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It sounds like a prisoners dilemma type of question - no 'correct' answer. If acting purely in self interest you should choose 300.
Further - you have no way of knowing how many iterations the average person will make. But you could assume that the probability of x iterations decreases as x increases.
So the 1st iteration is that everyone would pick 300 as it is bigger.
The 2nd iteration is that 200 is a better bet if 6 or more people pick 300.
The 3rd iteration is that 300 is a better bet if 5 people get to iteration 2.
Etc etc.
The extra size of the kitty for voting 300 makes it the better choice in a large enough sample I think
I think there must be an error in the question. One prisoner has to risk getting it wrong - surely it should be that one prisoner has to be right for them all to go free.
E.g. Prisoner 1 writes x if he can see odd number of colour 1, y if he can see even number of colour 1.
Prisoner 2 does the same with colour 2, 3 with 3, 4 with 4.
Prisoner 5 can therefore ascertain what is on his head. Not sure how you do it simultaneously though
Has anyone got anywhere with the Omnichronicon puzzle that's been doing the rounds? I can get that the numbers stand for letters (I.e 1 = A) and then you get PRIME MOVER on the right hand side, but after that I'm stumped. Seems like I'm not the only one!
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