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.
Let E1 be the expected time from being one side away; E2 from two sides away and E3 from three sides away.
E1 = 5/3 + (2+E1)/3 + (E2+5)/3
E2 = (E1+5)/3 + (2+E2)/3 + (E3+5)/3
E3 = 2 * (E2+5)/3 + (2+E3)/3
The spider starts three sides away so the expected time is E3.
Using Wolfram Alpha (using x, y and z for E1, E2 and E3):
x = E1 = 32
y = E2 = 52
z = E3 = 60
The answer is 60 seconds or 1 minute.
This assumes the spider makes his decision the instant he arrives at the vertex. That is probably the correct interpretation of the wording of the puzzle.
Variation:
If, however, the 2 second time was the time to make a decision--even the one to stay put, and that time was not included in the five seconds travel time, then all the 5's in the equations need changing to 7's to allow for the thinking time before the actual move, and the respective times are 42, 68 and 78, with the latter being the answer in seconds.
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Posted by Charlie
on 2019-09-12 18:49:10 |
my solution
