I'm wondering about this probability problem: Three Erupting Geysers.
TL;DR = There are $3$ geysers $A$, $B$ and $C$ which erupt at exact intervals of $2$, $4$ and $6$ hours respectively. Note that these intervals are fixed, so the time between two eruptions of $A$ is always exactly $2$ hours, et cetera. You don't know the phase of each geyser's cycle.
Question: when you arrive at a random moment, what are the probabilities for each geyser to erupt first?
The correct answer is [spoiler] (see youtube link above for explanation).
Now what I'm actually looking for: is there a way to solve this with integrals?
It seems to me that the $3$ geysers' phases are uniformly distributed over $2$, $4$ and $6$ hours. Note that I'm not talking about the time between subsequent eruptions for each geyser, as these occur in fixed $2$/$4$/$6$ hour intervals. Rather I mean the phase or offset within each geyser's eruption cycle.
Put differently: at the moment you arrive, the time it will take for geyser $A$ to erupt is a uniformly distributed random time between $0$ and $2$ hours, and likewise $0$ to $4$ hours for $B$, and $0$ to $6$ for $C$.
If we consider two hours as one 'unit of time', we have three Uniformly distributed variables $A$~Uni$(0,1)$ and $B$~Uni$(0,2)$ and $C$~Uni$(0,3) $ where $A$, $B$ and $C$ are the time it takes for that geyser to erupt after you arrive. Now the question is: what is the probability that $A$ is the smallest of these three, and similar for $B$ and $C$.
We only have to observe the first two hours (i.e. time from $0$ to $1$) because after that at least $A$ will have erupted for sure.
For any time $t$ between $0$ and $1$, we have the following probabilities:
- $P(A≤t) = t$
- $P(B≤t) = t/2$
- $P(C≤t) = t/3$
- $P(A>t) = 1-P(A≤t) = 1-t$
- $P(B>t) = 1-P(B≤t) = 1-t/2$
- $P(C>t) = 1-P(C≤t) = 1-t/3$
- note that for $t>1$ we have $P(A≤t)=1$ and $P(A>t)=0$
Now for example to calculate the probability that $B$ is the smallest, i.e. P(B<A and B<C) I would say we can integrate P(A>t)•P(B≤t)•P(C>t) over t=0..1:
$$ \int_0^1 (1-t)\cdot\left(\frac{t}{2}\right)\cdot\left(1-\frac{t}{3}\right) \mathrm dt = \frac{5}{72} $$
Likewise, the result for A would be 19/72 and C would be 3/72.
These answers are incorrect (obviously, as they don't even add up to $1$). Why is this wrong? What is the mistake in this approach?
Note that this is NOT a distribution of the minimum of multiple uniform distributions. This is about which of multiple uniform variables turns out to be the minimum.
You're integrating the probability that at time $t$ $B$ has erupted and $A$ and $C$ haven't. That can't be the probability for $B$ to erupt first, since it depends on when this happens. For instance, if all three geysers were very likely to erupt in the first few seconds after your arrival, this integral would be very small, whereas the probability of $B$ erupting first could still be anything between $0$ and $1$. So you're measuring something like the extent of time over which something is likely to happen, not a probability.
Instead of the probability of $B$ not having erupted yet before time $t$, you need the probability density of $B$ erupting at time $t$, i.e. the derivative of $P(B\le t)$ with respect to $t$. That yields
$$ \int_0^1(1-t)\cdot\frac12\cdot\left(1-\frac t3\right)\mathrm dt=\frac29\;, $$
as expected.