> Sloppy Joe <[EMAIL PROTECTED]> wrote:
> >Greetings -
>
> >Suppose I have a method such as a fair 6-sided die. I roll the die 10
> >times and get the following trial history:
>
> >3-5-1-3-6-4-6-2-1-5
>
> >From what I can recall from probability, I cannot predict the next roll
> >of the die based on the previous rolls. Each number on the die will
> >have an equal probability of occuring on the 11th roll no matter what
> >has occurred previously: 1/6. My question is this ...
>
> >How can this be proven mathematically?
It can't be. It's just the conventional definition of "fair".
There is no reason why a die cannot be weighted so that the probability
of a certain number appearing on any roll is *not* 1/6; and there is no
reason why a die cannot be constructed (say, using some cunning
clockwork, or lead shot in a very viscous liquid) so that the
probabilities for one roll depend on the rolls before it. Neither of
these is ruled out by the observation described above (though they
provide some evidence against the "repeating" die last described).
Moreover, a die may have equal probability without independence, or
independence without equal probability. "Fairness" is generally
understood as implying both properties.
(This may not quite correspond to the general meaning. If I say "I have
a weighted die in my pocket that only rolls one number. I bet you 5 to 1
that you cannot guess what it will roll!" that is a fair bet - once!
Moreover, it is a fair bet even if I do not tell you that the die is
loaded, provided its faces only have conventional numbers of spots (1
through 6, though not necessarily one of each.))
-Robert Dawson
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