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Imagine that you have a very long (finite) line and that you are asked to mark a number of points in sequential fashion on the line, subject to a few conditions:

(1) For each new point that is marked on the line, there is a region of length L either side of that point within which no consecutive points are allowed to be marked. In other words, new points may only be marked if they are at least a distance of L from all other existing points.

(2) The location of each new point is chosen at random from any of the available locations along the line.

(3) Once there are no more possible locations, no new points are marked.

What would be the final average spacing of the points? The final spacing must be between L and (2)L, and the most obvious guess would be (3/2)L. However, I have done thousands of computer simulations of the problem and the average spacing seems to approach (4/3)L - but I have no way of proving this. Any thoughts?

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    What does "very long" mean in this context? Must it be finite? Does it relate to $L$ in some way? – 2017-01-12
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    "Very long" may be taken to be infinite. But in my simulation, I've taken the total length of the line to be in the order of 100 times L. – 2017-01-12
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    Correction: the line can't be infinitely long, because it could never fill up with enough points. So assume that the length of the line is finite but at least a few orders of magnitude longer than L. – 2017-01-12
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    Also, it doesn't make obvious sense to say 'select a random real number' if the line is 'infinitely long'. So maybe, the equivalent would be: Imagine you have the interval $(0,1)$, and that $L$ is such that $0 < L < 1$. Same question I think. – 2017-01-12
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    See [The parking problem riddle](http://math.stackexchange.com/questions/376236/the-parking-problem-riddle) This is not quite a duplicate as the disallowed regions are allowed to overlap here, but are not in the linked problem. – 2017-01-12
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    Great, that seems to answer the question - at least in part. Thank you. Edit: after a more rigorous simulation of the problem, the spacing of the points looks as though it's exactly what is predicted by the parking problem, so perhaps the issue of overlap is unimportant. I'm still not quite able to explain this though. – 2017-01-12
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    I think the reason the final spacing is less than $\frac {3L}2$ is that you can put points just over $2L$ apart, forcing the final spacing there close to $L$. This will bias the average downward. – 2017-01-12

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