A telephone wire stretched tight between two poles placed ten meters apart is a favorite resting spot for a flock of crows.
Suppose one morning two crows land on the wire, each at a random spot (the probability is uniformly distributed). With a bucket of paint and a brush you mark the stretch of wire between them. A certain length of wire will have been painted.
On average, what length of wire would you expect to have painted? Assume that each bird is a single point along the line, and so has no width.
Suppose instead that a dozen crows landed on the wire, each at an independent, random location, and you painted the stretch of wire between each bird and its nearest neighbor. On average, what total length of wire would you expect to have painted now?
And if a thousand crows landed?
A computer-generated solution could be found, but bonus points will be awarded for a formal proof!
Im not sure if this is right at all but maybe someone can let me
know. The misleading thing about the problem is that you assume
that the birds will pair up and there will be 6 connecting lines.
Actually if you put the birds on the wire in the series,
1.5+(.4+.1N) or 0,1.5,1.9,2.4,3 ,for example, each will connect with
the bird before it and the bird on the end will connect with the bird
next to it no matter the length. The least number of lines is 6 and the
most is 11, with every bird being connected. To summarize the rest:
#b= number of birds
avg#lines=((#b-1)+(#b/2))/2 or for 12 birds- the avg of 6&11 or 8.5
and the average length of each line=(10/#b) or for 12 birds=.83
Then to get the total you would multiply these two numbers by eachother (total length=(avgL of line)(avg# of lines)
x=(8.5)(.83)=7.055 meters of total painted line
for 1000 birds 7.495=(1499)(.01)
Let me know if this makes any sense
Posted by Aaron
on 2004-07-01 21:16:35