I've modeled the following problem with monte-carlo and my simulations give an answer that's usually just barely lower than my calculation-- I'm wondering if it's myself or my model that's wrong.
The problem is as follows:
We have 5 planes, each labeled $A,B,C,D,E$. They are leaving an airport in a random, uniformly distributed order. Two people arrive together at the airport. They do not know what planes have left yet. It is equally likely that they've arrived at the airport before any of $0,1,2,3,4,5$ planes have left.
Both of the two individuals will take the first plane available that takes them to their destination.
Planes A and B will take both persons to their desired destinations. Planes C and D will only take one of the two people to their desired destination. Plane E takes neither of the two to their desired destinations.
If we know that the two people boarded the same plane, what is the probability that they arrived at the airport before the first plane left?
My logic is as follows:
The probability of them leaving together is $\frac{1}{2}$.
The probability of them leaving together on the first plane they see is $\frac{2}{5}$, with associated probability $\frac{1}{5}$ that they came before the first plane left.
The chance of plane E being the first plane they see is $\frac{1}{5}$. The chance of the second plane being either A or B is $\frac{1}{2}$, with associated probability $\frac{1}{4}$that they arrived before the first plane left.
$$P(Arrived\ before\ first\ plane |Left\ together) = \frac{P(Arrived\ First \cap Left\ Together)}{P(Left\ Together)}$$
$$P(Arrived\ before\ first\ plane |Left\ together) =\frac{\frac{2}{5}\cdot \frac{1}{5}+\frac{1}{5}\cdot \frac{1}{2}\cdot \frac{1}{4}}{\frac{1}{2}} = \frac{21}{100}$$
Is this reasoning / answer proper?
Thanks!
The probability of them leaving together is $\frac{1}{2}$ only if you condition on there being at least one of $A,B,C,D$ remaining.
That will not happen if they arrive after the last plane leaves (probability $\frac16$) or if they arrive after the second last plane leaves and last plane is $E$ (probability $\frac16\times \frac15$).
So the probability of them leaving together is $\frac{1}{2} \times \left(1-\frac16 - \frac16\times \frac15\right) = \frac25 $.
The probability they arrived before the first plane and left together $\frac16 \times \frac12 =\frac1{12}.$
So the probability that they arrived at the airport before the first plane left, given they left together, is $\dfrac{\frac1{12}}{\frac25}=\frac{5}{24}$.
This is $0.208333\ldots$, barely lower than your $0.21$.