The Cube is shown above, where A is aman and M is the monkey. Let p be the probability that Aman feeds the monkey given that he is holding the banana. For each non-monkey neighbor of aman, let q be the probability that aman feeds the monkey

given that the neighbor is holding the banana. (By symmetry, it is the same probability for each neighbor.) For each non-aman neighbor of the monkey, let r be the probability that aman feeds the monkey given that the neighbor is holding the banana. (Again, by symmetry, it is the same probability for each of the monkey's

other neighbors.) Finally, let s be the probability that aman feeds the monkey given that the person opposite from aman is holding the banana. Note also that if the person who is opposite the monkey has the banana, then the probability that aman feeds the monkey is

1 \over 3 , by symmetry, since each of the monkey's three neighbors is then equally likely to be the feeder. These probabilities are labeled on the diagram above. When aman has the banana, he has probability

1 \over 3 of immediately feeding the monkey, and probability

2 \over 3 of passing it to a neighbor. Therefore,

p =

1\over 3 +

2 \over 3 q

When one of aman's non-monkey neighbors has the banana, she has probabilty

1\over 3 of throwing

the banana back to aman, probability

1 \over 3 of throwing the banana to one of the monkey's other neighbors, and probability

1 \over 3 of throwing the banana to the person opposite the monkey. Therefore,

q=

1 \over 3 p +

1 \over 3 r +

1 \over 9
When one the monkey's other neighbors has the banana, he has probability

1 \over 3 of feeding the monkey (in which case aman does not feed the monkey), probability

1 \over 3 of throwing the banana to one of aman’s neighbors, and probability

1 \over 3 of throwing the banana to the person

opposite aman. Therefore,

r =

1 \over 3 q +

1 \over 3 s.

Finally, when the person opposite aman has the banana, she has probability

2 \over 3
of throwing it to one of the monkey’s non-aman neighbors, and probability

1 \over 3
of throwing it to the person opposite to the monkey. Therefore,

s =

2 \over 3 r +

1 \over 9
Thus, we have the system of equation

p=

1 \over 3 +

2 \over 3 q

q=

1\over 3 p +

1 \over 3 r +

1 \over 9
r=

1 \over 3 q +

1 \over 3 s

s=

2 \over 3 r +

1 \over 9
Solving this system of equation gives p=

7 \over 12 and q=

3 \over 8 . So the probability that aman feeds the monkey is

7 \over 12
Thanks.