Nice brainteaser

[partygirluk]

100 male utilitarian logicians are told that tomorrow morning they are to be lined up and a hat placed on their head. Meanwhile they can communicate with each other. The hat will be either black or white with the probability of black for each individual 2/3 (edited to clarify: Each logician has a 2/3 chance of getting a black hat independent of what colour hat the other logicians have).

They will be asked from the backmost logician forwards what colour their hat is. If they guess correctly they are freed, if they get it wrong they die. Each man can only see the colour of the hat of the men in front of him, but they known what has happened beforehand (i.e. they know what each guy has guessed and whether they were freed or killed).

What is the best strategy for the logicians to devise in order to have the highest EV of saved lives? What about the highest number of guaranteed saved lives?

If you have heard this question (and answer) before please do not list the solution in your post!

[tbach24]

Obviously they should just guess black until the odds of black/white change into the favor of white, if they ever do.

[Macdaddy Warsaw]

Wouldn't everyone guessing black guarantee the most lives?

And wouldn't saying black until 33 people who had said black had survived be the most +LV value? EDIT: 33 people who said black and lived or said white and died but nobody should be saying white until this point anyway.

I have no idea.

[gaming_mouse]

If the original decision of black is just 2/3, and all the decisions are independent, then it makes no difference at all (EDIT: no difference what other people have done). If you are saying that only 2/3 of this logicians can have black hats, then how many have black hats, 66 or 67?

[1800GAMBLER]

[ QUOTE ]
until the odds of black/white change into the favor of white, if they ever do.

[/ QUOTE ]

[ QUOTE ]
The hat will be either black or white with the probability of black for each _individual_ 2/3

[/ QUOTE ]

Unless i'm being stupid, which i must be because otherwise this seems like a dumb brainteaser; 2/3 chance black here does not meean there are 67 black hats and 33 white hats. It means if the people giving out the hats start off using 2 black and 1 white and the first person gets a white hat, they then replace the white hat and move on to the next guy.

Meaning the best strategy would be for everyone to pick black and 2/3s will live.

[Duke]

[ QUOTE ]
Obviously they should just guess black until the odds of black/white change into the favor of white, if they ever do.

[/ QUOTE ]

You're obviously wrong.

I heard a similar one before so I won't answer.

~D

[Thythe]

To me it sounds like the probability is always 2/3 for a black hat. In this case then it is possible for all 100 hats to be black or for all 100 hats to be white, though either case is unlikely. Is this correct or is there an exact amount of each color hat (i.e. 33 white 67 black)?

[jason_t]

After you posed this to me last night, I thought of a rather interesting variant.

We place no assumption on the distribution of the hats. Instead of the logicians being in a line, they are all in a room and can see every hat but their own. They do not submit answers verbally. They do not know the results of anyone else's guess. So, the only information that each logician has is the colour of every other logician's hat but their own.

Same questions. What is the best strategy for the logicians to devise in order to have the highest EV of saved lives? What about the highest number of guaranteed saved lives?

The answer is surprising.

[tbach24]

[ QUOTE ]
[ QUOTE ]
until the odds of black/white change into the favor of white, if they ever do.

[/ QUOTE ]

[ QUOTE ]
The hat will be either black or white with the probability of black for each _individual_ 2/3

[/ QUOTE ]

Unless i'm being stupid, which i must be because otherwise this seems like a dumb brainteaser; 2/3 chance black here does not meean there are 67 black hats and 33 white hats. It means if the people giving out the hats start off using 2 black and 1 white and the first person gets a white hat, they then replace the white hat and move on to the next guy.

Meaning the best strategy would be for everyone to pick black and 2/3s will live.

[/ QUOTE ]

They can see the 100th one ahead of them so that they can adjust their count. Also, they know who the count of who is dead/alive behind them and their guesses, so they can adjust their guesses.

[Duke]

Why'd you even include the 2/3 thing. That's getting everyone thinking about it incorrectly.

~D

[mostsmooth]

i have no time to think about this(not that it matters), but if they can all see all the hats in front of them, the last guy can say the color of the first guys hat, the second to last guy say the color of the second guys, etc, until they get to guy number 50, makes a gues of his own, and then guys numbered 49-1 will know their color (assuming they can remember what was told to them, and if it meant their life id guess they could remember).
that alone gives you about 50% survival, plus a percentage of numbers 50-100 who may have guessed their own color while announcing the guys color in the front half of the line
i dont know if the fact they are all men matters, or what would a good survival rate is.
was i even close?
/images/graemlins/cool.gif

[tbach24]

[ QUOTE ]
After you posed this to me last night, I thought of a rather interesting variant.

Assume the hats are distributed uniformly at random. That is each hat is black or white with probability 1/2 and each logician's hat colour is independent of all the other hats. Instead of the logicians being in a line, they are all in a room and can see every hat but their own. They do not submit answers verbally. They do not know the results of anyone else's guess. So, the only information that each logician has is the colour of every other logician's hat but their own.

Same questions. What is the best strategy for the logicians to devise in order to have the highest EV of saved lives? What about the highest number of guaranteed saved lives?

The answer is surprising.

[/ QUOTE ]

Flip a coin to decide black or white since there is no possible way to decifer which color you will have.

[jason_t]

[ QUOTE ]


[ QUOTE ]
The hat will be either black or white with the probability of black for each _individual_ 2/3

[/ QUOTE ]

Unless i'm being stupid, which i must be because otherwise this seems like a dumb brainteaser; 2/3 chance black here does not meean there are 67 black hats and 33 white hats. It means if the people giving out the hats start off using 2 black and 1 white and the first person gets a white hat, they then replace the white hat and move on to the next guy.

[/ QUOTE ]

No. People have some fundamental misunderstandings of probability.

[ QUOTE ]
Meaning the best strategy would be for everyone to pick black and 2/3s will live.

[/ QUOTE ]

The EV in this strategy is 2/3 but it does not guarantee that 2/3 will live.

[jason_t]

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Why'd you even include the 2/3 thing. That's getting everyone thinking about it incorrectly.

~D

[/ QUOTE ]

Agreed. It's a red herring.

[jason_t]

[ QUOTE ]
i have no time to think about this(not that it matters), but if they can all see all the hats in front of them, the last guy can say the color of the first guys hat, the second to last guy say the color of the second guys, etc, until they get to guy number 50, makes a gues of his own, and then guys numbered 49-1 will know their color (assuming they can remember what was told to them, and if it meant their life id guess they could remember).
that alone gives you about 50% survival, plus a percentage of numbers 50-100 who may have guessed their own color while announcing the guys color in the front half of the line
i dont know if the fact they are all men matters, or what would a good survival rate is.
was i even close?
/images/graemlins/cool.gif

[/ QUOTE ]

Each solution should have three components. A strategy, a proof that the strategy guarantees x people live and a proof that there is no strategy guaranteeing that more than x people live.

[mostsmooth]

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[ QUOTE ]
i have no time to think about this(not that it matters), but if they can all see all the hats in front of them, the last guy can say the color of the first guys hat, the second to last guy say the color of the second guys, etc, until they get to guy number 50, makes a gues of his own, and then guys numbered 49-1 will know their color (assuming they can remember what was told to them, and if it meant their life id guess they could remember).
that alone gives you about 50% survival, plus a percentage of numbers 50-100 who may have guessed their own color while announcing the guys color in the front half of the line
i dont know if the fact they are all men matters, or what would a good survival rate is.
was i even close?
/images/graemlins/cool.gif

[/ QUOTE ]

Each solution should have three components. A strategy, a proof that the strategy guarantees x people live and a proof that there is no strategy guaranteeing that more than x people live.

[/ QUOTE ]
i think i gave a strategy, but i cant give you any further math, because i have no training. this is the best i can do.
sorry /images/graemlins/frown.gif

[partygirluk]

Each logician has a 2/3 chance of getting a black hat independent of what colour hat the other logicians have.

[jason_t]

[ QUOTE ]
i think i gave a strategy, but i cant give you any further math, because i have no training. this is the best i can do.
sorry /images/graemlins/frown.gif

[/ QUOTE ]

You gave a strategy, you argued that it guarantees that 50% survive but you didn't show that there is no better strategy.

[mostsmooth]

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i think i gave a strategy, but i cant give you any further math, because i have no training. this is the best i can do.
sorry /images/graemlins/frown.gif

[/ QUOTE ]

You gave a strategy, you argued that it guarantees that 50% survive but you didn't show that there is no better strategy.

[/ QUOTE ]
sorry, its the best i can give you. maybe somebody else can prove it for me?

[mmbt0ne]

I think this will do it, but I'm not sure it's legal:

<font color="white"> Place hand on left shoulder if guy in front of you has a black hat. Place your hand on his right shoulder, if he has a white hat. I can't think of anyway that the back man isn't just a straight up guess though. Either way, 99 lives should definitely be saved, with an EV of 99.67 if backmost guesses black.</font>

[mostsmooth]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
i think i gave a strategy, but i cant give you any further math, because i have no training. this is the best i can do.
sorry /images/graemlins/frown.gif

[/ QUOTE ]

You gave a strategy, you argued that it guarantees that 50% survive but you didn't show that there is no better strategy.

[/ QUOTE ]
sorry, its the best i can give you. maybe somebody else can prove it for me?

[/ QUOTE ]
then again its probably way wrong, so people shouldnt bother i guess /images/graemlins/cool.gif

[gumpzilla]

I haven't read any responses.

My initial thought was that you can guarantee 50 saved lives, and have an EV of 50 + (5/9)*50. The trick to doing this is to have every odd numbered person say the color of the hat in front of them. 50 people are guaranteed to survive, you can easily work out the rest of it.

But you can do much better . . .

<font color="white">
You can guarantee that 99 live, with an EV of 99.5. How? Agree ahead of time that black caps = 1, white caps = 0. Now the person in back adds up the numbers he gets from all of the caps. IF it's even, he says "White." If it's odd, he says "Black." Now half of the time he'll get his color right and live, the other half he dies. The next guy in line adds up the caps ahead of him. If the parity is the same as what the last guy suggested (the guy said "White" and number 2 gets an even number) he knows his cap is white. If it has changed, he knows it is black. So he can deduce the color of his cap. Each man, being a perfect logician, can keep track of all of this. So they will all live for sure, except for the first dude, who lives half of the time. </font>

Sweet problem.

[jason_t]

[ QUOTE ]
I think this will do it, but I'm not sure it's legal:

<font color="white"> Place hand on left shoulder if guy in front of you has a black hat. Place your hand on his right shoulder, if he has a white hat. I can't think of anyway that the back man isn't just a straight up guess though. Either way, 99 lives should definitely be saved, with an EV of 99.67 if backmost guesses black.</font>

[/ QUOTE ]

Not legit, as I understand it. The only way to communicate is through the answers starting from the back.

[jason_t]

Apparently you didn't read two aspects of the problem. The distribution is 2/3 black (this affects the answer you gave) and please do not list the solution in your post!

[gaming_mouse]

[ QUOTE ]
Each logician has a 2/3 chance of getting a black hat independent of what colour hat the other logicians have.

[/ QUOTE ]

Then knowing what happened to the other logicians does not change the chance that you have a black hat yourself. You should always black, since you have a 2/3 chance of being right. I don't get it. What am I missing?

[jason_t]

[ QUOTE ]
[ QUOTE ]
Each logician has a 2/3 chance of getting a black hat independent of what colour hat the other logicians have.

[/ QUOTE ]

Then knowing what happened to the other logicians does not change the chance that you have a black hat yourself. You should always black, since you have a 2/3 chance of being right. I don't get it. What am I missing?

[/ QUOTE ]

With this strategy you can not guarantee that anyone survives.

[gumpzilla]

Eh. I can go back and remove it, if you like, but I figured whiting it out would make it so that you could only see it if you wanted to see it. The post also specified that you shouldn't post the solution if you've seen this one before, and I figured it out in ten minutes in the shower, so I thought I was safe.

I'll now get back to your mathematical point, in more white text:

<font color="white"> Yes, it just occurred to me that my EV will be slightly off, but I don't think it will be off by much. We would have to work out the distribution of odd and even sums of caps and find out how often the guy in back says black or white. But they should be quite close to evenly distributed, and I'm pretty sure that if you flip a coin to decide whether to call black or white, you'll be right 1/2 of the time. There are four outcomes in the coinflip scenario: wear black/guess black, wear white/guess white, wear black/guess white, wear white/guess black. These have probabilites 1/3, 1/6, 1/3, 1/6 respectively, so the EV for that dude would be about 1/2. This isn't quite correct because his guessing distribution isn't 50/50, but it's going to be pretty damn close and this is clearly the best you can do. Way to go, utilitarian logician in back. </font>

[Thythe]

[ QUOTE ]
Apparently you didn't read two aspects of the problem. The distribution is 2/3 black (this affects the answer you gave) and please do not list the solution in your post!


[/ QUOTE ]

If we aren't listing our guesses for the solution, what are doing here?

[gaming_mouse]

[ QUOTE ]
[ QUOTE ]
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Each logician has a 2/3 chance of getting a black hat independent of what colour hat the other logicians have.

[/ QUOTE ]

Then knowing what happened to the other logicians does not change the chance that you have a black hat yourself. You should always black, since you have a 2/3 chance of being right. I don't get it. What am I missing?

[/ QUOTE ]

With this strategy you can not guarantee that anyone survives.

[/ QUOTE ]

Explain what I'm missing. The other answers are talking about communication, but as I understood the problem you are not allowed to communicate or agree on an overall strategy to begin with. If you can, then this changes things. Can the backmost guy see the colors of all the hats in front of him?

[partygirluk]

Sorry if my original post was unclear.

This is the kind of answer I am looking for:

The first 98 all vote black. The 99th votes for whatever the colour of the 100th. The 100th is now guaranteed survival whilst the 99th survives only when he has the same colour hat as the 100th which happens 5/9 (2/3*2/3 + 1/3*1/3) of the time.

So the EV of this is

98*2/3 + 5/9 + 1 = 66.888888 which is higher than the 66.66666 who survive if all guess black.

[gaming_mouse]

So they all CAN agree on a global strategy.

Just for clarity (not my answer), a strategy COULD be, you guess the color of the hat of the guy in front of you, but only if the guy in front of HIM also has that same color hat. Otherwise you guess the opposite of his color.

[partygirluk]

Exactly.

[Scotch78]

Hint: there is a way to guarantee 99 lives.

Scott

[fluxrad]

[ QUOTE ]
Apparently you didn't read two aspects of the problem. The distribution is 2/3 black (this affects the answer you gave) and please do not list the solution in your post!


[/ QUOTE ]

FWIW - Gumpzilla is the only one that's got it right so far.

<font color="white"> The specific number of black hats to white hats is completely extraneous to the solution.</font>

[jason_t]

[ QUOTE ]
[ QUOTE ]
Apparently you didn't read two aspects of the problem. The distribution is 2/3 black (this affects the answer you gave) and please do not list the solution in your post!


[/ QUOTE ]

If we aren't listing our guesses for the solution, what are doing here?

[/ QUOTE ]

Trust me, you will know when you have the correct answer. If you are unsure of your answer then you don't have the right answer. Post it and receive comments.

[fluxrad]

It looks like this is the same as the "Monty Hall" problem in that some people with the wrong answer are still pretty sure it's the right one ;-)

[mmbt0ne]

</font><blockquote><font class="small">En réponse à:</font><hr />
</font><blockquote><font class="small">En réponse à:</font><hr />
I think this will do it, but I'm not sure it's legal:

<font color="white"> Place hand on left shoulder if guy in front of you has a black hat. Place your hand on his right shoulder, if he has a white hat. I can't think of anyway that the back man isn't just a straight up guess though. Either way, 99 lives should definitely be saved, with an EV of 99.67 if backmost guesses black.</font>

[/ QUOTE ]

Not legit, as I understand it. The only way to communicate is through the answers starting from the back.

[/ QUOTE ]

I don't see where it says that, but whatever. My way would still save lives. /images/graemlins/smile.gif

[jason_t]

[ QUOTE ]
Hint: there is a way to guarantee [censored] lives.

Scott

[/ QUOTE ]

I bet you ruin movies for your friends too. Hint: Edit your post.

[jason_t]

[ QUOTE ]
It looks like this is the same as the "Monty Hall" problem in that some people with the wrong answer are still pretty sure it's the right one ;-)

[/ QUOTE ]

No.

[jason_t]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
I think this will do it, but I'm not sure it's legal:

<font color="white"> Place hand on left shoulder if guy in front of you has a black hat. Place your hand on his right shoulder, if he has a white hat. I can't think of anyway that the back man isn't just a straight up guess though. Either way, 99 lives should definitely be saved, with an EV of 99.67 if backmost guesses black.</font>

[/ QUOTE ]

Not legit, as I understand it. The only way to communicate is through the answers starting from the back.

[/ QUOTE ]

I don't see where it says that, but whatever. My way would still save lives. /images/graemlins/smile.gif

[/ QUOTE ]

This is outside of the box thinking rather than a mathematical or logical solution. The intent of the puzzle is obviously mathematical/logical, not OOTBT.

[partygirluk]

[ QUOTE ]
After you posed this to me last night, I thought of a rather interesting variant.

We place no assumption on the distribution of the hats. Instead of the logicians being in a line, they are all in a room and can see every hat but their own. They do not submit answers verbally. They do not know the results of anyone else's guess. So, the only information that each logician has is the colour of every other logician's hat but their own.

Same questions. What is the best strategy for the logicians to devise in order to have the highest EV of saved lives? What about the highest number of guaranteed saved lives?

The answer is surprising.

[/ QUOTE ]

Interesting. I think I have the highest EV method, which at first glance also appears to be the highest guaranteed method, but the possibility of a 50-50 split complicates things somewhat.

[Scotch78]

I see, so you don't enjoy putting together jigsaw puzzles because you already know what the picture will look like?

Scott

[jason_t]

[ QUOTE ]
I see, so you don't enjoy putting together jigsaw puzzles because you already know what the picture will look like?

Scott

[/ QUOTE ]

I don't enjoy putting jigsaw puzzles together because they are boring and offer no challenge. This puzzle offers more of a challenge and more of a reward when you discover the solution for yourself without a hint.

[purnell]

My brain hurts.

[tolbiny]

So the math guys lose to anyone who bothered to take cosi 100- as the rear person can guess "black" in either a high pitched voice, or a low pitched voice, indicating the color of the hat infront of them (high for white, low for black)- and each man continues knowing the color of his hat, and giving the color of the hat infront of him.
Only the poor sack of crap at the back, or the guy standing infront of dead (as dead is totally the type of guy to screw it up on purpose just for laughs) have a chance at dying.

[jstnrgrs]

If the chance of getting a black hat is independent of the other hats, then the information about what color hats the others have is as useless as the history pole on a roluette wheel. Maximize EV by everyone gessing Black. There is no way to guarentee that any lives are saved.

[ethan]

[ QUOTE ]
[ QUOTE ]
Hint: there is a way to guarantee [censored] lives.

Scott

[/ QUOTE ]

I bet you ruin movies for your friends too. Hint: Edit your post.

[/ QUOTE ]

Heh. I had a solution written out in white text, but I'll hold off on posting it. I like this problem, it's clever.

[Tboner7]

Couldn't the guy behind just tell the guy in front? Only the back person could get it wrong. So 99% is gauranteed, and the last guy will guess it right 2/3 of the time.

[Duke]

You guys are embarassing me with your solutions.

Think, McFly, think!

~D

[ethan]

[ QUOTE ]
Couldn't the guy behind just tell the guy in front? Only the back person could get it wrong. So 99% is gauranteed, and the last guy will guess it right 2/3 of the time.

[/ QUOTE ]

No. Each person gets to say one word. If he says the color of the hat in front of him, he's not necessarily going to be saying the color of his own hat.

My solution:
Best guaranteed: <font color="white">99</font>. Best EV: <font color="white">at least 99.5, probably slightly higher</font>

Explanation: <font color="white">The important concept here is parity (odd vs even). The person here in the back of the line says "black" if he can see an odd number of black hats in front of him, and "white" if he can see an even number. Say he says "black". If the person in front of him can see an even number of black hats, he knows his hat is black. If he sees an odd number, his hat is white. And so on. (If the guy in back says he can see an odd number of black hats, then the next guy says "black" because he can see an even number, guy #3 knows he'll see an odd number if his hat is black and an even number if his hat is white.) Thus everyone but the guy in back is guaranteed to live.

The reason the EV is probably over 99.5 is that the probability of an odd/even number of black hats in front of the first guesser is unlikely to be exactly 50%/50%. It'll be close though. You could figure out whether an odd number or even number was more likely, and have him say "black" with the more likely case. Were it exactly 50/50, he'd be guessing black half the time and white half the time, thus surviving 50% of the time.</font>

[Duke]

It's about damn time someone figured it out. Nice work ethan.

~D

[ethan]

[ QUOTE ]
It's about damn time someone figured it out. Nice work ethan.

~D

[/ QUOTE ]

The modified problem jason_t posted (they can all see each other, but write down their answers simultaneously) is interesting. I don't think the solution's quite as elegant, but I have something plausible for that one too.

[youtalkfunny]

[ QUOTE ]
100 male utilitarian logicians are told that tomorrow morning they are to be lined up and a hat placed on their head.

[/ QUOTE ]

They have to be male? OK, the hat goes "on their head". Each guy puts it on the head of his prick, and we have 100% success rate.

[ QUOTE ]
They will be asked from the backmost logician forwards what colour their hat is.

[/ QUOTE ]

A teammate will ask the question? And they can set up their codewords ahead of time?

I didn't read this thread, but how can a question this simple generate so much discussion?

("What color hat are you wearing?" = White
"What color hat do you have on your head?" = Black)

[Dead]

Plop.

[jason_t]

[ QUOTE ]
[ QUOTE ]
100 male utilitarian logicians are told that tomorrow morning they are to be lined up and a hat placed on their head.

[/ QUOTE ]

They have to be male? OK, the hat goes "on their head". Each guy puts it on the head of his prick, and we have 100% success rate.

[ QUOTE ]
They will be asked from the backmost logician forwards what colour their hat is.

[/ QUOTE ]

A teammate will ask the question? And they can set up their codewords ahead of time?

I didn't read this thread, but how can a question this simple generate so much discussion?

("What color hat are you wearing?" = White
"What color hat do you have on your head?" = Black)

[/ QUOTE ]

They can only answer in turn "white" or "black."

[jason_t]

[ QUOTE ]
[ QUOTE ]
It's about damn time someone figured it out. Nice work ethan.

~D

[/ QUOTE ]

The modified problem jason_t posted (they can all see each other, but write down their answers simultaneously) is interesting. I don't think the solution's quite as elegant, but I have something plausible for that one too.

[/ QUOTE ]

The answer is rather surprising, I believe. I agree that the strategy is not as elegant however the proof that you can do no better seems interesting.

[DavidC]

[ QUOTE ]
100 male utilitarian logicians are told that tomorrow morning they are to be lined up and a hat placed on their head. Meanwhile they can communicate with each other. The hat will be either black or white with the probability of black for each individual 2/3 (edited to clarify: Each logician has a 2/3 chance of getting a black hat independent of what colour hat the other logicians have).

They will be asked from the backmost logician forwards what colour their hat is. If they guess correctly they are freed, if they get it wrong they die. Each man can only see the colour of the hat of the men in front of him, but they known what has happened beforehand (i.e. they know what each guy has guessed and whether they were freed or killed).

What is the best strategy for the logicians to devise in order to have the highest EV of saved lives? What about the highest number of guaranteed saved lives?

If you have heard this question (and answer) before please do not list the solution in your post!

[/ QUOTE ]

Well, if it weren't independent, this would be a simplified no-two-birthdays-on-the-same-day solution, I think. /images/graemlins/smile.gif

That being said, the answer is to always pick black, as it's the most likely, because they're independent.

Regardless of independents, the correct first guess is always black.

[LSUfan1]

[ QUOTE ]
Each logician has a 2/3 chance of getting a black hat independent of what colour hat the other logicians have.

[/ QUOTE ]

If this is the case, since it's independant of the others there is always a 2/3 chance, and black must always be selected.

Let's try this though. Since you know there will ultimately be 2/3 black and 1/3 white, you should always decide black, unless more than 1/3 of the existing hats are black which would turn the numbers into the favor of white.

[xorbie]

I haven't read any other posts, but here's my answer:

<font color="white"> The last guy guesses black if he sees an even number of black hats, and guesses white if he sees an even number. Because there is a 2/3 chance of it being black and (I'm guessing) the highest probability is that the last guy sees 66 black hats out of the 99 men in front of him (as opposed to seeing 67 or 65), this should leave him with the highest probability of living.

If he sees an odd number though.. oh well. Being last is -EV.

Based on what the guy behind him has says, the second person should be able to get it right easily enough. For example, let's say guy in the back says "black" because he sees 66 black hats. Whether he lives or dies doesn't matter. If the second guy to guess sees 65 black hats, then he knows his is black (to make it even) so he guesses black. He lives.

Let's say next guy has a white hat. He would then still see 65 black hats, but knows that the guy who just guess previously was black, so he knows his own must be white. He guesses white and lives. And so on and so forth.

Thus you are gauranteed 99 to live, and have a whatever chance of getting 100 to live (I don't know what the probability of getting an even number of out 99 flipping a 2/3 weighted coin is).</font>

[youtalkfunny]

[ QUOTE ]
They can only answer in turn "white" or "black."

[/ QUOTE ]

But the person asking can SEE the answer, right?

All he has to do is word his question properly, to tip the answer. The man in the hat gives his (correct) one-word answer, and we move on.

[gumpzilla]

[ QUOTE ]

We place no assumption on the distribution of the hats. Instead of the logicians being in a line, they are all in a room and can see every hat but their own. They do not submit answers verbally. They do not know the results of anyone else's guess. So, the only information that each logician has is the colour of every other logician's hat but their own.

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Some potentially spoilish questions inside:

<font color="white">How are they allowed to transmit information to each other? Can they, for example, organize themselves in interesting ways within the room? </font>

[ethan]

[ QUOTE ]
[ QUOTE ]

We place no assumption on the distribution of the hats. Instead of the logicians being in a line, they are all in a room and can see every hat but their own. They do not submit answers verbally. They do not know the results of anyone else's guess. So, the only information that each logician has is the colour of every other logician's hat but their own.

[/ QUOTE ]

Some potentially spoilish questions inside:

<font color="white">How are they allowed to transmit information to each other? Can they, for example, organize themselves in interesting ways within the room? </font>

[/ QUOTE ] Comments: <font color="white">My understanding is that they can't transmit information to each other. They can't move, no eye contact, no standing on one leg...nothing.</font>

[purnell]

Bingo. All that extraneous info sent me down a rabbit hole...