarantee.
Regardless of how many visits you assume for each prisoner, there will always remain a probability that the Counter's visits will be clustered early in the visitations. In such a scenario, the Counter's role becomes useless and all prisoners will die in prison.
The answer: http://www.cartalk.com/content/prison-switcharoo-0?answer