How to choose whether to quit the bus queue or stay there using probability theory? I've been thinking of something for some time now, and since I am not very proficient in probability theory I thought this could be a good place to ask this question. This is something that came up to me in the long queues of the public transport.
Suppose that you're in a bus station, and you know that a bus (or several buses) will certainly come in the future (during the day,) but you don't know the exact moment. You imagine a probability that the bus will arrive within five minutes. So you wait five minutes. But the bus doesn't arrive. Is now the probability less than or greater than the original one you imagined?
The question is because if you're using the past to predict the future, maybe you won't be very optimistic about the bus arriving. But maybe you could also think that it actually makes the event more likely: since the bus hasn't arrived yet, there are less minutes available in the day and thus the probability is higher.
Think of the last five minutes of the day. You've been there the whole day and no buses have come. So, judging solely from the past, you can't predict that the bus is going to arrive within the next five minutes. But since you're sure that a bus will arrive before the day ends, and there are only five minutes for the day to end, you can be 100% sure that the bus wil arrive within five minutes.
So, the question is, if I'm going to calculate the probability and drop out of the queue, which method should I use? It's because sometimes I quit and suddenly the bus arrives, but sometimes I wait and wait and wait and the bus doesn't come. Or maybe this whole question is nonsense and that is simply terribly random?
 A: It depends on how near to a schedule your buses are coming at.

*

*If they were actually keeping to a regular schedule, every minute you wait is a minute closer to a bus arrival, and on average you wait half the inter-bus interval.


*If the buses were to arrive at varying inter-bus times, at a certain average rate per hour, you're more likely to arrive at the bus stop in a long gap than a short one. Indeed, if they arrive "effectively at random" (according to a Poisson process) it doesn't matter how long you wait, your expected remaining wait is the same.


*If things get worse than that (gappier/burstier than "random" arrivals, perhaps because of traffic issues) then you could be better off not waiting.
A: great question!
From a probability perspective, waiting may certainly make the odds go up.  That will be true of Gaussian and Uniform distributions.  It would not, however, be true for exponential distributions - the neat thing about exponential distributions being "memoryless" in that sense as the probably for the next interval is always the same.
However, I think a more interesting thing might be to generate some cost function.  What is the cost of the alternate transportation (taxi, ueber)?  What is the cost of being late?  Then you can dust off the calc book and minimize the cost function.
To convince myself that the odds always increase for Gaussian distributions, I wrote a bit of matlab, but I will try to come up with something more mathematically pure.  I think for uniform it is obvious, as the numerator is constant (until nothing) and the denominator is always decreasing towards nothing.
A: I think you answered your own question. Suppose you are sure that n buses will arrive by the end of the day (which is h hours away) but are not sure when in those h hours they will arrive, you can use a poisson distribution with rate equal to n/h and compute the probability of a single bus arriving in the next ten minutes, say. As you wait for the bus and h starts to reduce, the rate n/h begins to increase and the chance that a buss will arrive in the next ten minutes increases. So with every passing moment, it makes less and less sense for you to quit the queue (assuming the bus will have space for you when it arrives). 
A: If you drop the restriction that the bus must arrive at some point during the day then it can be argued that the longer you wait, the longer you expect to have still to wait.  The reason?  The longer you wait, the greater your belief that the Poisson rate parameter is small.  See question 1, here.
