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
Let p be the probability that we're looking for. In order for the colony to die out, either the amoeba must die in the first minute, or whatever amoebas are remaining must all die out. If n amoebas are alive at one instant, then they can all be considered different independent colonies, so the odds of all of them dying out is p^n. This gives the framework for the answer.
For each case, weight the probability of the colony dying out with the probability of the case occurring. Since each case is equally likely, use a weight of 1/4. Thus:
p = 1/4 + p/4 + p²/4 + p³/4
4p = 1 + p + p² + p³
p³ + p² - 3p + 1 = 0
p is the solution to a cubic equation. I don't have the analytic solution for a cubic handy, but I can probably generate a numerical one quickly enough unless someone else wants to kill this one off.
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