==> 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.