Why is the throughput (performance) of the Aloha protocol ≈ 0.18?
This number is reached first by assuming that messages will be produced and sent according to a Poisson process.

$$P[(N(t + \tau) - N(t)) = x]  = \frac{{e^{ - \lambda\tau } (\lambda\tau) ^x }}{{x!}}$$

This means, that the probability of $x$ messages arriving for the given interval $[t, t+\tau]$ where $\lambda$ messages are expected to arrive, is equal to right side of the equation. In other words, if we are expecting 10 messages to arrive per hour, the chances of 6 messages arriving in an hour is roughly 6.3%.

Now, messages sent by ALOHA need to be alone in order to make it to all other the nodes undisturbed. In other words, no other nodes can be sending messages at the moment we want our message to be sent and must stay silent until our message is completely sent.

This means we want to find: $$P(exactly\space1\space\space message\space arriving) * P(exactly\space0\space\space messages\space arriving\space in\space next\space t)$$

The odds of exactly one message arriving is

$$P[(N(t + 1) - N(t)) = 1]  = \frac{{e^{ - \lambda\tau } (\lambda\tau) ^ 1 }}{{1!}}  = \lambda\tau e^{ - \lambda\tau} = \lambda e^{ - \lambda}$$ 

The odds of 0 messages arriving for $t$ is:

$$P[(N(t + 1) - N(t)) = 0]  = \frac{{e^{ - \lambda\tau } (\lambda\tau) ^ 0 }}{{0!}}  = e^{ - \lambda\tau} = e^{ - \lambda}$$ 

and together they make 

$$\lambda e^{ - \lambda} \times e^{ - \lambda} = \lambda e^{ -2 \lambda}$$.

This function tells us the throughput of the Aloha for all values of $\lambda$. So which $\lambda$ offers the highest throughput? To find out, we take the derivative of $\lambda e^{ -2 \lambda}$ and get:

$$ e^{ -2 \lambda} - 2\lambda e^{  -2 \lambda}$$ 
set to 0 and solve for $\lambda$ and we get

$$\lambda = \frac{1}{2}$$

This means that in order to have the highest throughput (i.e. be most efficient) we should have an expected rate of incoming messages to be $\frac{1}{2}$ per timeslot. 

And after all that, what is the maximum throughput that we get when we have an expected rate of $\frac{1}{2}$? Simply, plug it into our equation  $\lambda e^{ -2 \lambda}$ and we get:

$$\frac{1}{2} e^{ -1} = 0.18393.... $$