Your mother gifted you a chocolate bunny on your birthday. But this is a
magic bunny. On the first night, the bunny either disappears or splits into two identical magic bunnies, with equal probability. The next night, if you have two bunnies, they each (independently) either disappear or split in two. And so it continues, each night any remaining bunnies each independently either disappear or split into two bunnies.
What is the probability that you will eventually be left with no magic
bunnies?
At the beginning you have and odd number of bunnies, ie 1. After the first move, there will never again be an odd number since each bunny turns into either 0 or 2, both of which are even numbers.
If you have zero bunnies, the probability of ending up with zero bunnies is 1.
p(0) = 1
p(1) = 1/2 + p(2)/2
p(2) = (1 + 2p(2) + p(4))/4
p(3) = 3 or any odd number can never occur
p(4) = (1 + 4p(2) + 6p(4) + 4p(6) + p(8))/16
p(1) = (1 + p(2))/2 = 1/2 + p(2)/2
p(2) = (1 + 2p(2) + p(4))/4
(1/2)p(2) = (1 + p(4))/4
p(2) = (1 + p(4))/2 = 1/2 + p(4)/2
p(4) = (1 + 4p(2) + 6p(4) + 4p(6) + p(8))/16
p(4) = (1 + 4(1/2 + p(4)/2) + 6p(4) + 4p(6) + p(8))/16
p(4) = (1 + 2 + 2p(4) + 6p(4) + 4p(6) + p(8))/16
(3/8)*p(4) = (3 + 4p(6) + p(8)) / 16
p(4) = (3 + 4p(6) + p(8)) / 6
p(4) = 1/2 + (2/3)p(6) + (1/6)p(8)
p(1) = 1/2 + 1/4 + p(4)/4
p(1) = 1/2 + 1/4 + 1/8 + p(6)/6 + p(8)/24
It looks like if continued the pattern will be:
p(1) = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...
Which, in the limit, approaches 1.
All the chocolate goes away.
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Posted by Larry
on 2024-08-04 15:27:46 |
Analytic plus slight hand waving.
