Waiting for two buses Coming back from work today I had the option to take one of two buses arriving one after another, both of the same line (i.e., going to the same place).
The stop where I get on is relatively early on the route, so both buses were fairly empty by the time they pulled up - so I had no way of knowing at that point which one would be full of people first.
Some assumptions for an idealized scenario based on my commute:


*

*I'm the only person at the stop - and so don't see other people's choices of bus

*It's more desirable to be on a bus with fewer people

*This bus goes through the downtown core of a large city at rush hour before getting to my destination - so one of the buses will be packed with people by the time I want to get off
Speaking as a layman - someone with only a very basic knowledge of game theory - does game theory allow me to determine which bus would be emptier?
 A: The only criteria you've supplied for determining the boarding rate is the qualitative assessment that its more desirable to board a bus with fewer people.
The bus you get on will have more people on it than the one you don't so the next passenger to arrive will find the other bus more desirable to board.  If that passenger decides to choose the more desirable option, the buses will again be equally desirable and should be expected to fill at the same rate.  However if that passenger chose the less desirable (and thus less likely)option , the other bus instead becomes even more desirable to the next passenger after that.
Let's get some ballpark numbers and say that the probability of choosing a bus is determined by the ratio of people-plus-ones on the buses; such that, if bus $A$ contains $m$ people and bus $B$ holds $n$ then let the probability of choosing bus A given these numbers be $p_{m,n} = \frac{1+n}{2+m+n}$.  
So $p_{0,0} = p_{1,1} = 1/2$ and $p_{1,0} = 1/3$, etc.
So if you choose bus A and two other people arrive, let $X$ be the number of people on bus $A$.  $$\begin{align}
P(X=3) & = p_{1,0}p_{2,0} \\[1ex] & = \frac{1}{3}\frac{1}{4} \\[1ex] & = \frac 1 {12}\\[2ex] P(X=2) & = p_{1,0}(1-p_{2,0})+(1-p_{1,0})p_{1,1} \\[1ex] & = \frac{1}{3}\frac{3}{4}+\frac{2}{3}\frac{2}{4} \\[1ex] & =\frac 7 {12} \\[2ex] P(X=1) & = (1-p_{1,0})(1-p_{1,1}) \\[1ex] & = \frac{2}{3}\frac{2}{4} \\[1ex] & =\frac 4 {12} \\[2ex] \mathsf E(X) & = 3\times\frac 1 {12} + 2\times \frac 7 {12} + 1\times\frac 4 {12} \\[1ex] & = \frac {9}{4} \\[1ex] & = 1.75 \end{align}$$
While on the other bus: $$\mathsf E(Y) = 3-\mathsf E(X) = 1.25$$
This is a rough indication, but it would seem that whatever bus you board is expected to fill slightly more than the other - because you are on it.
