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Re: The Six Dice Puzzle
I don't think it is correct, every time you lift a cup and it is not a 6 you have a higher probability of having a six...but to do the probability calculus you need to what is the probability of having a six in the first place (which is less than 1/6 since you know that 1 or zero dice are six (while to have the probability of 1/6 you would need to know that 1 dice is 6), hence there is a probability given in advance...).
For example if we have the probability of 75% to have one six then:
cupA: p(6|A)=prob(A)prob(throw 6)prob(external75%)=1/6*1/6*3/4
cupB: p(6|B)=1/5*1/6*3/4, since now only five ways to choose the order of cups
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cupF: p(6|F)=1*1/6*3/4
-->hence it seems to me that in F it should be the most probable one. Now, if I understood wrong and you meant that there is the probability of 1/6 any dice gives 6, than they same argument holds without the term 3/4=75%...
Just note, I don't know if this is correct, but I think it is the way to think.
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