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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