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Let p(n) be the probability the colony will die out if there are n amoebae currently in it. We seek p(1).
First, p(1) = 1/4 + p(1)/4 + p(2)/4 + p(3)/4, as this is the total of the probabilites of going into any one of the states 0, 1, 2, 3 in the next minute multiplied by the probability of ultimate colony demise from that state.
But from the state of having 2 amoebae, the likelihood of 1 particular one of those two have all its progeny ultimately die off is just p(1) again. In order for the 2-amoeba colony to die out, each of its two members, considered as separate colonies must die out, and this is independent, so p(2)=(p(1))^2. Similarly the 3-amoeba state can be considered as three independent colonies (again, there is no sexual reproduction). Therefore p(3)=(p(1)^3)
Thus p(1)=1/4 + p(1)/4 + (p(1))^2/4 + (p(1))^3/4.
Calling p(1) x, we need to solve the cubic
x^3+x^2-3x+1=0
There are various ways to solve cubics, but it's easiest to use the solver add-in in an Excel spreadsheet, to find a value for x that makes x^3+x^2-3x+1 zero. It comes out to .41421356 as the required probability
As you might recognize this is √2 - 1, which it is. For an analytic solution with this same result, see friedlinguini's solution starting at comment 1.
The solution assumes the colony has unlimited room to grow affording it the greater safety of possible huge size, whereas a practical case would limit the colony to some maximum size, precluding the greater safety to the colony afforded by expanding beyond a practical limit. It's a particular example of a gambler's ruin problem, where the house has unlimited wealth and so cannot be broken.
A more mathematical generalization can be found at the braingle site's answer at http://www.braingle.com/teaser.php?op=2&id=8705&comm=1
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