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## Question

I DID NOT MAKE UP THIS PROBLEM (but I'm putting it in my own words) I did a search, and didn't find it, so I'm posting it ;D

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An amoeba population starts out with 1 amoeba.

At every generation, each ameoba has a 1/4 chance it will die without producing offspring, otherwise (a 3/4 chance) it divides into 2 new amoebas.

What is the probability of the amoeba population lasting forever?

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It has a 1/4 chance... I think...

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Zero passed on failure gene

it is to high a rate of failure!

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I think it could keep going though... as long as it makes it past the first 2 generations... so maybe a 1 in 2 chance?

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Zero. Time is infinite; the odds that they will die out is 100%. Entropy.

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Surely... well... surely there are infinate possibilities for the probability of it surviving for forever?

So unless the answer's 1/4 I dont know... Although im guessing theres a trick; there always is!

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it's not a simple sequence sum of 1/4 + 1/16 + 1/64 + 1/4etc... cuz that's implying either they all die or they don't... what if only a few die, and the rest double? etc

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Computer simulation gave me a little more than a 68% success rate...I assumed that a population would last forever if the size reached 30, since the expected size of the next generation is 1.5 times the size of the current one, and I figured it was pretty safe to assume that a population would grow indefinitely at that point.

I wrote a Java program and attached it for anyone who wants to experiment.

*Just kidding* It doesn't let me upload a .java file, so I've uploaded it as a .txt. Download it and change the file extension to .java if you want to run it.

I could do this on paper, but it would get messy really quickly, and I don't know a better way.

AmoebaPopulation.txt

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I'm with Mr. Heroux on this one.

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is there a 1/4 chance it wll be a unic? is that wat u mean

Edited by pieman
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I still believe a 1/4 chance or zero is the answer... I guess at some point though, the population might just get unlucky and all the amoeba die off...

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it's not a simple sequence sum of 1/4 + 1/16 + 1/64 + 1/4etc... cuz that's implying either they all die or they don't... what if only a few die, and the rest double? etc

You ask the chance of not dying out.

That's 1.00 - P[extinction].

P[extinction] = P[ext1] + P[ext2] + ... + P[exti] + .... where P[etxi] is probability of all the amoebae of the ith generation dying.

P[ext1] = 1/4. The others are quickly decreasing but positive.

The sum P[ext] is thus kind of but not much larger than .25 -- perhaps .3 or .33.

Thus P[eternal survival] = 1.00 - something between say .25 and .35 = something between .65 and .75

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You ask the chance of not dying out.

That's 1.00 - P[extinction].

P[extinction] = P[ext1] + P[ext2] + ... + P[exti] + .... where P[etxi] is probability of all the amoebae of the ith generation dying.

P[ext1] = 1/4. The others are quickly decreasing but positive.

The sum P[ext] is thus kind of but not much larger than .25 -- perhaps .3 or .33.

Thus P[eternal survival] = 1.00 - something between say .25 and .35 = something between .65 and .75

No fair getting here first!!

1-(1/4) -(1/4^2)-(1/4^3)-(1/4^4)....

P(extinction) is a geometric series of the form b*(p^k) where k = 0....inf, p = 1/4 and b = 1/4

sum(b*p^k) for k = 0...inf = b/(1-p) = 0.25/0.75 = 0.33...

Cheers!

--

Vig

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it s been since i passed out of high school(think my probability and limits are correct).i guess this is the answer

or the altrernative answer could be

n->infinity ((3/4)^n log n)

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The answer is indeed 2/3, bonanova, but I meant that the chances of the whole population dying out doesn't depend primarily on what step it's at- it's not "either the whole population survives, or the whole population dies"- it's each individual amoeba. An example:

1 (survives) -> 2 (both survive) -> 4 (only 2 survive to double) -> 4 (3 survive) -> 6 (1 survives) -> 2 (both survive) -> 4 (all four survive) -> 8 (etc)

see what I mean? But still, the chance that the population will survive FOREVER works out to be 2/3 exactly (ignoring speculation about infinity and heat entropy and all that stuff, this is purely mathematical )

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You ask the chance of not dying out.

That's 1.00 - P[extinction].

P[extinction] = P[ext1] + P[ext2] + ... + P[exti] + .... where P[etxi] is probability of all the amoebae of the ith generation dying.

P[ext1] = 1/4. The others are quickly decreasing but positive.

The sum P[ext] is thus kind of but not much larger than .25 -- perhaps .3 or .33.

Thus P[eternal survival] = 1.00 - something between say .25 and .35 = something between .65 and .75

Yeah, I think Unreality is right about it not being a simple 1-(1/4)^n geometric series...the answer should be an infinite series of conditional probabilities.

[spoiler='Conditional Probability

']Like, you can't just sum the P(ext), as you call them, of the ith generations. You need to sum the P(exti)P(survi-1), where P(survi-1) is the probability of survival to the i-1 th generation. Put in less mathematical terms, it means you need to weigh the probability of extinction of that generation with the probability that the amoeba will actually survive to that generation. So it should be something like P(survival to infinity) = 1-sum(P(exti)P(survi-1)), and in the P(survi-1), you need to take into account the different cases where 1 out of 2 survives or both survive...

I haven't done the math yet, but it wouldn't surprise me it that series came out to sum to 2/3 as well...once I have time, I'll try to figure out the exact mathematical expâ€‹ression for the series

Edited by Yoruichi-san
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Err, my notation was off in my earlier post...sorry, it's been while since I took Probability Theory...

Anyways, hopefully this will make things clearer:

Okay, first off, for those who are unfamiliar with conditional probabilities, here a short explanation:

P(A|B) is the conditional probability of A given B. This means it is the probability that A will occur if B occurs.

The probability of A is equal to the conditional probability of A given B times the probability of B plus the probability of A given C times the probability of C and so on for all possible events (B,C,D, etc.)

So Bonanova's earlier formula was correct:

P[extinction] = P[ext1] + P[ext2] + ... + P[exti] + .... where P[etxi] is probability of all the amoebae of the ith generation dying.

However, what my point is, and what I think Unreality's point is is that P(exti) is NOT simply equal to (1/4)^2i. (1/4)^2i = P(exti|isurv), that is, the probability of extinction in the ith generation given that all the previous generations have survived, hence, there are 2i amoeba which each now have a 1/4 chance of extinction.

So to get P(exti), we need to take P(exti)=P(exti|2isurv)P(2isurv)+P(exti|2i-1surv)P(2i-1surv)+P(exti|2i-2surv)P(2i-2surv)+...+P(exti|2surv)P(2surv)+P(exti|1surv)P(1surv)

for each i. To get the total probability of extinction, we then need to sum these for all the ith generations from 0 to infinity...

I think it probably simplifies to something nice, which someone may have suggested already, but I'm still working on that...

Okay, hmm...well I had hoped this would clarify the matter, but it may just make things more confusing...but I'm pretty sure the logic is correct.

Oh well, thanks for the puzzle!

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yeah with this equation:

P = (2p - 1)/p

plug in 3/4 for p:

P = (6/4 - 1)/(3/4)

P = (1/2) / (3/4) = 1/2 * 4/3 = 2/3

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for each generation, where p=population, the odds of total extinction are 1/4^p. Given infinite time, even the most unlikely of scenarios will eventually occur, and the entire population is doomed to extinction. The paradox of this is that any time we are capable of experiencing is finite, and therefore defies this concept, so that no proof is possible.

Isn't the probability of an event reliably manifested in infinite tests, so that a 50% probability will occur in 50% of infinite (whatever that means) tests, but could vary within standard deviation for a finite number of tests? Given that, extinction is inevitable.

Otherwise, I'm with the folks who say 33ish%

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Nope. It's not 1/4^p. The amoebas die or separate as individuals, not as a whole.

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