At the center of espionage in Kznatropsk one is thinking of a new method for sending Morse telegrams. Instead of using the traditional method, that is, to send letters in groups of 5 according to a Poisson process with intensity $1$, one might send them one by one according to a Poisson process with intensity $5$. Before deciding which method to use one would like to know the following: What is the probability that it takes less time to send one group of $5$ letters the traditional way than to send $5$ letters the new way (the actual transmission time can be neglected).
I tried to solve this problem for a very long time, but I don’t understand how to find this probability.
Let $N(t)$ be a Poisson process with intensity $5$ and $M(t)$ a Poisson process with intensity $1$. For a fixed time $t$, the probability there are at least $5$ arrivals in $M(t)$ is $$ 1 - e^{-5t}\sum_{k=0}^4 \frac{(5t)^k}{k!}. $$ Let $T$ be the time of the first arrival in $N(t)$, then $T$ has $\mathsf{Exp}(1)$ distribution. So for each $t>0$ we have $\mathbb P(T>t) = e^{-t}$. Since $N(t)$ and $M(t)$ are independent, we may compute the probability that there are at least $5$ arrivals in $M(t)$ before the first arrival in $N(t)$ as follows: \begin{align} \int_0^\infty \mathbb P(M(t) \geqslant 5, N(t) = 0)\ \mathsf dt &= \int_0^\infty \left(1 - e^{-5t}\sum_{k=0}^4 \frac{(5t)^k}{k!} \right)e^{-t}\ \mathsf dt = \frac{3125}{7776}. \end{align}