Places in the restaurant

Question

My question is:

Johnny goes to the restaurant every evening to see if there's a place, and if there is he eats a meal. If there's no place he goes home and tries the day after. Johnny is going to go there everyday until there's a place and he will eat a meal (and then he will stop to go). It is known that the places in the restaurant distributes by Poisson with 0.4 rate of free space. What is the $V(x)$ of the number of times Johnny is going to go to the restaurant until he will eat a meal?

If I do the Poison distribution as a geometric (with Poisson variables), and then from that I tried to calculate the V(x) by the formula $(1-p)/p^2$ where $p$ is $(1-e^.4)$, I get a negative number - any ideas?

Answer 1

If $X$ represents the number of tables available on a given day, then $$X \sim \operatorname{Poisson}(\lambda = 0.4),$$ with $$\Pr[X = x] = e^{-0.4} \frac{(0.4)^x}{x!}, \quad x = 0, 1, 2, \ldots.$$ Therefore, a table is available if $X \ge 1$, or $$p = \Pr[X \ge 1] = 1 - \Pr[X = 0] = 1 - e^{-0.4}.$$ The number of days $Y$ that he will visit until a table is available is a geometric random variable $$Y \sim \operatorname{Geometric}(p = 1 - e^{-0.4})$$ with PMF $$\Pr[Y = y] = (1-p)^{y-1} p, \quad y = 1, 2, \ldots.$$ The variance of $Y$ is $$\operatorname{Var}[Y] = \frac{1-p}{p^2} = \frac{e^{-0.4}}{(1-e^{-0.4})^2}.$$

Answer 2

Obviously Johnny needs only one place, so if the number of free places is Poisson($0.4$), the probability to have no free spaces is $P(S) = e^{-0.4}$. This can be seen as a parameter in Geometric distribution, $p$, so the probability of $n$ trials (n-1 failed and 1 last successful) is $$ P(S) = e^{-(n-1)0.4} (1-e^{-0.4}) $$