Riddle

[Shadowless]

Of three(3) prisoners in a certain jail, one had normal vision, the second had only one eye, and the third was totally blind. The jailor told the prisoners that, from three white hats and two red hats, he would select three and put them on the prisoners’ heads. None could see what color hat he wore. The jailor offered freedom to the prisoner with normal vision if he could tell what color hat he wore. To prevent a lucky guess, the jailor threatened execution for any incorrect answer. The first prisoner could not tell what hat he wore. Next the jailor made the same offer to the one-eyed prisoner. The second prisoner could not tell what hat he wore either. The jailor did not bother making the offer to the blind prisoner, but he agreed to extend the same terms to that prisoner when he made the request. The blind prisoner said:

I do not need to have my sight;
From what my friends with eyes have said,
I clearly see my hat is ________!

How did he know?

[Chuckers]

He had a white hat. In any case that he would have a red hat, one of the others would have been able to figure out their own hat color.

1w 2r 3r = one would know he has a white hat because he sees both red hats so he would have guessed correctly
1r 2w 3r = two would know he has a white hat because he sees both red hats so he would have guessed correctly
1w 2w 3r = 2 would know that he has a white hat because he knows that 1 didn't see two reds (or else 1 would know that his is white), and two sees that 3 does have a red hat, so he would know that he must have a white one.
Since neither 1 or 2 guessed correctly, they 3 knows that he must have a white hat.

[Shadowless]

Can you explain that again? I am finding it hard to follow. sorry. :confused:

[Chuckers]

I know, I had a hard time putting it to words. Just work it out from each person's point of view using the different combinations. The confusing one is when 1 and 2 have white hats and 3 has a red hat.

In this case 1 looks at 2 and 3 and sees a white and a red (he can't tell based on that info). Then 2 looks at 1 and 3 and also sees a white and a red, but since the red hat is on 3 then he (2) knows that if he himself also had a red hat on then 1 would have known that he himself had a white one (because he would have seen 2 reds) therefore in this circumstance (1w 2w 3r) 2 would have guessed correctly. Since he didn't this combination is not possible.

Is that better?

[Shadowless]

Hahaha, wow now that i think about it, it's very easy lol. Thank you! hilarious

[gcalvin]

Here's another way to look at it -- some might find this easier...

If the first man doesn't see at least one white hat, then he knows his own hat must be white. Since he didn't know his own hat was white, he saw at least one white hat.

If the second man sees a red hat on the third man, he knows his own hat has to be white, because he knows that the first man had to see at least one white hat. But he sees a white hat on the third man, so he doesn't know whether his own hat is red or white.

The third man doesn't have to see anything by this time to know that his own hat is white.

-Gary-

[Basket Weaver]

Man, does this bring back bad Statistics studying scenarios or what?