Genius Challenge Question

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-e

A = Front genius

B = Middle genius

C = Rear genius

A and B can't both be wearing white hats because, as there are only two white hats, C would then be able to deduce that his hat is red. But as C does not know the color of his hat, one or both of the forward hats must be red.

If A's hat is white, then B's hat must be red because C has indicated that the two forward hats are not both white. But as B is unable to deduce the color of his hat, A's hat must be red.

Therefore A's hat is red.

Well done. ;-)

1st off (= I'm just back to work) happy & healthy New Year to all

Great solution. Wow. I'd seen this riddle several years ago and just couldn't figure it out...

Best Regards

RF_g

If the last genius does not know then:

The two in front cant be each wearing a white hat. Because two whites would leave three red. Therefore the two in front are either wearing a red hat each or one is wearing a red hat and the other white?

If the second genius does not know then:

He knows that the last genius does not know, therefore he also knows that he and the first genius is wearing either a red hat each or white and red hat. And if the genius in front is wearing a red hat, he does not know if his hat is red or white.

Therefore if the second and last genius cant deduct what colour they are wearing then the first genius must be wearing a red hat. If he had been wearing white then the second genius would have know that he was wearing the red hat.

Regards JD.

Well done and a good job describing your logic.

Yes. I liked your diatribe that followed your deductions, too!

There are only four ways to arrange the middle and front hats (the only ones which may be seen by anyone). They are:

1. WW
2. WR
3. RR
4. RW

If the rear man sees WW he knows his hat is red. He doesn't so 1. is not possible.

If the middle man sees W (combination 4.) he knows his hat must be red, otherwise the genius behind him would have known his hat was red. He doesn't so 4. is not possible.

Knowing how clever the guys behind are, the front man knows combination 2. or 3. are the only ones possible. Either way, he knows his hat is red. Needless to say, he does not know the color of the hat behind him.

WHITE! right?

Pretty easy. #1 knows it is red.

The guy in the rear would have known he had on a red hat had he seen two white hats. So he's uncertain.

The guy in the middle could see either a red hat or a white hat and still have no idea what he had on because of rear guys uncertainty.

The guy in front knew right away because of the degree of uncertainty on each of the other guys leading him to the obvious answer...White Hat on my head!

You are close! Your reasoning is mostly correct, but your conclusion is still wrong. Can you see why?

red

This little puzzle was around fifty years ago and I presented it to a nimber of people back then when I taught school. Only one person figured out the answer instantly.. Most had a hard time with it.

The rear person does not know the hat he wears. He sees either two red hats or a red and a white. the middle person also does not know so the front person knows the middle person does not see a white hat on him. If the middle person sees a white hat on the first person he would know the color of his hat. Therefore the front person has a red hat on.

Chazl

Red color!

Explanation: If last genius in the row, who see two hats in front of him dont know the color of his hat, that means he see either hats of different color in front of him, or two red hats, as if he would see two white hats, then surely his hat should be red one. Middle genius can take this into account, but since he see just one color of hat in front of him, if this hat is red, he cannot know what first genius see, hats of same color or two different colors, while in case hat in front of him is white, he could be sure that his hat is red as it has greater probability to be chosen AND genius behind him dont see two white hats. Based on this, foremost genius can know for sure that his hat is RED...........

Hmm. The puzzle doesn't state that each genius can hear the others answers. In fact those geniuses, at a venerable age of 109 years, were deaf, and they had not any communication possibility but with the puzzle creator who wasn't allowed to make any comments. So the answer of the front genius would be: I DON'T KNOW!

Now, because he knew a little statistics, the probability of RED being a good answer was 60% so this is why he said, with aplomb (he was a genius, wasn't he?): RED!

Cute! ;-)

Flattered

Hm. Puzle also dont state they were able to see hats before they were blindfolded, so how they would know how many colors there are, even if blindfolding them tells they probably were not also BLIND :-))

If so, then neither of them could know which color of hat they wear, so corect answer would be: "I dont know!"........

Puzle also did not state that blindfolds were taken off, so even if they have seen hats before, it would not help, and if use of probability is tought as solution, then RED is just 60% of times correct answer, but it need not BE correct answer!

Really, if they were not blindfolded, then since there was no rule againest it, they could have simply taken hat off, look at it and say correct answer :-))

Sorry I did not think of that, but I did not peek at my hat also :-)) comming late to this tread......

Henrik,

I per aspera, per aspera, per aspera...

Yeah, me too! Perhaps becaust I keep looking up to see which color of hat I am wearing.........

Now, puzzles of this kind are good, but we are all geniuses here, so one corect answer finish contest (it would be beter if one can see other answers ONLY after giving own answer, but that may be too much to ask from CR4 team to program) and this one I guess has got enough correct answers, even some >>Out of Box<< thinking, that somebody would call cheating if rule that it is alowed all that is not strictly forbiden should not be applied :-))

I have one little problem (beside big financiall ones) and I wonder if someone could help: I am inventor and would like to find CAD/CAE program with built in simulation capability that is multidomain, i.e. fluid motion, thermal, opticall and mechanicall and electromagnetic........

Closest thing I have found is SimulationX (there is couple of others, too), at least according to advertisement on WEB, but trial version look pretty badly cripled or they advertise just their wish list.........

I have planed to make such program myself, but for this I have to hire physicist and unfortunately I did not accrued enough funds as yet......

Also, what I see is that people making such programs go entirely wrong way, asking user to input formulas that apply, instead to decide which formulas are applicable from design itself, how I would make it.........

Good program like this that would work my way would be mighty helpful to every engineer, and I could make routine for superoptimization if everything is described by parameters accesible with program!

That would surely mean less "per aspera" and more "ad Astra" for any engineer, and specially for inventors like me!

If anybody is allready working on such program, we could cooperate, or if anyone use good program with such capabilities, please let me know!

Marijan Pollak, IT SE/SA 1st Class, Instructor and Team Leader

red - last does not see 2 white hats, must be a red hat or two - if middle saw a white hat on first he would know he had a red hat, but he doesn't know - which means front has a red hat.

red. The person in back would only know if the two in front were both white, so that's out. The person in the middle would know if the person in front had white, because he would know that his could not also be white. Only if the person in front were wearing red, would the person in the middle not know if he had either the other white hat or if they were both wearing red.

Fantastic.

If I ever need a bunch of geniuses to tell me what colour of hat I'm wearing, I know whom to call.

UFG

Im sorry but there is some information missing for this to be completed.... I am thinking that the persons know how many hats and colors are there, am I correct?? obviously because if not then it would not be possible.

Good catch! We should assume that the three genius gents do know the number of possible red and white hats.