Can you solve these two?

##### Three Hats

Three wise men are told to stand in a straight line, one in front of the other. A hat is put on each of their heads. They are told that each of these hats was selected from a group of five hats: two black hats and three white hats. The first man, standing at the front of the line, can’t see either of the men behind him or their hats. The second man, in the middle, can see only the first man and his hat. The last man, at the rear, can see both other men and their hats.

None of the men can see the hat on his own head. They are asked to deduce its color. Some time goes by as the wise men ponder the puzzle in silence. Finally the first one, at the front of the line, makes an announcement: “My hat is white.”

He is correct. How did he come to this conclusion?

##### One Hundred Hats

One hundred persons will be lined up single file, facing north. Each person will be assigned either a red hat or a blue hat. No one can see the color of his or her own hat. However, each person is able to see the color of the hat worn by every person in front of him or her. That is, for example, the last person in line can see the color of the hat on 99 persons in front of him or her; and the first person, who is at the front of the line, cannot see the color of any hat.

Beginning with the last person in line, and then moving to the 99th person, the 98th, etc., each will be asked to name the color of his or her own hat. If the color is correctly named, the person lives; if incorrectly named, the person is shot dead on the spot. Everyone in line is able to hear every response as well as hear the gunshot; also, everyone in line is able to remember all that needs to be remembered and is able to compute all that needs to be computed.

Before being lined up, the 100 persons are allowed to discuss strategy, with an eye toward developing a plan that will allow as many of them as possible to name the correct color of his or her own hat (and thus survive). They know all of the preceding information in this problem. Once lined up, each person is allowed only to say “Red” or “Blue” when his or her turn arrives, beginning with the last person in line.

Your assignment: Develop a plan that allows as many people as possible to live. (Do not waste time attempting to evade the stated bounds of the problem — there’s no trick to the answer.)

Ok, got the first one but don’t want to spoil it.

As for the second one, I think I could save about 75 of 100 people.

There is a 50 percent chance of saving 100 people.

There is a 50 percent chance of saving ??? people.

0?

99.

My plan guarantees 50 live and the other 50 get a 50/50 shot so I figure 75 live give or take.

I’m think that the answer is that there is a way to say the color of your own hat while at the same time signaling to a person in front of you the color of their hat as well, leaving only the first person with the 50/50 chance.

I suspect the answer includes the last person in line either lying or telling the truth about the color of the first hat in line and the signal of whether or not the gun goes off.

I was thinking the same thing, (my instinct was to have the first person sacrifice themselves, thus signaling the person in front of them) but hit a brick wall at the point where you try to figure out how to achieve 2 objectives with one word.

1. Save yourself

2. Signal someone of their hats color

Obviously, the gunshot would indicate whether the person was acting altruistically.

My final decision was to use the first 50 people as sacrificial lambs by signaling anothers color. #100 signals #1, #99 signals #2, that way 1-50 will live for sure while 51-100 will have a 50/50 shot leaving 75 alive.

This is the kind of stuff I would have slaved over all afternoon with graph paper and drawings when I was 14, but now I get tired of thinking about it after about 10 minutes. Plus, it helps to know what you’re shooting for. If someone told me 99 was definitely possible, I’d probably put another 10 minutes into it before googling it.

I couldn’t quite find an answer, but the attempted answers I saw were focussing around the idea of the last person in line counting the number of red hats in line and signaling whether they saw an even or an odd number. Each subsequent person would count how many reds in front of them, and if the count matched, they would say blue or if it didn’t then they would know they had a red hat on.

That it?

Yeah, you got it.

“The prisoners can use a binary code where each blue hat = 0 and each red hat = 1. The prisoner in the back of the line adds up all the values and if the sum is even he says “blue” (blue being =0 and therefore even) and if the sum is odd he says “red”. This prisoner has a 50/50 chance of having the hat color that he said, but each subsequent prisoner can calculate his own color by adding up the hats in front (and behind after hearing the answers [excluding the prisoner in the back]) and comparing it to the initial answer given by the prisoner in the back of the line. The total number of red hats has to be an even or odd number matching the initial even or odd answer given by the prisoner in back.”

Now, if the following setup had been given (as pulled from Wikipedia), I might have actually figured it out (but probably not)….

“…..A friendly guard warns them of this test one hour beforehand and tells them that they can formulate a plan where by following the stated rules, 9 of the 10 prisoners will definitely survive, and 1 has a 50/50 chance of survival. What is the plan to achieve the goal?”

Certainly, if this setup had been given to you, me and Stephany Bruecken at the back table in Hilbergs math class, we could have had it figured out in about 2 minutes. 🙂

Here is the final answer from the guy who posed the question: http://tierneylab.blogs.nytimes.com/2009/03/30/solution-to-100-hat-puzzle/

He also gives us this new one:

My first thought was that Oppenheimer tries to make two dice with equal probabilities of winning. I think its significant that we are given the chance to make three dice however. It leads me to believe that there would be a way for Oppy to create a rock/paper/scissors of dice. I.e., each of the dice would, probability wise, be likely to beat one dice and likely to lose to the other dice. Then, since Alberto has to choose first, Oppy picks whichever of the two remaining dice has the odds of beating the other.

Sidebar: I posted this a while back. Great article about Einstein’s relationship with Kurt Godel later in life. http://www.newyorker.com/archive/2005/02/28/050228crat_atlargeSo, it’s not clear in the description, does JRO know what they’re going to be doing with the dice at the time he makes them? I’m guessing by your comment that you think he does.

That’s my assumption but I have no more information than I’ve given. You can find the original here: http://tierneylab.blogs.nytimes.com/2009/03/30/the-god-einstein-oppenheimer-dice-puzzle/

Also, I read the link about the hat solution and I actually like one of the reader submitted solutions better. By the use of silence, you can save the person in front of you. An immediate answer tells the person in front of you their hat is red. A short delay means the hat is blue. The guy writing the article didn’t like this (due to extraneous reasoning not present in the setup, like the guards might figure it out, but honestly, that’s a weak argument since if the guards cared, they wouldn’t let you solve it with ANY solution), but I think it solves the puzzle even more simply than the binary solution.