Thread: Another probability question

A colony of rare spherical cocci bacteria lives deep under lunar ice. After thousands of years of slowly reproducing (a process that often results in abject failure and death) in this harsh climate, only one lonely bacteria named John remains.

Suppose that the probability of an individual bacteria dividing successfully into two bacteria is 3/4, and that a bacteria must attempt the division process within two months of its immediately prior division experience.
John's time is up. It is time for him to attempt a division. What is the probability that poor John will be the proud predecessor of never-ending successive generations of his bacterial species?

Originally Posted by chickeneaterguy

A colony of rare spherical cocci bacteria lives deep under lunar ice. After thousands of years of slowly reproducing (a process that often results in abject failure and death) in this harsh climate, only one lonely bacteria named John remains.

Suppose that the probability of an individual bacteria dividing successfully into two bacteria is 3/4, and that a bacteria must attempt the division process within two months of its immediately prior division experience.
John's time is up. It is time for him to attempt a division. What is the probability that poor John will be the proud predecessor of never-ending successive generations of his bacterial species?

I think the general way to approach this is to take 1 and subtract the probability that the race goes extinct.

For the first generation, the probability of extinction is 1/4.

In the second generation, the probability is (3/4)(1/4)(1/4), which is the probability that both children fail given that the first generation was successful.

For the third generation, it is (3/4)^3*(1/4)^4 + (3/4)(1/4)(3/4) * (1/4)^2 + (3/4)(3/4)(1/4) * (1/4)^2.

Since it's getting messy, there's probably a better way to organize things, and possibly you can get an exact result somehow. (I'm reminded of those infinite sums that turn out to have in the answer, although I have no idea what it would be in this case.) But if you only need an approximation, you can get a pretty small error quickly this way, just on paper. And with a computer program you can get to a high number of generations quickly using dynamic programming.

Oh and since the bacteria reproduce asexually, we don't need to consider the possibility that they take varying times to decide to reproduce. Whether they wait a minute or two months, the overall result is the same. But I am assuming that the bacteria have limited lifespans, and so they all attempt to reproduce.

Originally Posted by undefined

I think the general way to approach this is to take 1 and subtract the probability that the race goes extinct.

For the first generation, the probability of extinction is 1/4.

In the second generation, the probability is (3/4)(1/4)(1/4), which is the probability that both children fail given that the first generation was successful.

For the third generation, it is (3/4)^3*(1/4)^4 + (3/4)(1/4)(3/4) * (1/4)^2 + (3/4)(3/4)(1/4) * (1/4)^2.

I assume the 2nd generation the 3/4 is coming from the chance that the other survived and the other 2 could die?

I don't understand the 3rd gen one.

I chose to double-post instead of editing my first post, so that people will get email notification if desired. If this is not the proper etiquette, please let me know.

I think it's possible to get an exact answer using Markov chain theory. Some literature is here (Chapter 11 of Introduction to Probability by Grinstead and Snell, 2nd ed, freely distributable under a GNU license.) Extinction can be set up as a sole absorbing state, and then the matrix inverse can be found as described in the .pdf file. I haven't actually used this method before, but I believe it applies here.

Originally Posted by chickeneaterguy

I assume the 2nd generation the 3/4 is coming from the chance that the other survived and the other 2 could die?

I don't understand the 3rd gen one.

Oh, we posted at almost the same time, haha.

Yes, the second generation probability I gave is p(B|A) where A = the first reproduction was successful, and B = the race goes extinct at that generation.

For the 3rd generation, we consider cases.

Case 1: John was successful, and both of his two children were successful. So there are 4 bacteria, and the probability of extinction right now is (1/4)^4 times the probability of getting to this state, or (3/4)(3/4)(3/4).

Case 2: John was successful, and Child A failed, but Child B succeeded. There are two bacteria now, so we have (1/4)^2 times the probability of getting there, (3/4)(1/4)(3/4).

Case 3: Same as case 2 but with Child A and Child B reversed.

Originally Posted by undefined

Oh, we posted at almost the same time, haha.

Yes, the second generation probability I gave is p(B|A) where A = the first reproduction was successful, and B = the race goes extinct at that generation.

For the 3rd generation, we consider cases.

Case 1: John was successful, and both of his two children were successful. So there are 4 bacteria, and the probability of extinction right now is (1/4)^4 times the probability of getting to this state, or (3/4)(3/4)(3/4).

Case 2: John was successful, and Child A failed, but Child B succeeded. There are two bacteria now, so we have (1/4)^2 times the probability of getting there, (3/4)(1/4)(3/4).

Case 3: Same as case 2 but with Child A and Child B reversed.

Okay, this makes a whole lot of sense. I'll try it out. Thanks.

Actually, I think there should only be one child out of the first split. I think it it splitting into two, and not creating two extra.

Originally Posted by chickeneaterguy

Actually, I think there should only be one child out of the first split. I think it it splitting into two, and not creating two extra.

My understanding is that after a successful division, the parent no longer exists, and two children exist in its place.

For unsuccessful division, the parent dies and there are no children.