## Probability of seeing a stage in a concert.

January 1, 2009 by dberenstein

It seems that the readers of this blog like puzzles. There is circumstantial evidence for that, so being a theorist, I will declare that to be true.

I thought I would start the year with one such puzzle. Here is a typical interesting probability problem that is posed for amusement. Assume that you have a stage (I’ll use cartesian coordinates and put it at (0,0)) and that people can be seated anywhere on a rectangular grid (the locations with integer entries). It could also be a honeycomb grid. In fact, any lattice will do. You can also assume that the arena where this is taking place has radius N where N is large. An important question that you might ask, is if you can see the stage when you are seated at some location. For simplicity, let us assume that you can see the stage if the straight line from your location to the origin hits no other head (another integer lattice point). If the show is packed, meaning all locations are full, what is the probability that you will have an unobstructed view?

At first sight, it might seem like that probability is very low when N is large,

but the answer is actually finite in the large N limit.

I will give you a non-standard physics-like `proof’ of the correct answer in a subsequent post that will generalize to higher dimensions immediately. I happened on this proof one morning in late December while I was in the shower and this problem popped in my head (I’m not sure why). Since I couldn’t remember the answer precisely, I ended up proving it again. This is also a famous problem in number theory if you look at it the right way. In the meantime, I will leave the puzzle on to make you think about it for a while (in case you don’t already know the answer).

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on January 2, 2009 at 12:12 amRobertImagine you have a dyon of charge (m,n). What is the chance it is stable and cannot decay?

on January 2, 2009 at 12:31 amcarlbrannenGiven p and q integers, what is the probability that they are relatively prime? To do this, they must have no common factor of 2, or 3, or 5, or 7, or …

Sounds like Riemann zeta function stuff.

on January 2, 2009 at 2:18 amonymousFinite head size in the real world complicates matters. Also the necessity for super-tall hipster dudes to stand next to their super-petite hipster girlfriends. The emergent organization at small venues seems highly non-optimal.

on January 2, 2009 at 9:08 ammeichenlMy intuition is the same as Carl’s. I think the answer has to do with the exponential of -zeta(2).

As he mentioned, the condition is that (x,y) coordinates must be relatively prime. 3/4 of pairs have no common factor 2, 8/9 have no common factor 3, 24/25 have no common factor 5, etc. For large numbers these “probabilities” are “independent”, because, for example, 3/4 of all integer pairs have no common factor 2, and also 3/4 of pairs that do have common factor 3 have no common factor 2.

P(can see stage) = 3/4*8/9*24/25*48/49*…

log(P) = log(3/4) + log(8/9) + log(24/25) +…

log(1+ x) = x for small x

log(P) = -1/4 – 1/9 – 1/25 – 1/49 – … = -zeta(2) = -1.65

P = 20% approx.

on January 2, 2009 at 9:22 ammeichenlah, that’s not actually zeta(2) i had there. I wanted zeta(2) – 1, which gives something more like 52%

on January 2, 2009 at 3:16 pmonymousmeichenl: right idea, but you want the “Euler product formula” to get an exact answer.

on January 2, 2009 at 4:04 pmdberensteinDear Onymous:

A little work shows that Meichenl’s calculation is in the end the correct Euler product formula. And yes, the answer is .

on January 4, 2009 at 11:53 pmProbability of seeing a stage, part II « Shores of the Dirac Sea[…] 4, 2009 by dberenstein In the previous post in this series, I described a problem dealing with a lattice with an origin, and asked what is the probability […]