A prison warden is responsible for 10 prisoners on Death Row.
One day, the warden offers the prisoners one chance at freedom.
Each of the prisoners will be placed in solitary confinement,
(in completely soundproof cells) with absolutely no way to communicate with one another.
The warden will randomly take one prisoner at a time to another room containing two light switches side by side.
The switches are not connected to anything.
The warden makes it clear that, due to the random process, prisoners may be taken to the room any number of times.
At the beginning, both switches will begin in the “Off” or “Down” position.
The rule is each time one of the prisoners enters the room he must flip exactly one of the switches.
If one of them correctly announces that all 10 prisoners have been in the room, then they will all be released.
If one of the prisoners incorrectly claims that all 10 have been in the room, then in comes the electric chair!

The prisoners are given some time to come up with a plan before placed in solitary confinement.
How do they ensure their freedom?

Are the switch positions the only thing that can signal the inmates in some manner?

Or, are you being devious making us think that there's ten inmates when in fact there's only two because the number you've written is binary?

Like your style Cue. You know CD all too well to know he is very likely to throw a twist like that into the problem

Well surprise surprise guys...had a moment of weakness:
absolutely no "tricks" or "deviousness" !

ONLY the switch positions can be used as signals.

AND:
prisoners = 10 is a number I picked; the puzzle could
be worded like: there are P prisoners.
In other words, any number greater than 1.

So ya'll may now proceed being assured that the
solution is FULLY logical ... and btw quite "cute".

There don't seem to be any time constraints in this problem so .... If the prisoners wait long enough, the odds of not having all ten prisoners enter the room will become vanishingly small.
I think they should wait until some arbitrarily large number of room entries would have been made and then claim all have entered. At least they will all be alive during the waiting period.

I'd go for 100,000 or more entries as being sufficient.

Nope Robert...there is a perfectly valid/logical solution.

Hints:
1: in their plan, one of the prisoners is chosen to be the "counter"
2: call the switches A and B: on 1st visit by a prisoner,
swith A flipped; any furter visits, this prisoner flips switch B

CD,
I don't see that your hint can provide sufficient information. The "counter" might think that if he stops seeing switch A being changed that all prisoners have been in the room but that would be incorrect.
Random entry could have several prisoners never go into the room which would leave switch A subsequently unchanged but unrelated to the number of different prisoner entries. All the counter could then tell is if there is a change in switch B which is also insufficient because even numbers of entries leave switch B unchanged.
For the sake of argument, let random selection exclude two prisoners with the "counter" being the "last" entry. The total in would be odd (7), switch A would be Up which the "counter" would turn Off/Down, and the counter would then only see switch B being changed. How is this different from all other nine having entered?

What is certain is with switch A being Up, an odd number of non-repeat entries have been made. With switch A Down, an even number of non-repeats have been made.
The "counter" would eventually need to see switch A Down if all 10 prisoners have entered but that can't be distinguished from any other even number of unique prisoners.

To make matters worse, the "counter" could be the first in the room and never reenter. Then what would be counted?

Robert, the solution must assume that at one
point, everybody has entered the room at least
once, and the counter at least as often as
required "to do his job".

Next hint:
when a prisoner (not including the counter) enters
room for FIRST time, he flips "ON" switch A,
IF that switch is in the "OFF" position; if not,
he flips switch B (up or down, depending; position
of switch B is not important).
And that prisoner needs to come back...

I hope this is logical - the reasoning made sense to me at the time!
One prisoner is counter and he is the only prisoner allowed to flick switch A into the off position. The others must flick the switch A up on precisely one of their visits only. They must look at both switches when they enter the first time. If A is in the off position the prisoner must put it in the on position, otherwise he flicks switch B, but flicks on switch A on the first visit he finds it in off. On subsequent visits they always flick switch B. The counter will always switch A into the off position when he enters the room (if it’s already off, he uses the other switch.) When he has counted nine switches to the off position on switch A he will know the other nine have been through.

PERFECTO, Jill!!!!!!

Apologies if the "wording of the puzzle" created
some confusion. I did my best to word it as clearly
as possible; here's how the darn thing was worded
when I first saw it:
.................................................
Switches off to on
Freedom or Alligators
Once upon a time, a prison warden was responsible for 22 prisoners on Death Row. These prisoners were students who had done terrible things: some illegally downloaded movies and music, some texted during classes, some were addicted to Facebook, and some watched Glee.
One day, the warden offers the prisoners one chance at freedom. After a brief discussion period in which they could plan their strategy, each of the prisoners will be placed in solitary confinement (in completely soundproof cells) with absolutely no way to communicate with one another. The warden will arbitrarily take one prisoner at a time to another room containing two light switches side by side. The switches are not connected to anything, but the warden tells the prisoners that at the beginning of the entire process both switches will begin in the “Off” or “Down” position. The rules are that each time one of the prisoners enters the room he or she must flip exactly one of the switches.
The warden tells the prisoners that if one of them ever correctly announces that all 22 prisoners have been in the room that they will all be released. However, if any of the prisoners ever incorrectly claims that all 22 have been in the room, then all 22 will be fed to the warden’s pet alligators.
The warden makes it clear to that prisoners may be returned to the room any number of times, but there is no way for the prisoners to communicate with one another other than by way of the two light switches. The room with the switches will be thoroughly cleaned after each prisoner leaves.
The prisoners are given some time to come up with a plan before they are to be placed in solitary confinement.
How do they ensure their freedom?
.................................................

So I re-worded that mess !!!!

Ah, those bought up on text adventures have a logical mind!!

Bravo, Jill!

But as Robert wrote, who's to say the person they picked as counter is even going to be able to go in enough times to count. Of course that is the logical answer for the counter to use the one switch, but if he only goes in 5 times, how do the rest get counted?

It's "probability", Flo.
We know the probability of rolling a 2 (6-sided dice) is 1/6...
But it's possible not to roll a 2 in 1st 1000 rolls!

Same with this puzzle/problem: possible that it takes years before the counter gets 9.

Well there ya go. Thanks