A spider has its web in the shape of a regular hexagon. A fly is stuck to the web at a vertex that is diametrically opposite from the vertex at which the spider is. The spider can walk freely along the edges of the hexagon. At each vertex, it randomly chooses between walking on one of the two adjacent edges or staying at the vertex, all three choices with equal probability. If the time it takes to travel an edge is 5 seconds, while the waiting time at a vertex is 2 seconds, find the expected time it will take the spider to get to the fly?
Note: At the end of a waiting period at a vertex, a new random decision to stay at the vertex, or move along one of the two edges is made, with equal probability for the three choices.
(In reply to
my solution by Charlie)
Going over my entry into Wolfram Alpha, I must have miskeyed it, as
x=5/3+(2+x)/3+(5+y)/3,y=(x+5)/3+(2+y)/3+(5+z)/3,z=2*(y+5)/3+(2+z)/3
results in
x = 30, y = 48, z = 54
in agreement with the Larry's and Steven's results.
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Posted by Charlie
on 2019-09-12 19:34:38 |
re: my solution
