Waiting For The Bus

Relevant wiki: Convergence Tests

Bus 1 will arrive at some (unknown) fraction $t$ of a minute. We will show that, unless $t=0$ , there is zero chance no other bus will arrive by then. The chance that no other bus arrives by $t$ is the infinite product of the chances, for $i \geq 1$ , that Bus $i$ does not arrive by $t$ :

$p = \prod_{i=1}^\infty (1-\frac{t}{100i})$

Where $\exp(x)$ means $e^x$ ,

$p = \exp \Big( \ln \Big( \prod_{i=1}^\infty (1-\frac{t}{100i}) \Big) \Big) = \exp \Big( \sum_{i=1}^\infty \Big( \ln (1-\frac{t}{100i}) \Big) \Big)$ Since $\ln(x) \leq x-1$ , $p \leq \exp \Big( \sum_{i=1}^\infty \Big( -\frac{t}{100i} \Big) \Big) = \exp\Big(-\frac{t}{100}\sum_{i=1}^\infty \Big(\frac{1}{i}\Big) \Big)$ Since the harmonic series diverges to infinity, $p \leq e^{-\infty} = 0$

Bus 0, of course, also has zero probability of arriving at time 0. So there is $\boxed{0}$ probability that Bus 0 will arrive before other buses have arrived.

In fact, with probability 1, no bus will be the first to arrive. Every bus will arrive, but each bus will be preceded by infinitely many others. So you have no wait at all, but you will be unable to board the first bus, and indeed whatever bus you board, there will be infinitely many earlier ones you could have boarded!