MT, I agree that each prisoner has a 1/2 chance of wearing a black hat (in the absolute).
But let's say I'm prisoner #9 and that 7 out of the 8 prisoners in front of me are wearing a black hat. Then I would choose to say white, because the cumulative odds of 8 prisoners out of 9 wearing a black hat are lower than the odds of 7 black hats out of 9.
Still I would prefer to use words to save my life.
Pamela Anderson's Left Nipple wrote:
I was correct in stating that the solution, the assertion with 100% accuracy that all prisoners had been in the room, could never come to fruition.
You are confusing "accuracy" with "probability."
u r wrong wrote:
Pamela Anderson's Left Nipple wrote:I was correct in stating that the solution, the assertion with 100% accuracy that all prisoners had been in the room, could never come to fruition.
You are confusing "accuracy" with "probability."
not only does he have that word wrong, but he's talking about the wrong assertion. the assertion in the riddle, that all 100 prisoners have been in the room, will of course be possible. the assertion that cannot be made is that there will, with 100% probability, be an opportunity to make that assertion. and it's not a question of that assertion "ever coming to fruition", it is just plain wrong to make.
Alice Runner wrote:
Then I would choose to say white, because the cumulative odds of 8 prisoners out of 9 wearing a black hat are lower than the odds of 7 black hats out of 9.
Steve Wynn became a billionaire off of schmucks who thought that way.
Alice runner is right. With no other options available, playing the odds is the smartest way to go. I'm pretty sure Steve Wynn never made money off of prisoners on death row choosing hats.
Alice Runner wrote:
MT, I agree that each prisoner has a 1/2 chance of wearing a black hat (in the absolute).
But let's say I'm prisoner #9 and that 7 out of the 8 prisoners in front of me are wearing a black hat. Then I would choose to say white, because the cumulative odds of 8 prisoners out of 9 wearing a black hat are lower than the odds of 7 black hats out of 9.
Still I would prefer to use words to save my life.
The cumulative odds have no effect on your own odds, as the color of each hat is entirely independent. Yes, the probability that 8 out of 9 hats are black is very low. However, even if you count 7 black hats in front of you, the odds that your hat is black still remain at 50%. This method can only give you some psychological sense of control, because your odds will not change.
Similarly, every time you roll a die, you odds of rolling a 1 are 1/6. It is highly unlike that you can roll 1000 times without coming up with a 1, but even on your 99th roll, the probability that the NEXT roll will be a 1 remains 1/6.
Sagacious wrote:
Alice runner is right. With no other options available, playing the odds is the smartest way to go. I'm pretty sure Steve Wynn never made money off of prisoners on death row choosing hats.
But you can't play the odds in this scenario: every person has a 50% chance of wearing white or black, regardless of the other colors.
When an event is truly random, its probability never changes. You cannot be "due." This is how gamblers are duped: just because your cards have not come up 100 times in a row, they are no more likely to come up the next time around.
Pamela Anderson's Left Nipple wrote:
In other words, even though an accurate plan had been designed, it is theortically possible that the plan could never be implemented.
math teacher wrote:
Agreed there.
What makes you guys think it's theoretically possible the plan could not be implemented?
dukerdog wrote:
What makes you guys think it's theoretically possible the plan could not be implemented?
Because there is a theoretical possibility that all prisoners never even make it into the room. The prisoners are chosen at random, so the same prisoner could be chosen day in, day out--there is no guarantee that says every 5 years, for example, each prisoner will have been into the room.
As time extends to infinity, the limit of the probability that all prisoners have been in the room DOES reach 1, but we can't take this limit to infinity, since the prisoner's declaration requires an end date.
Okay, I haven't gone through the whole thread, but it doesn't seem too hard. Designate one person as chief, and use the following strategy. When a person enters the room for the first time ever, they turn the light bulb on if it is off. If it was on, they leave it on.
Whenever the chief enters, he turns the light off if it is on, leaves it off otherwise. Once he has had to turn it off 99 times, then everyone has been in the room.
Is it that simple?
Already suggested.
kartelite wrote:
Okay, I haven't gone through the whole thread, but it doesn't seem too hard. Designate one person as chief, and use the following strategy. When a person enters the room for the first time ever, they turn the light bulb on if it is off. If it was on, they leave it on.
Whenever the chief enters, he turns the light off if it is on, leaves it off otherwise. Once he has had to turn it off 99 times, then everyone has been in the room.
Is it that simple?
Chief, as well as many of his fellow inmates, would be dead by the time he turns it off 99 times. Even so, mathematically speaking, zero reward divided by zero risk is indeterminate.
malmo wrote:
Chief, as well as many of his fellow inmates, would be dead by the time he turns it off 99 times. Even so, mathematically speaking, zero reward divided by zero risk is indeterminate.
Not sure what that last bit meant--but according to most calculatoins I have seen, this solution, on average, takes about 30 years to complete.
i can think of a way to save seven guaranteed
the first three guys sacrifice themselves. they each count up the number of black hats among guys 1-7. then they respond like this
if 0 black hats, they answer "white, white, white"
if 1 black hat, they answer "white, white, black"
if 2, answer "w,b,w"
3, "w,b,b"
4, "b,w,w"
5, "b,w,b"
6, "b,b,w"
7, "b,b,b"
then the other seven guys will be saved. here is how:
the seventh guy knows how many total black hats there are among the first seven, but he can see the first six. so if he knows the first seven have four black hats total, but he sees only three, then he knows his hat is black. if he knows there are four black hats and sees all four, his hat is white.
then the sixth guy does the same thing. he can "see" the hat color of the seventh guy, who was behind him, because he knows the seventh guy's hat was whatever color he said it was. so the sixth guy pulls the same trick as the seventh, and the fifth pulls the same trick as the sixth, "seeing" two hats behind him, etc. up to the first guy
that way, you are guaranteed to save at least seven.
can you do eight?
Sagacious wrote:
Without any codes, #'s 10,8,6,4,2 would say the color of the hat in front of them. This way #'s 9,7,5,3,1 would guarantee survival. Plus average odds would say another two or three of the even numbered group lived.
With codes, Alice runner's solution is a good way to save 9 at least and leave the tenth at the mercy of chance.
I've done some sleuthing and from what I can tell, there is a solution, without coding, that allows 9 prisoners to survive, with the last prisoner left to chance. I haven't quite figured this out yet, but obviously the first prisoner, with his one word of "white" or "black," has to communicate enough information for the rest of the 9 prisoners to know their colors. The rest of the prisoners are locked into what they are saying by their own color.
So we've got to work out some mechanism by which more information can be conveyed than just the color of the person in front. I think I've worked out a mechanism that can be guaranteed to save a few more prisoners:
Prisoner 1 is assigned to save 2 and 3. If they have the same color, then he says the color of prisoner 10. If they have different colors, then he says the OPPOSITE of the color of 10. Now prisoner 2 looks at 10: if he hears the right color, then he says the color of prisoner 3. If he hears the WRONG color, he says the opposite of prisoner 3. Prisoner 3 can also work out his color; if 1 said the color of 10, then he says what 2 said. If 1 said the opposite color from 10, he says the opposite from 2.
I think I've worked out the possibilities, and this can also save 10, who listens for what 1, 2, and 3 say. If their sequence goes BBB, WWB, BWW, or WBW, he says black. If the sequence goes WWW, BBW, WWB, or BWB, then he says white.
Then 4 can do the same thing, indicating to 5 and 6 their colors by using the color of 9. Then 9 can use the same set of sequences to determine his own color. Finally 7 just tells 8 his color.
So there are 3 sacrifices and 7 saves.
I thinking the solution to save all 9 has to be along these lines.
oh wait, you can do nine.
the first guy just says whether he sees an odd or even number of black hats. if he sees an odd number, he says "black". otherwise he says "white"
suppose he says "black", so there are an odd number of black hats among the first nine guys. then the ninth guy looks and if he sees an odd number of black hats, he says "white", otherwise he says "black". he lives.
the eighth guy can "see" the ninth guy's hat color because he knows the ninth guy lived. he also knows whether the total number of black hats among the first nine is odd or even. so he pulls the same trick as the ninth guy did. this carries up the line until the first guy, who knows the color of everybody behind him's hat, and therefore because he also knows if the total number of black hats was odd or even, he can surmise his own hat's color.
you can save nine, with 50% chance of saving all ten.
haha math teacher you posted first but i got the answer. do i get an "A" now?
Good shit, I hoping I could've come around to that one myself eventually. Yes, you would get an A in my class. I teach in one of the worst schools in America; some of my high schoolers don't even know what "even" or "odd" means.
OK...sorry math teacher, but you don't seem to know much about math, logic or probabilities.
Most posters (who are not math teachers I assume) came up with much better answers, explanations and reasoning.
You need a new handle wrote:
OK...sorry math teacher, but you don't seem to know much about math, logic or probabilities.
Most posters (who are not math teachers I assume) came up with much better answers, explanations and reasoning.
Huh? I'll readily admit that I'm not much of a math teacheer, but only one other person proposed a solution to the second problem that allowed for more than 5 people to be saved. And what's wrong with my reasoning?