| Topic Name: |
Brainteaser |
| Message Name: |
nope. here is the answer to the 3 |
| Date Posted: |
03/23/2002 |
| In Reply To: |
1-1/{2^(n-1)} where n is the number of people for n>1. If n=1 then it's 1/2.
The idea is for all of them to line up and then starting at one end, each person passes unless all the people in front of them have alternating hat colours in which case they name the hat colour of the person standing in front of them. In the case where this is true for the first person he can just guess, but the probability is the same as if he just names the hat colour of the person in front of him.
The idea is that if all the people behind you have passed, it's because somewhere in front of them there is a set of identically coloured hats in sequence. So if you can't see such a sequence in front of you, you know that you're the second last person in the sequence.
As for the probability it's probably easiest to see like this- There's a half chance that the end hats are of the same colour. If not there's a half x half (=1/4) chance that the second and third last hats are of the same colour and so on. Thus, totalling the probabilities, the more people, the greater the chance that someone will be able to answer with certainty.
As far as not being allowed to wait and react is concerned I don't see how you can improve the odds from 1/2 if nobody's allowed to know whether anyone else is going to pass. In such a case, there's no additional information given to anybody and since the colour of your own hat is independant of everyone else's presumably the best that can be done, is for one person to be nominated as a guesser and everyone else to pass, which would give odds of 1/2. |
| Message: |
You can beat 1/2 and you don't have to wait.
all three agree to answer only if they see that the other two have the same color hat. So the lights come on, if any of them see that the other two have the same color hats they say the opposite color. If any sees that the other two have different color hats, he passes. There are 8 total combinations only in the combinations where all three are the same color will they all be wrong and lose. They are right the other 6 times so the answer is 3/4.
I have no idea how to do the situation for 7 but the theoretical correct answer is 7/8. It involves using some complex switching coding math I don't understand. I found this question originally in the Times. search for "Hat Game" if you are interested.
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