Chance of meeting in a bar

Two people have to spend exactly 15 consecutive minutes in a bar on a given day, between 12:00 and 13:00. Assuming uniform arrival times, what is the probability they will meet?


I am mainly interested to see how people would model this formally. I came up with the answer 50% (wrong!) based on the assumptions that:


- independent uniform arrival
- they will meet iff they actually overlap by some \(\displaystyle \epsilon > 0\)
- we can measure time continuously


but my methods felt a little ad hoc to me, and I would like to learn to make it more formal.


Also I'm curious whether people think the problem is formulated unambiguously. I added the assumption of independent arrival myself for instance, because I think without such an assumption the problem is not well defined.

Rather than measure time continuously, lets start with discrete time to try to get a handle on an approach. Divide the time into n equal minutes: d, 2d, 3d, ...,nd where nd= 1 hr. = 60 min. The first person arrives at time jd. Ignoring end points, the probability that the next person arrives within 15d of that time is (j-15)d to (j+15)d or 30d. Since j can be any of n numbers we have, ignoring end points, the probability is 30d/60d = 50%.

That gives us an approach, but we have to take into account the end points, i.e. the time restriction. Well 1/n times the arrival will be at 1d and the sum will be restricted to d to 16d or just 15d. Next is arrive at 2d and sum is 16d, etc. So, instead of a constant function of 30d, we have a ramp function starting at 15d going up to 30d, staying there for awhile and going back down to 15d. Add those up with the proper weights.