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Suppose we have a "particle" which can be at some position x∈N (where N={0,1,2,...}). The probability that the particle is at position x can be written as:

P(x) = 1/(2

Now suppose we have two particles p1 and p2.To keep things simple, assume that the individual probability distribution for each particle is the same as above (that is, if the other particle was absent).

The main condition is that we don't want to allow both particles being at the same position (note that we are talking about either (i) at the same time or (ii) assume both particles to be still).

The question is that what would be the probability that p1 is at position a and p2 is at position b (where a≠b) at some given time?

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Here is my own attempt for the answer:

The probability should be:

C*P(a)*P(b)

where C is a constant that we have to determine.

We define a constant c to be:

c=(1/2)

Now we define:

C=1/(1 - c)

==========

The main idea I had in mind was something along these lines. Few months ago I was reading about "spaceships" in Conway's game of life. I was thinking that suppose we have a spaceship that launches from the origin and is moving along the positive y-axis. At some point along the way there is a "dust storm" (say initially composed of thousand or so cells that stretch over a strip of finite vertical length but infinite horizontal length -- in terms of probability distribution). We want to make something like a computer simulation to determine the probability that the spaceship will not be "destroyed" by the dust storm. We would need to specify precisely a more precise definition of the term "destroyed" though it seems.

P(x) = 1/(2

^{x+1})Now suppose we have two particles p1 and p2.To keep things simple, assume that the individual probability distribution for each particle is the same as above (that is, if the other particle was absent).

The main condition is that we don't want to allow both particles being at the same position (note that we are talking about either (i) at the same time or (ii) assume both particles to be still).

The question is that what would be the probability that p1 is at position a and p2 is at position b (where a≠b) at some given time?

-----

Here is my own attempt for the answer:

The probability should be:

C*P(a)*P(b)

where C is a constant that we have to determine.

We define a constant c to be:

c=(1/2)

^{2}+(1/4)^{2}+(1/8)^{2}+(1/16)^{2}+......Now we define:

C=1/(1 - c)

==========

The main idea I had in mind was something along these lines. Few months ago I was reading about "spaceships" in Conway's game of life. I was thinking that suppose we have a spaceship that launches from the origin and is moving along the positive y-axis. At some point along the way there is a "dust storm" (say initially composed of thousand or so cells that stretch over a strip of finite vertical length but infinite horizontal length -- in terms of probability distribution). We want to make something like a computer simulation to determine the probability that the spaceship will not be "destroyed" by the dust storm. We would need to specify precisely a more precise definition of the term "destroyed" though it seems.

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