Specifically, assume infinitely many hosts (this is a technical assumption required by the analysis). Collectively the hosts try to send new and retransmitted packets at a rate described by a Poisson distribution with a mean of N frames per frame time. For anything useful to happen, N must be less than one.
The throughput is the expected value of the offered load, that is, Np where p is the probability of a successful transmission. A host is successful if it “reserves” the channel for two frame times, that is, if no other host tries to send for up to two frame times.
Pr[k], the probability that k hosts will transmit during a frame time, is given by \[ \frac{N^ke^{-N}}{k!} \] For two frame times the mean doubles and the probability is given by \[ \frac{(2N)^ke^{-2N}}{k!} \] The probability that no host sends for two frame times is p = Pr[0] = e-2N, and the calculated throughput is Ne-2N.