Q: 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 and does not leave it. 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.

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.