Heptagonal Spider Web
A spider has its web in the shape of a regular heptagon (seven sides) . A fly is stuck to the web at a vertex that is three edges away from the vertex at which the spider is. The spider can walk freely along the edges of the heptagon. At each vertex, it randomly chooses between the two adjacent edges, with equal probability, then proceeds to walk along the chosen edge. On average, how many edge lengths will the spider walk before getting to the fly?
The answer is 12.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Suppose the spider is walking on the sides of a regular N -gon. Label the vertices 0 , 1 , 2 , . . . , N − 1 , with 0 the vertex where the fly is. Let E j be the expected number of edges that the spider has to move along to get to the fly, given that the spider is currently at vertex j . Then E j = 1 + 2 1 ( E j + 1 + E j − 1 ) 1 ≤ j ≤ N − 1 with E 0 = E N = 0 . Thus we deduce that E j = F j − j 2 for 0 ≤ j ≤ N where F j = 2 1 ( F j + 1 + F j − 1 ) 1 ≤ j ≤ N − 1 and hence F j = A + B j for 0 ≤ j ≤ N for some constants A , B . Thus E j = A + B j − j 2 . Since E 0 = E N = 0 we deduce that E j = j ( N − j ) for 0 ≤ j ≤ N .
In this case N = 7 , and we want E 3 = 1 2 .