### Brain Teasers

# The Dwarvish Feast

In a forest somewhere in Scotland lives a group of 100 dwarves. Each night they meet in the middle of the forest for a grand feast. When morning comes, they all go home. Each dwarf is wearing either a red hat or a blue hat. Curiously, there are no mirrors in the forest, so no dwarf knows the color of his own hat. A dwarf would never take off his hat to see its color, and a major dwarf faux pas is to comment on the color of another dwarf's hat. The dwarves know, however, that there is at least one red hatted dwarf and one blue hatted dwarf. One day, the master dwarf announces that the nightly feast will only be intended for blue hatted dwarves, and as soon as a dwarf knows that he is wearing a red hat, he should not come back the next day, and he should never return.

How many days does it take before there are no dwarves left with red hats at the party? (Assume all the dwarves are equally capable of figuring it out, in other words, there are no smart dwarves, and no stupid dwarves...)

How many days does it take before there are no dwarves left with red hats at the party? (Assume all the dwarves are equally capable of figuring it out, in other words, there are no smart dwarves, and no stupid dwarves...)

### Hint

The number of days depends on the number of red hats.### Answer

It will take one more day than the number of dwarves with red hats.How this works is: assume that there is only one red-hatted dwarf. He would go to the party only to see 99 blue hats. He knows that there is at least one red hat. Since all he sees is blue, he knows it must be him, so he knows not to come back the next day. Meanwhile, the blue hatted dwarves can only see one red hat, so they think that if he comes back tomorrow, then he must have seen another red hat. Since each blue hatted dwarf only sees one red hat, he would assume that if the red hatted dwarf came back tomorrow, then he must be the other red hat that the first red hatted dwarf saw. So the second day, all the blue hatted dwarves would come expecting to see if the red hatted dwarf came back. Which he wouldn't, because he knows he has a red hat. So the first day there are no dwarves with red hats is day 2, or one day more than the number of dwarves with red hats.

If there were two dwarves with red hats, each red hatted dwarf would realize on the second day that he was wearing a red hat, because he would see that the other red hat had come back, and that means that he must have seen another red hat. Since the red hats would see 98 blue hats, they would know on the second day, and the third day would be the first day that no red hatted dwarves came to the feast. This same system works for any number of red hats, because each time a blue hatted dwarf is thinking that if the red hats come back tomorrow, then he must have a red hat, the red hatted dwarves already know, and so they don't come the next day.

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## Comments

What if each dwarf went up to another and said "See you tomorrow." The dwarves seeing the red hatted dwarf would say, "Oh are you coming then?" The red hatted dwarves would know from the others response that they were not expected. Anyway, I like the puzzle and the logical explanation seems good.

I guess the dwarves didn't notice the color when the put on the hat good logic puzzle though.

This sounds logical but I'm still not seeing it. The logic works for 1 or two red hats but what if there were say 5 - how does it happen?

I don't think it works even with two. Why would not a blue-hatted dwarf think, when he saw all of the red-hatted dwarves return the second day, that he was also red-hatted? Why would only another red-hatted dwarf think that, given essentially the same info the blue-hats have?

and what will happen if there are more than 50 red hat dwarfs or lets say 90 red-hat dwarfs.i dont think it will work then

If you follow the logic out, amazingly enough it actually works for any number of red hats.

As it's explained in the answer, 1 red hatter, seeing only blue hats and no red hats, would know he had the only red hat, so he wouldn't return the next day. So if I see one red hatter and he DOES return the next day (day two), he must have seen a red hat. Since everyone else I see has blue hats, I must have the other red hat. So I wouldn't return the next day, day three (and neither would he, having reached the same conclusion).

So I know that if there were 2 red hatters, each seeing only the one other red hat, then neither one would come back on day three. So if I see 2 red hatters, and they come back on day three, that means that they saw more than just the red hat on each other. Since everyone else I see has a blue hat, that means I must have the other red hat that they're seeing. So on day four all 3 of us red-hatters, having reached the same conclusion, would not return.

So if I see 3 red hatters and they're still here on day four, then they must not have reached the above conclusion, meaning they must each see more than just the other 2 red hats. Since everyone else I see has a blue hat, that means I must have the other red hat that they're seeing. So on day five all 4 of us red-hatters, having reached the same conclusion, would not return.

So if I see 4 and they're still there on day five, they must see a red hat on me, so I won't come back.

If I see 5 and they're still there on day six, they must see a red hat on me, so I won't come back.

And on and on ...

Presumably, since all the dwarves would understand this formula, they would each count the number of red hats they see (even if it's a large number) and wait that number of days +1. If all the red hatters are gone that day, they'd know they were safe, if not then they'd know they couldn't come back the next day.

So if you see 86 red hatters, you know that if you have a blue hat then each of them are only seeing 85 red hats and they'll leave after day 86 comes and goes, but if day 87 comes AAH MY BRAIN FELL OUT!!

As it's explained in the answer, 1 red hatter, seeing only blue hats and no red hats, would know he had the only red hat, so he wouldn't return the next day. So if I see one red hatter and he DOES return the next day (day two), he must have seen a red hat. Since everyone else I see has blue hats, I must have the other red hat. So I wouldn't return the next day, day three (and neither would he, having reached the same conclusion).

So I know that if there were 2 red hatters, each seeing only the one other red hat, then neither one would come back on day three. So if I see 2 red hatters, and they come back on day three, that means that they saw more than just the red hat on each other. Since everyone else I see has a blue hat, that means I must have the other red hat that they're seeing. So on day four all 3 of us red-hatters, having reached the same conclusion, would not return.

So if I see 3 red hatters and they're still here on day four, then they must not have reached the above conclusion, meaning they must each see more than just the other 2 red hats. Since everyone else I see has a blue hat, that means I must have the other red hat that they're seeing. So on day five all 4 of us red-hatters, having reached the same conclusion, would not return.

So if I see 4 and they're still there on day five, they must see a red hat on me, so I won't come back.

If I see 5 and they're still there on day six, they must see a red hat on me, so I won't come back.

And on and on ...

Presumably, since all the dwarves would understand this formula, they would each count the number of red hats they see (even if it's a large number) and wait that number of days +1. If all the red hatters are gone that day, they'd know they were safe, if not then they'd know they couldn't come back the next day.

So if you see 86 red hatters, you know that if you have a blue hat then each of them are only seeing 85 red hats and they'll leave after day 86 comes and goes, but if day 87 comes AAH MY BRAIN FELL OUT!!

JimShorts' explanation really helped! I wasn't very convinced by the written answer beyond 2 red hats..

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