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[Solved]: probability wheel, redistribution of probabilities

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Problem Detail: 

I have a contiguous ordered data structure (0 based index):

x= [1/3, 1/3, 1/3] 

Let's say I selected index 1 and increased the probability by 1/3. Rest of the probabilities each decrease by 1/6 and the total probability remains P = 1.

x= [1/6, 2/3, 1/6] 

Let's say I selected index 2 and increased the probability by 1/3. Rest of the probabilities in total need to decrease by 1/3 to make the total probability remain P= 1.

x= [1/10, 2/5, 1/2] 

Is there a name for this kind of data structure? I'd like to research that name and use a library instead of my custom rolled code if possible.

Asked By : pwned

Answered By : A.Schulz

This can be easily achieved by an array $A$ and an additional variable $m$. The entry $A[i]$ stores the probability of $i$ times $m$.

In your example you start with

A=[1,1,1], m=3 

The you increase the probability of the second element by 1/3. This can be carried out by setting $A[2]=4$ and $m=6$, which gives

A=[1,4,1], m=6 

In general you could set the probability of element $k$ to $p$ by solving the system $$\sum_i A[i] = m \text{ and } A[k]=p\cdot m,$$ for the unknowns $A[k]$ and $m$.

So all you need to do is to set $m$ to $$ m = \frac{\sum_{i\neq k} A[i]}{1-p},$$ and $A[k]=p\cdot m$ for the new $m$.

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Question Source : http://cs.stackexchange.com/questions/6741

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