### Brain Teasers

# Amoeba

Probability
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^(n-1) ( 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^(n-1)(x))*d/dx(f^(n-1)(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

I think the probability may be closer to 100%. Eventually the universe gets cold and dark and the amoebas die.

This is very complex!!!!!!!!!!!!!!!!!!

Ummmm.... I'm looking at the answer but I don't see an answer...

If you understand the teaser then you know it's a little more complicated than that.

i found this exact question in my text book

So what is the probability then?

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???

its obvious you just coppied it straight out of a book or another website

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

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

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%

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%

was there even an answer?

Jul 02, 2007

There was no answer.

Why wasn't there an answer?

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

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.

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.

4p = 1 + p + p^2 + p^3;

=> p = 2^(1/2) - 1;

or aprox 41%

=> p = 2^(1/2) - 1;

or aprox 41%

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/(1-0.75))

= 1.

So it is certain that each amoeba will eventually die, with certainty.

(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/(1-0.75))

= 1.

So it is certain that each amoeba will eventually die, with certainty.

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/(1-0.75))

= 1.

So it is certain that each amoeba will eventually die, with certainty.

========================

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/(1-0.75))

= 1.

So it is certain that each amoeba will eventually die, with certainty.

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