piter, please mind the yelling-caps. Why don’t you show you’re right, rather than telling at full volume?
It wasn’t that hard of a puzzler, and if you can go 1, 2 you can keep track of odd and even. And EVERYONE, even number 30 is acting in their own best interest and giving the information the next person needs to get out of jail.
No matter what they decided #30 will only have a 50/50 chance, it doesn’t matter to him what color hat he says because his answer will be luck, so he’s dead if he guesses wrong, and walk if he guesses right, but it’s still a guess. So by following strategy, he is no worst oft and everyone else walks.
Even if the guy behind you can’t keep track of even/odd you can figure out what hat you have by what happened to him, assuming you know what happened to him right away.
"piter, please mind the yelling-caps. Why don’t you show you’re right, rather than telling at full volume? "
You have a problem with capitalizing one word at a time? Might want to take it up with MikeinNH.
Just because a bunch of geeks in the 90s decided it means yelling doesn’t make it so, any more than it would on the paper page.
I haven’t taken the time to determine how many prisoners die/live with Ray’s solution using a random hat distribution, but I can save 25 out of 30 on average with a random distribution of hats and no less than 20 will live on a very bad day.
The end prisoner (first to guess hat color) and every third one thereafter (10 in all) will have a 50% chance because they will say 'Black if the two in front of him are the same and white if they are different. The next man (woman?) can see the one in front of him and will know if his is the same or different; he should never be wrong. The next one knows if he is the same or different than the one behind him and now knows the correct color on the man behind, so he is never wrong. Then the sequence starts over. So, all 20 of the number 2 and number 3 prisoners live. Assuming everyone actually cooperates, the biggest chance for error is forgetting if you are a 1, a 2, or a 3.
“I haven’t taken the time to determine how many prisoners die/live with Ray’s solution using a random hat distribution”
Yet you apparently think you know that “at least 29 survive” is not correct, without “taking the time”.
“but I can save 25 out of 30 on average with a random distribution of hats and no less than 20 will live on a very bad day.”
Well, the average doesn’t matter since you only get one trial of your method, so your method claims that at least 20 will survive.
marks…
By using Ray’s solution…you GUARANTEE to save all but 1…and the 1 has a 50/50 chance of being saved…
And that’s why when this Puzzler was used in 2002 they drew straws for the last position.
“I haven’t taken the time to determine how many prisoners die/live with Ray’s solution using a random hat distribution”.
Thees ees not ze real world, piter. Zees is ze jungle, where ze guerillas do as zey pleeze. Du hast recht.
Isn’t this almost the same as a previous puzzler from this year?
No. It is the same as a previous puzzler from 2002.
Blowing on the neck is not allowed, and it is in no one’s interest to guess a color that may be wrong. Ray’s solution is too complicated for prisoners to figure out, but this one isn’t, and it will guarantee a correct answer by all but the first to answer, who will have only a 50% chance. The last prisoner only needs to see the color of the hat on head immediately in front of him, and he can call out either white or black. If the hat in front of him is black and he chooses to guess black for himself, he says “black” in a distinctive manner (e.g., b-l-a-c-k or in a high-pitched voice). This will signal to the next man what the color of his hat is. If he chooses to guess the opposite color of the man in front of him, he will say “white” in a different distinctive manner (e.g., a clipped “white” or in a low-pitched voice). That way will also signal the color of the hat, i.e., the opposite color. The next man will then say the color of his hat (which he now knows for sure) using the distinctive speech code decided on so as to give the next man the signal about the color or his hat, and so on down the line.
Piter… I agree… rolling parity bit is all they need to track, and compute the color of their own hat by what they see. What I was trying to point out is that it’s very challenging to track the rolling bit. As evidenced by those in this discussion that have trouble figuring it out even when shown. I’m not in Mensa, but not dumb either; when I run it mentally, I lose track. As you pointed out it was a long way to go, and I made a mistake (I didn’t check it, I trust you that it is wrong - which makes my point of how difficult it is to track). I guess my only point is that the STATED answer seems so simple, but they did not reveal just how challenging it is as the rolling bit changes from ODD=BLACK to ODD=WHITE.
Perhaps you have stated the MOST informative thing about the darn puzzle:
They don’t have an hour to practice, they have an hour to get 30 people to agree on a strategy and practice it.
From my experience with experiential learning puzzles - you can’t get 30 highly motivated people to agree on any strategy in one hour. 8 is like the max! Let alone practice time…
I think you’re all putting way too much into Ray’s VERY SIMPLE solution.
All you have to do is keep track of “Is the number of white hats odd or even and what the person behind you said.” This is NOT difficult. You put too much emphasis on the fact that they’re prisoners and they’re not smart…Just change that to College Professors and if they get the answer right they get tenure. If the guy behind you says “White” (which means he sees an even number of white hats)…and you see an odd number of white hats…that tells you that you MUST have a white hat. If you also see an even number of white hats…then it means you MUST have a black hat.
The confusing part is - that sometimes white means odd count, and sometimes black means odd count. That is hard to keep track. College Professors notwithstanding. Each one has to think, “He said [black/white] which means in this case odd. I see even, therefore I must be wearing one of the ones he saw, so I say that color, which NOW means even.” Each prisoner has to track this thinking for each one behind him, without making an error.
The confusing part is - that sometimes white means odd count, and sometimes black means odd count.
So you’re saying that it’s difficult to keep track of the number white hats you see, plus what the person behind you said???
Sorry…but I still don’t find this a difficult problem…or the solution complicated.
It’s not keeping track of what the person behind said, it’s keeping track of what it means (odd or even). If you are prisoner number 20, and we all agree that prisoner number 30 says WHITE if he sees an EVEN number of WHITE hats. You hear them count off behind you, “white white black black white black white white white black.” To me it’s not easy to know what that last “black” means. Odd or even? Black doesn’t always mean odd, and white doesn’t always mean even. So there’s no simple rule to follow.
Or I could be slow… :-{D
I think you’re getting stuck in thinking that you have to keep track of what everyone says. You only have to pay attention to the last person to speak and depending on what he says, in your head you change “the last person saw an even number of white hats” to “the last person saw an odd number of white hats” or not. You keep one thing in your head, “even” or “odd” and change it based on what the last person said. When the last person is right behind you, you know what color your hat is because you always know the even/oddness that the last person saw. You ignore what everyone before him saw.
Put ten quarters in your hand. Shake them up. Pour them out and put them in a row. Heads is white, tails is black. Try it.
It’s not keeping track of what the person behind said, it’s keeping track of what it means (odd or even).
At the beginning you establish that person #30 will say white if he sees odd number of white hats…
And lets say he says black (and it’s correct). And then next person says black…meaning that #30 and #29 are seeing and odd number of white hats…
I use to ref basketball and we had to keep track of which team had possession for a jump ball situation. All we did as refs was keep a small card one of our front pockets…If it was in the right pocket it meant it was home team…Once there was a jump ball and the home team got possession…then I’d move the card from my right pocket to my left (which meant that the visitors had possession. This same principle can be applied in this situation…Me as prisoner 20…I start out as if I hear white it means odd…so I close my right fists…And I keep it closed until I hear white again…then I’ll open my fist…And when I hear white again…I close my fist…And continue to do this until it’s my turn…When it’s my turn…If my fist is closed then it means odd…If I see an even number of white hats then I know I have a black hat…if my hand is open (even number of white hats)…and I count an even number of white hats then I know I have a black hat.
I’m sorry I still don’t think it’s difficult…
Ahh… I see what you are saying. Just track whether the remaining number of hats is odd or even, by flipping it every time someone says WHITE - as one WHITE hat is removed from the visual line at that point. Cool, elegant, clever. Except I don’t think you can assume ODD (close fist) to start. #30 always tells the truth, so close it if he says WHITE, open it if he says BLACK. If you start out with closed fist, and reverse it every time you hear WHITE, you will be wrong from the start. So I agree with you that this solution is simple. Ray did not give enough detail to show the simplicity. However… you must admit that figuring it out is tough. I made one mistake in my explanation, and you made two. (start with closed fist, and you said “black” hat for both odd and even counts above). 
You can assume that Odd is the start…but if #30 is WRONG then you know it’s even.
#30 has a 50/50 chance of being right…So it’s agreed that he’s going to start out as White = Odd and Black = Even…If he’s wrong…then there were an even number…