Amoeba
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
A jar begins with one amoeba. Every minute, every amoeba turns into 0, 1, 2, or 3 amoebas with a probability of 25% for each case (dies, does nothing, splits into 2, or splits into 3). What is the probability that the amoeba population eventually dies out?
Answer
If p is the probability that a single amoeba's descendants will die out eventually, the probability that N amoebas' descendants will all die out eventually must be p^N, since each amoeba is independent of every other amoeba. Also, the probability that a single amoeba's descendants will die out must be independent of time when averaged over all the possibilities. At t=0, the probability is p, at t=1 the probability is 0.25(p^0+p^1+p^2+p^3), and these probabilities must be equal. Extinction probability p is a root of f(p)=p. In this case, p = sqrt(2)1.
The generating function for the sequence P(n,i), which gives the probability of i amoebas after n minutes, is f^n(x), where f^n(x) == f^(n1) ( f(x) ), f^0(x) == x . That is, f^n is the nth composition of f with itself.
Then f^n(0) gives the probability of 0 amoebas after n minutes, since f^n(0) = P(n,0). We then note that:
f^(n+1)(x) = ( 1 + f^n(x) + (f^n(x))^2 + (f^n(x))^3 )/4
so that if f^(n+1)(0) > f^n(0) we can solve the equation.
The generating function also gives an expression for the expectation value of the number of amoebas after n minutes. This is d/dx(f^n(x)) evaluated at x=1. Using the chain rule we get f'(f^(n1)(x))*d/dx(f^(n1)(x)) and since f'(1) = 1.5 and f(1) = 1, we see that the result is just 1.5^n, as might be expected.
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Comments
Quax
Dec 09, 2002
 I think the probability may be closer to 100%. Eventually the universe gets cold and dark and the amoebas die. 
SPRITEBABE44
Dec 27, 2002
 This is very complex!!!!!!!!!!!!!!!!!! 
Dazza
Jan 07, 2003
 Ummmm.... I'm looking at the answer but I don't see an answer... 
something
Feb 10, 2003
 If you understand the teaser then you know it's a little more complicated than that. 
snoopdogg
Mar 11, 2003
 i found this exact question in my text book 
jimbo
Mar 13, 2003
 So what is the probability then? 
Codammanus
Mar 26, 2003
 Just when I thought I had mustered up the courage to tackle a probability teaser, I ran into this. Hats off to you! (I guess I better get that old math textbook I forgot to return to school). You're SOMETHING else. By the way, Was this an Original??? 
kimballchilcott
May 18, 2003
 its obvious you just coppied it straight out of a book or another website 
gogogo1
Jun 07, 2003
 I have a different answer and a different eplaination.
We can join the wo probabilities that they split into 2 or 3 since the question does not ask how many there will be if they do survive. We now have 3 possibilities. they split 50%, they stay or they die 25% each. If they stay, they will be wiped out eventually anyway so we can say that either they split or they die. Thus giving a 50% chance to both. If you find something wrong with this comment, don't blame me. I'm only 12 
NomadShadow
Nov 15, 2005
 This teaser is so easy except for one point at which I could find no proper explanation.
It was hard for me to imagine the life cycle of an amoeba, so i stated it in a matter of fomulas instead.
The probability of an ameoba tree of desendants eventually dying is P, where 1 >= P >= 0.
You get the equation
P = 0.25 + 0.25*P +0.25*P^2 + 0.25*P^3
or P^3 + p^2 3*P + 1 = 0 which yields 3 solutions
P = 1 , P = 1 + SQRT(2) , P = 1  SQRT(2)
The last solution is discarded because it falls out of our accepted range
Now the second solution is the right one, but still how can we discard the first? After all, a probabilty of an event can be 100% !
In other words, how can we prove the probability of all ameobas dying is less than 100% 
triskit
May 06, 2006
 was there even an answer? 
(user deleted)
Jul 02, 2007
 There was no answer. 
javlad27
Sep 22, 2007
 Why wasn't there an answer? 
cm_mcginty
Sep 12, 2008
 this should be in the trick category, the answer is 100% because the question says "eventually" which means at some point, every single ameoba must die because it will go on indefinitely, we could easily be talking in terms of x google centuries away but it's inevitable, the same sort of idea as an infinite number of monkeys typing on an infinite number of typewriters for an infinite amount of time will eventually write all shakespeare's works 
javaguru
Dec 08, 2008
 I came up with the same answer and reasoning as cm mcginty: the answer is 100% because "eventually" it must be the case that the terminal state in which every amoeba is dead occurs.
This must happen because regardless of the size of the population, there is a finite probability that the entire population dies in a single generation. "Eventually" that probability will occur because "eventually" is infinite and the probability of the generation dying is finite. 
Abhi8
Aug 12, 2010
 4p = 1 + p + p^2 + p^3;
=> p = 2^(1/2)  1;
or aprox 41% 
(user deleted)
Aug 30, 2010
 The probability of the amoeba DYING in the next minute is 0.25 and that of the amoeba NOT DYING is 0.75. The question asks the probability of the amoeba population eventually dying  that means we have to take the cases that the amoeba:
(a)dies in the next minute => Prob = 0.25
(b)stays alive in the next minute and dies in the following minute => (0.75)*(0.25)
(c)stays alive in the next two mins and then dies in the third minute => (0.75)^2 * (0.25)
... and so on.
Probabilities of all the events when added leads to:
0.25 + (0.25)*(0.75) + 0.25) * (0.75)^2 + (0.25)*(0.75)^3 + ... upto infinity
= (0.25) * (1/(10.75))
= 1.
So it is certain that each amoeba will eventually die, with certainty. 
(user deleted)
Aug 30, 2010
 CORRECTING TYPO  REPOSTED
========================
The probability of the amoeba DYING in the next minute is 0.25 and that of the amoeba NOT DYING is 0.75. The question asks the probability of the amoeba population eventually dying  that means we have to take the cases that the amoeba:
(a)dies in the next minute => Prob = 0.25
(b)stays alive in the next minute and dies in the following minute => (0.75)*(0.25)
(c)stays alive in the next two mins and then dies in the third minute => (0.75)^2 * (0.25)
... and so on.
Probabilities of all the events when added leads to:
0.25 + (0.25)*(0.75) + (0.25) * (0.75)^2 + (0.25)*(0.75)^3 + ... upto infinity
= (0.25) * (1/(10.75))
= 1.
So it is certain that each amoeba will eventually die, with certainty. 
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