Friday puzzle -- red and blue cap reprise

just off the top of my head

expected 75, guaranteed 50

That's an improvement

but they can do still better.

P.s., I presume you got to that by having each even-numbered wise man tell the one in front of him what color his cap was? If you came to this answer by another route, that would be interesting.

Re: That's an improvement

I think I see a way to get 66% guaranteed, 83 expected.

getting closer

Go ahead and state your method for achieving those results.

Re: getting closer

The first wise man looks at the two immediately in front of him. If they both have the same color hat he says "blue". If they have different color hats he says "red". Of course he has only a 50% chance of being correct about his own hat.

The second wise man remembers this and looks at the one immediately in front of him. If the first man said "blue", then he knows his hat is the same color as the one in front of him and says whatever color that is. Otherwise, he knows and says the opposite.

The third wise man remembers what the first two said. He knows what color the hat behind him is and knows if it is the same or different than his. So he can also speak his color with confidence.

The process repeats with each group of three until there is one left who tosses a coin.

On the right track

Got it.

Unless you know of a way to get better than 99.5.

Excellent!

Let's leave it open for a while in case anyone else wants to try.

Hint -- the goal

As Bill said in his post titled "got it", it is possible to get expected survival of 99.5. It is not possible to give the wise man at the back of the line any better chance than 50-50, since he has no opportunity for knowledge to improve his guess. But it is possible to assure survival of the other 99!

I think I have it also

hmmm... or maybe not ;-(

Re: Hint -- the goal

I think even I got it, thanks to Bill and the hints from Alex.

or maybe so ;-).....

But not the way I thought of originally. (and maybe never without some of the hints.)

Bill, do the honors?

You got it with the least hinting (I didn't get it at all).

After you give your explanation, I'll show the "solution" given where I saw this one, which I found almost as difficult as the puzzle!

My solution

The first wise man can see the other 99 caps. Among these caps there must be either an odd number of red caps or an odd number of blue caps. He will say the color with the odd count. For this example we will assume it is blue.

The second wise man will count the number of blue caps that he can see. If it is odd, then his cap must be red. If it is even, his cap must be blue. So he can say the right thing.

Each subsequent wise man will count the number of blue caps in front of him plus the number of blue caps reported behind him (excluding the report from the first wise man). If the total is even, his hat is blue. Otherwise it is red.

That is a clear explanation

Just for comparison, here is the "solution" I saw, which left me pretty befuddled. It is correct, but sounds more like directions for someone to code the solution in a programming language than an explanation:

The first wise man counts all the red hats he can see (Q) and then answers "blue" if the number is odd or "red" if the number is even. Each subsequent wise man keeps track of the number of red hats known to have been saved from behind (X), and counts the number of red hats in front (Y).

If Q was even, and if X&Y are either both even or are both odd, then the wise man would answer blue. Otherwise the wise man would answer red.

If Q was odd, and if X&Y are either both even or are both odd, then the wise man would answer red. Otherwise the wise man would answer blue.

Re: That is a clear explanation

Just check to see if the parity bit is set or not ;-)