Very nice indeed, thanks!
-- Owen
On Aug 18, 2008, at 2:19 PM, Roger Critchlow wrote:
> On Mon, Aug 18, 2008 at 11:19 AM, Owen Densmore
> <[EMAIL PROTECTED]> wrote:
>
>> 1 - The probability for each path is calculated by looking at the
>> possible choices at each point in the path. If you see a "3" at a
>> node, for example, the probability assigned to the next move is 1/3.
>> The total probability for a given path is the product of the
>> individual ones.
>> Question: how can I show that this forms a true probability
>> space? .. i.e. the sum of the probabilities for all possible paths is
>> 1.0.
>>
>
> Given the paths of length N, i, and their probabilities, p(i), we
> form the
> paths of length N+1 by one of the following steps:
>
> 1) if there is only one way to extend path i, then the probability
> of that
> extension is p(i) * 1.
>
> 2) if there are two ways to extend path i, then the probabilities
> of those
> two extensions are each p(i)/2.
>
> 3) if there are three ways to extend path i, then the probabilities
> of
> those three extensions are each p(i)/3.
>
> There cannot be four or more ways to extend path i.
>
> If the sum(i, p(i)) == 1, then the sum(i, p(i)*1 or 2*p(i)/2 or
> 3*p(i)/3) ==
> 1.
>
> Starting from the lower left corner, we have 2 choices, each assigned
> probability 1/2. 2 * 1/2 == 1.
>
> If it's true for the paths of length N, then it's true for the paths
> of
> length N+1; and it's true for the paths of length 1.
>
> -- rec --
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