The only strategy I can think of using the light switch is this: the prisoners decide that when each one goes into the room for the first time he notices the light position (on or off); then, when he leaves he flips the switch. If the prisoner returns to the room the second or more times he notices the light position but does not flip the switch when leaving.
So when a prisoner comes into the room for his first time, if the light is off he knows that an even number of prisoners have been in the room, and if the light is on he knows that an odd number of prisoners have been in the room.
If he also knows when the process started and knows how many times some prisoner has been in the room then he can start to make inferences using the probability approach that David has outlined.
If there are only two prisoners instead of 100, this approach would allow the second prisoner to enter the room for the first time to know that the other prisoner had already been in the room one or more times (but in this case he also would know by counting the 15 minute intervals). If there are three prisoners and prisoner #3 enters the room after 2 or more selections and he notices the light is off he would know that the other two have been in the room. If the light is on he would know that one of the others has been in the room twice and the remaining prisoner has not been in the room. Probabilities enter the equation as the number of prisoners increase.
And in any event, if there are n prisoners each prisoner must wait for at least n/4 hours to elapse before there is even a chance for all prisoners to have been in the room.
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- Late Friday Puzzler -- 100 Prisoners