A colony of amoebae is started by placing one amoeba in a petri dish. In any given minute, each amoeba present at the time may do one of four things with equal probability (i.e., 1/4):
1. Die, leaving no progeny.
2. Continue to live, but not split.
3. Split into 2.
4. Split into 3.
What is the probability that the colony will die out, rather than take hold and continue forever?
adapted from braingle at http://www.braingle.com/teaser.php?id=8705&op=&comm=1
(In reply to
Most of a Solution by friedlinguini)
So that nobody has to click back, the equation that I'm trying to solve is:
p³ + p² - 3p + 1 = 0
One obvious solution for p is 1. Is this the answer we're looking for? I don't think so. If you start with one amoeba, the expected number of amoebae (is that right) in the next minute is:
0/4 + 1/4 + 2/4 + 3/4 = 3/2
This means that the colony wants to increase in size. If you have a large number of amoebae, then the tendency is for the colony to keep growing geometrically. I don't think that the colony's dying out is a certainty given the rules of the problem.
Fortunately, cubics can have multiple solutions; we just have to look for another one that makes sense. If 1 is a solution to the equation, then (p - 1) is a factor of the polynomial. I'll spare everybody the details of synthetic division (If anybody wants an explanation of this, let me know and I'll post it in the Reference forum) and just say that
(p³ + p² - 3p + 1) / (p - 1) = p² + p - 1.
Applying the quadratic formula to p² + p - 1 = 0 gives
-1 ±√2.
The probability can't be negative, but -1 + √2 is a plausible probability.
Simplifying that a bit, I'd say that the solution is √2 - 1, or roughly 0.414.
re: Most of a Solution
