# 419 probability/reactor.p

## Description

This article is from the Puzzles FAQ,
by Chris Cole chris@questrel.questrel.com and Matthew Daly
mwdaly@pobox.com with numerous contributions by others.

# 419 probability/reactor.p

There is a reactor in which a reaction is to take place. This reaction

stops if an electron is present in the reactor. The reaction is started

with 18 positrons; the idea being that one of these positrons would

combine with any incoming electron (thus destroying both). Every second,

exactly one particle enters the reactor. The probablity that this particle

is an electron is 0.49 and that it is a positron is 0.51.

What is the probability that the reaction would go on for ever?

Note: Once the reaction stops, it cannot restart.

probability/reactor.s

Let P(n) be the probability that, starting with n positrons, the

reaction goes on forever. Clearly P'(n+1)=P'(0)*P'(n), where the

' indicates probabilistic complementation; also note that

P'(n) = .51*P'(n+1) + .49*P'(n-1). Hence we have that P(1)=(P'(0))^2

and that P'(0) = .51*P'(1) ==> P'(0) equals 1 or 49/51. We thus get

that either P'(18) = 1 or (49/51)^19 ==> P(18) = 0 or 1 - (49/51)^19.

The answer is indeed the latter. A standard result in random walks

(which can be easily derived using Markov chains) yields that if p>1/2

then the probability of reaching the absorbing state at +infinity

as opposed to the absorbing state at -1 is 1-r^(-i), where r=p/(1-p)

(p is the probability of moving from state n to state n-1, in our

case .49) and i equals the starting location + 1. Therefore we have

that P(18) = 1-(.49/.51)^19.

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