Gitt Information from Evolution FairyTale

Well, it doesn't look like any of the thread's participants are interested in resuming here, so I'll describe the last issue being discussed in case anyone here wishes to comment.Deadlock posed this problem: And when I gave the "wrong answer" he responded with this solution: Where Deadlock writes "Cn,y" he means combination, i.e., "n things taken y at a time".As part of his solution Deadlock claims that the probability of a mutation at a specific point in the genome is (1/3)¹⁰⁷ (nice to be back in a land where HTML in messages is legal). When the thread was closed I was attempting to explain to Deadlock why this was incorrect, that this value is many thousands of orders of magnitude smaller than a goolgol-th and is ridiculous. But he stood by it, then the thread was closed.Anyone have any idea where Deadlock might have gotten the idea that this was the correct probability? What's the correct value, and how would you persuade Deadlock that it is the correct value?--PercyEdited by Percy, : Grammar.

I think he may really have meant (1/3)¹⁰⁷, because 10⁷ is the number of base pairs in his genome. I would argue that he was thinking, "There are three wrong bases for any position, and the probability of getting one of those three wrong bases is 1/3. So the probability of the first base pair in the genome being wrong is 1/3, and the probability of the second base pair being wrong is 1/3, and so forth, so multiple 1/3 by itself 10⁷ times."Which is, course, completely bogus.What he really wanted to do if he was determined to use that style of approach was properly figure out the probability of an error for a base pair, and 10^-8 seemed to be an acceptable figure to him. This means that the probability of all the base pairs being right would be: This happens to be around .9, which means there's about a .1 probability of one or more mutations.This is so similar to his (1/3)¹⁰⁷ that it makes it seem likely that he was trying to use this particular approach, but what he did was both wrong (the 1/3) and too simplistic (this has to be calculated, it isn't something simple that you write down off the top of your head).So the probability of three mutations at specific predesignated positions in a single reproductive event is: I'm on the edge of my competency here, so let me know if I've gone wrong. Anyway, it is indeed a very small number. As I told Deadlock, predesignating the positions isn't the way evolution operates. Natural selection doesn't sit around waiting for beneficial mutations, it just works on what it has. And if you apply 10⁶ Bernoulli trials like this you still get a very small number: This is unsurprising because of the unlikelihood of three mutations in predesignated locations in a single reproductive event. As I explained to Deadlock several times, this isn't the way evolution works because of multiple offspring and generations, and because of the post facto fallacy (is that the proper name for this fallacy, calculating the odds of what really happened, like winning a lottery, when all the outcomes were equally unlikely but one of them has to happen?).Anyway, that's my attempt at things, not the same answer you got, but I had a different interpretation of this problem. What do you think?--Percy

Yep, that's the same answer I get, 9.08*10^-19 (I counted decimal places wrong, the exponent of 10 in my previous message should have been -19, not -20).I was trying to use the approach Deadlock insisted on using for Bernoulli trials, so it sounds like you're saying I did it properly? I'm amazed!I used to be more intuitive with probabilities. I was able to follow the approach you used pretty easily, but I'm not sure I would have got there on my own. I use that skill very rarely these days and it is seriously rusty. Thanks for the help!--Percy