Luck's monster problem (Edit)

[Stephen]

http://wordplay.blogs.nytimes.com/2013/12/30/luck-2/?_r=0An

[AllYouNeedIsLuck]

nothing posted brah

[Warhorse]

Jello!

[HarassedOffTheSite]

TMTSF disorder.

 

Too Much Time Spent On The Forums

[JoKeR]

It worked for me.

The puzzle question seems to be missing some parts like if the number of black and whites hats are equal.

[Jason_]

I didn't realize Luck was the creator of "Lost"

[OffensivelyPC]

It worked for me.

The puzzle question seems to be missing some parts like if the number of black and whites hats are equal.

I thought the same thing at first, but you can make it work.  Check in the comments section, but essentially, the first person must call out the color of the odd number of hats in front of him.  He has a 50-50 chance of living, but everyone else is guaranteed to survive.

[Superman]

Mr. Luck also happens to enjoy the puzzles we explore each week on Numberplay. I asked him if he had a suggestion for a puzzle, and how he would go about solving it: if he would draw it out, or use equations or do something else. “The puzzle with the monster on the island is a good one. The monster who says he’ll eat everyone but gives them a chance to survive. As for solving it — I wouldn’t really draw out anything or use equations. I would just talk it through like I’m talking to you right now. I don’t write down very much when I’m solving problems. Even back when I was a student. I think about problems a lot more clearly if I can discuss them.”

 

[Superman]

I thought the same thing at first, but you can make it work.  Check in the comments section, but essentially, the first person must call out the color of the odd number of hats in front of him.  He has a 50-50 chance of living, but everyone else is guaranteed to survive.

 

The problem doesn't say how many hats of each color there are. There could be 9 black hats and 1 white hat. There's not necessarily any pattern to how the hats are placed. And the wording is ambiguous. "The monster will place either a black or white hat on each person's head." Doesn't say that he'd use both; he could choose to use all black or all white.

 

This monster is a big jerk.

[Coffeedrinker]

I thought the same thing at first, but you can make it work.  Check in the comments section, but essentially, the first person must call out the color of the odd number of hats in front of him.  He has a 50-50 chance of living, but everyone else is guaranteed to survive.

That is assuming there are 5 white and 5 black.  That is not stated anywhere.  If that were the case then it would be easy and even the first person wouldn't be at risk.  If there are 5 white and 5 black and he sees and odd number of white hats then his/her would have to be black.  Then it's just a matter of simple subtraction.

[OffensivelyPC]

The problem doesn't say how many hats of each color there are. There could be 9 black hats and 1 white hat. There's not necessarily any pattern to how the hats are placed. And the wording is ambiguous. "The monster will place either a black or white hat on each person's head." Doesn't say that he'd use both; he could choose to use all black or all white.

 

This monster is a big jerk.

It doesn't matter the configuration of hats.  There could be 3 black hats and 7 white hats.  As long as the first person honestly calls out which one is the odd number, everyone in front of him lives (if he lies, then #2 also has a 50/50 chance, but everyone else lives, assuming he tells the truth). 

 

Lets say there are hats in the following configuration (#1 sees all in front of him and guesses first).

 

W-B-B-W-B-B-B-B-B-B

 

#1 sees only 1 white hat, so he calls out "white."  Lucky for #1, he was wearing a white hat, so he lives.  #2, knowing that there is an odd number of white hats in front of him knows that there is still an odd number of white hats.  This must be true because when #1 saw an odd number of white hats, there must be an even number of black hats in front of him (which number two also knows) because there are 9 hats in front of #1.  You cant have both an even or odd number of white and black hats when there are 9 total hats in front of #1 (i.e. odd#+odd#=even# and even#+even#=even#) With 9 total hats remaining the only way you can add an odd number of white hats to get an odd number of total hats is to add another even number of black hats.  Therefore, if #2 sees 1 white hat in front of him and 7 other black hats, that means that #2's hat must be black, because at the time, #1 saw an even number of black hats. 

[OffensivelyPC]

That is assuming there are 5 white and 5 black.  That is not stated anywhere.  If that were the case then it would be easy and even the first person wouldn't be at risk.  If there are 5 white and 5 black and he sees and odd number of white hats then his/her would have to be black.  Then it's just a matter of simple subtraction.

See my above post.

[cjwhiskers]

You guys are thinking too much into this. You just have to listen to what color hats he's eating and which ones he's not. Then you say that color.

[matt986]

It doesn't matter the configuration of hats.  There could be 3 black hats and 7 white hats.  As long as the first person honestly calls out which one is the odd number, everyone in front of him lives (if he lies, then #2 also has a 50/50 chance, but everyone else lives, assuming he tells the truth). 

 

Lets say there are hats in the following configuration (#1 sees all in front of him and guesses first).

 

W-B-B-W-B-B-B-B-B-B

 

#1 sees only 1 white hat, so he calls out "white."  Lucky for #1, he was wearing a white hat, so he lives.  #2, knowing that there is an odd number of white hats in front of him knows that there is still an odd number of white hats.  This must be true because when #1 saw an odd number of white hats, there must be an even number of black hats in front of him (which number two also knows) because there are 9 hats in front of #1.  You cant have both an even or odd number of white and black hats when there are 9 total hats in front of #1 (i.e. odd#+odd#=even# and even#+even#=even#) With 9 total hats remaining the only way you can add an odd number of white hats to get an odd number of total hats is to add another even number of black hats.  Therefore, if #2 sees 1 white hat in front of him and 7 other black hats, that means that #2's hat must be black, because at the time, #1 saw an even number of black hats. 

 

This.

 

Number 1 can impart enough information into the system to allow the others (if they follow the strategy) to survive.

 

Of course if anyone wants to hose the people in front of them they can.  :shifty: 

[OffensivelyPC]

This.

 

Number 1 can impart enough information into the system to allow the others (if they follow the strategy) to survive.

 

Of course if anyone wants to hose the people in front of them they can.  :shifty:

Actually, only #1 can hose the people in front of them by throwing off the system.  It took me a while to figure it out, but I posted this on the original website above.

 

If #1 lies and said that there are an odd number of whites in front of him, #2 will not know that number one is lying, he will only know that #1 said "white" and either survived or died.

Take the following example:

W-B-B-B-W-W-B-B-B-B

#1, lies and says white. He lives. Now #2 sees only 2 whites, but cannot deduce that #1 was lying because he couldn't see his hat. Yes he lived (or died, it doesn't matter), but that fact does not prove whether #1 told the truth, so acting in reliance on #1, he calls out white and dies; based on  the assumption that #1 was telling the truth, him seeing 2 white hats in front of him would mean that he was wearing a white hat because #1, lying, indicated he saw 3 white hats (the two in front of #2, making the third hat the one on #2's head).

 

At this point, #3 (and everyone in front of him)can deduce that #1 lied, because #2 died when he had a 100% guarantee that he'd live. Thus, now #3 knows that (sorry if this is hard to follow) (a) #1 saw an odd number of black hats, which was proven true when #2 died for calling out white (b) at the time of #2's turn, he would have seen an even number of blacks and whites in front of him, © had #2 called out black, there would now be an odd number of blacks again, and (d) because there are still an even number of whites in front of #3, his hat must be black.

After #3-#10 live, they all beat #1 to death for being a liar liar pants on fire and they live happily ever after.

 

The reason that no one else can lie and confuse the poeple in front of him is because if they were to lie in the middle, he would be killing himself, number one, and number two, everyone in front of him would know he lied because they could "hear the result" which I presume is the sound of the monster making a piecemeal out of the liar.

[LockeDown]

I didn't realize Luck was the creator of "Lost"

 

Did you say "Lost"?

[FanFromtheWasteland]

It doesn't matter the configuration of hats.  There could be 3 black hats and 7 white hats.  As long as the first person honestly calls out which one is the odd number, everyone in front of him lives (if he lies, then #2 also has a 50/50 chance, but everyone else lives, assuming he tells the truth). 

 

Lets say there are hats in the following configuration (#1 sees all in front of him and guesses first).

 

W-B-B-W-B-B-B-B-B-B

 

#1 sees only 1 white hat, so he calls out "white."  Lucky for #1, he was wearing a white hat, so he lives.  #2, knowing that there is an odd number of white hats in front of him knows that there is still an odd number of white hats.  This must be true because when #1 saw an odd number of white hats, there must be an even number of black hats in front of him (which number two also knows) because there are 9 hats in front of #1.  You cant have both an even or odd number of white and black hats when there are 9 total hats in front of #1 (i.e. odd#+odd#=even# and even#+even#=even#) With 9 total hats remaining the only way you can add an odd number of white hats to get an odd number of total hats is to add another even number of black hats.  Therefore, if #2 sees 1 white hat in front of him and 7 other black hats, that means that #2's hat must be black, because at the time, #1 saw an even number of black hats. 

 

 

   

[Superman]

It doesn't matter the configuration of hats.  There could be 3 black hats and 7 white hats.  As long as the first person honestly calls out which one is the odd number, everyone in front of him lives (if he lies, then #2 also has a 50/50 chance, but everyone else lives, assuming he tells the truth). 

 

Lets say there are hats in the following configuration (#1 sees all in front of him and guesses first).

 

W-B-B-W-B-B-B-B-B-B

 

#1 sees only 1 white hat, so he calls out "white."  Lucky for #1, he was wearing a white hat, so he lives.  #2, knowing that there is an odd number of white hats in front of him knows that there is still an odd number of white hats.  This must be true because when #1 saw an odd number of white hats, there must be an even number of black hats in front of him (which number two also knows) because there are 9 hats in front of #1.  You cant have both an even or odd number of white and black hats when there are 9 total hats in front of #1 (i.e. odd#+odd#=even# and even#+even#=even#) With 9 total hats remaining the only way you can add an odd number of white hats to get an odd number of total hats is to add another even number of black hats.  Therefore, if #2 sees 1 white hat in front of him and 7 other black hats, that means that #2's hat must be black, because at the time, #1 saw an even number of black hats. 

 

You're assuming where there's no definite information. Your system only has a chance of working if there are both black and white hats. 

[matt986]

You're assuming where there's no definite information. Your system only has a chance of working if there are both black and white hats. 

 

I think that if all were black, #1 would die for sure, but the others would still have enough information to survive.

 

Edit:  Actually #1 would say black and live. MY mistake above.

[Stephen]

It will be interesting to see what they post friday as the official answer.

[Jason_]

It will be interesting to see what they post friday as the official answer.

 

the correct answer is -8πa

[Kyle]

Ok.  I got it.  You take your hat off, look at it, then answer.

 

The monster never stated this is not allowed.

[Superman]

I think that if all were black, #1 would die for sure, but the others would still have enough information to survive.

 

Edit:  Actually #1 would say black and live. MY mistake above.

 

That would work. I'm making my head hurt trying to figure it all out, but it would. Everyone else would just have to deduce what's going on. #1 has a 50/50 chance of living, and then everyone else should get it based on what happens to him.

 

You guys are geniuses. 

[Superman]

Ok.  I got it.  You take your hat off, look at it, then answer.

 

The monster never stated this is not allowed.

 

Each person in line can see all hats in front but not his or her own hat or the hats behind them.

[OffensivelyPC]

You're assuming where there's no definite information. Your system only has a chance of working if there are both black and white hats. 

No, because even if it is all black hats (or white):

 

B - B - B - B- B . . . .

 

By calling out an ODD number of hats, #1 will call black.  #2, looking at what is in front of him, sees 8 black hats.  If #1 called black because there were an odd number of black hats, then #2's hat could not be white, because that would mean that there was only one white hat, and you cannot add an odd number of black hats with an odd number of white hats (1 in this case) to come up with 9 total hats. 

 

What this system depends on is #1 telling the truth.  If #1 lies, it serves him no purpose or benefit, because no matter how yo ucut it, he has a 50/50 chance, so he'd just be making a * move.  In that instance, #2 would rely on #1's calling an the wrong odd number of colored hats, and now #2 will die.  #3, however, will have enough information to salvage himself and the rest of the islanders, provided they play by the same rules and assuming #3's honesty.

[Kyle]

Each person in line can see all hats in front but not his or her own hat or the hats behind them.

 

you can read it like a statement and not a rule

 

you cant see it because he placed it on your head and it is still on your head, not necessarily because it is against the rules.

BUT THEN you take it off, look at, and say what color it is.    

 

 

[Kyle]

i don't think anyone has actually said the answer.

 

The answer cant be that the first guy just has a 50/50 shot.  There is some way to do this so that all 10 people have a 100% chance of living.

[Superman]

you can read it like a statement and not a rule

 

you cant see it because he placed it on your head and it is still on your head, not necessarily because it is against the rules.

BUT THEN you take it off, look at, and say what color it is.    

 

 

 

Nice try.

[Superman]

No, because even if it is all black hats (or white):

 

B - B - B - B- B . . . .

 

By calling out an ODD number of hats, #1 will call black.  #2, looking at what is in front of him, sees 8 black hats.  If #1 called black because there were an odd number of black hats, then #2's hat could not be white, because that would mean that there was only one white hat, and you cannot add an odd number of black hats with an odd number of white hats (1 in this case) to come up with 9 total hats. 

 

What this system depends on is #1 telling the truth.  If #1 lies, it serves him no purpose or benefit, because no matter how yo ucut it, he has a 50/50 chance, so he'd just be making a * move.  In that instance, #2 would rely on #1's calling an the wrong odd number of colored hats, and now #2 will die.  #3, however, will have enough information to salvage himself and the rest of the islanders, provided they play by the same rules and assuming #3's honesty.

 

 

Yeah, your system works. I get it now.

 

The only problem is that the first guy only has a 50/50 chance. Of course, the question is what should they do, not what guarantees they all survive. 

[OffensivelyPC]

Yeah, your system works. I get it now.

 

The only problem is that the first guy only has a 50/50 chance. Of course, the question is what should they do, not what guarantees they all survive. 

Well, what they SHOULD do is spend the evening making a bunch of weapons and take the beast down. 

[pacolts56]

Yeah, your system works. I get it now.

 

The only problem is that the first guy only has a 50/50 chance. Of course, the question is what should they do, not what guarantees they all survive. 

That's the best I could come up with.

 

The first person calls out the color of the hat in front of him...hoping he has the same color on.

 

If he doesn't, then the survivors can hold a nice little memorial service.

[Coffeedrinker]

It doesn't matter the configuration of hats.  There could be 3 black hats and 7 white hats.  As long as the first person honestly calls out which one is the odd number, everyone in front of him lives (if he lies, then #2 also has a 50/50 chance, but everyone else lives, assuming he tells the truth). 

 

Lets say there are hats in the following configuration (#1 sees all in front of him and guesses first).

 

W-B-B-W-B-B-B-B-B-B

 

#1 sees only 1 white hat, so he calls out "white."  Lucky for #1, he was wearing a white hat, so he lives.  #2, knowing that there is an odd number of white hats in front of him knows that there is still an odd number of white hats.  This must be true because when #1 saw an odd number of white hats, there must be an even number of black hats in front of him (which number two also knows) because there are 9 hats in front of #1.  You cant have both an even or odd number of white and black hats when there are 9 total hats in front of #1 (i.e. odd#+odd#=even# and even#+even#=even#) With 9 total hats remaining the only way you can add an odd number of white hats to get an odd number of total hats is to add another even number of black hats.  Therefore, if #2 sees 1 white hat in front of him and 7 other black hats, that means that #2's hat must be black, because at the time, #1 saw an even number of black hats. 

You are still leaving the 1st person being eaten to chance.  The idea would be to save all 10 people.  You have all night, the only smart thing to do would be to spend that time and build a trap for the monster and capture he and his hats.  Why is everyone so much of a conformist and just line up to be eaten just because a monster says that is what you should do.

 

And even if the trap doesn't work, well at least then the monster is going to have to work for his food rather then everyone just being a lemming and lining up.

 

I see you just posted something similar above.

[Blue Horseshoe]

If you are willing to put prisoner 1 at risk then the solution to save any number of others is obvious.

Simply vary the pitch used to deliver each response to encode the color of the hat directly in front of each respondent.

So, if there is a white hat in front of you then you answer in a high pitch and if there is a black one then you answer with a low one. 

I haven't figured out how to save the first person.  The best I have come up with is to pick the color you see the least of unless the distribution is very unbalanced.  So, if you see 5 white and 4 black then answer black but if there are 9 white then answer white.  

[AntonMcG]

There is no possible way to save the first person... he can only save the rest.

 

I have seen a very similar problem before with the same concept.. it's a good one.

[AntonMcG]

I thought the same thing at first, but you can make it work.  Check in the comments section, but essentially, the first person must call out the color of the odd number of hats in front of him.  He has a 50-50 chance of living, but everyone else is guaranteed to survive.

 

Yup

[The Machine]

The solution is simple.

Person #1 is forced to guess whether the hat is white or black, therefore he only has a 50% chance to survive.

Everyone else will live because the person guessing right before them will either shout their guess if the hat in front of them is white or reply in a normal voice if the hat is black.

Everyone but the first person will know the color of their hat and survive.