whiskey the hedonist

The odds of picking the acids in a random order and the connecting them in the right order is the same as picking them in the right order in the first place, that is 1/20^32. I demonstrated that with the a,b problem.

lets review: You agree that the odds in the a,b, problem don't change for a one stage pick-connect ordered draw, or an unordered draw followed by a connect, Either way, the odds of one particular order are 1/16.

you agree that musgrave's calculation is for an ordered draw. It has to be, because he arrives at it the same way I arrive at the 1/16 number orignally, by simply choosing one out of every possible unique combination, which is calculated as 20^32.

therefore, the probability calculation of 1/20^32 is correct, and your original critique is a mistake, right?

lee_merrill

So then what is the probability of picking (say) the first three required amino acids, and then connecting them in the wrong order?

Regards,
Lee

whiskey the hedonist

Somewhere around 5/8000, I think. Who cares? To recap, we have shown that:

1. The basic laws of probability say that a unique ordered sequence has a certain probability of occuring from a range of possible unique ordered sequences, regardless of whether you "pick and connect" or "pick, then connect". That probability is 1/x, where x is the total number of possible unique sequences. This has actually been

2. There are exactly 20^32 ways that 20 amino acids can form a 32 acid long ordered sequence.

3. The peptide in question is one such ordered sequence.

4. The odds of it forming in an amino acid rich envirnonment are 1/20^32.

You have disputed 4, but you have accepted 1, 2, and 3. How do you explain this seeming departure from basic logic? Why won't you admit 4? Are you afraid?

wyzaard

Ummm... has anyone considered the fact that these probabilities are weighted? The dice are loaded depending on preferred arrangements and previous throws.

Nice loaded terms too. 'Correct'... heh.

wyzaard

Assuming of course that molecules are like evenly interchangable legos.... which they are not. What if the chance of 'a' connecting with 'b' is either higher than with c or d... or the only viable possibility given the natural properties of the combination? What if the large such a type molocule becomes, the stronger and more selective the possibilities will be? Given access to a large amount of material and energy in the system... it would seem that the abiogenesists have something there.

whiskey the hedonist

Hey Wyzaard,

Be very careful going there. Lee is now going to say:

We can comfortably make that assumption with amino acids, to my understanding. Perhaps someone more familiar with biochemistry could tell me differently.

The main point is that even with that assumption, we get pretty reasonable odds of 1/20^32 for the formation of our desired peptide. Given the number of simultaneous trials in the pre-life earth, it appears in about a week. The article in question can be found here.

Lee wants to contest that number by calculating the order of connection twice, once by "drawing" in order, and second by connecting. His original critique of the 1/20^32 number was that Musgrave had "forgotten the order". Now, he flat out refuses to admit that he was wrong about that, despite having to accept every argument for WHY he is wrong.

I've noticed this "big lie" trend with a lot of ID/Creation people. They make some grand statement and then if backed into a corner by facts they just gibber uncontrollably, utterly unable to admit they made a mistake.

That is why it seems reasonable to just declare victory, people like Lee are just incapable of acknowledging defeat even when it is mathmatically proven that they are simply wrong.

lee_merrill

Hi everyone,

The problem is that he did not address "connect" at all! Let's say we have the first two aminos acids picked and connected. Then what is the probability of picking the next correct amino acid in the sequence? 1/20. What is the probability of connecting it correctly? ½. What is the probability of connecting it incorrectly? ½ again. So this second factor was not included in the estimate.

I agree that it would have been better if I had said something different, such as "he forgot the connection step."

Let's try this on for size, how many ways can you get an A-B-A molecule? It's not a probability of 1/(20^3). So duplicates make a difference, and his molecule had duplicates, and the calculation did not factor that in, either.

Amino acids seem to be, though, they have molecule X on the left and molecule Y on the right and you connect an X to a Y to link them up, all the amino acids are the same in the way they are linked, and they have another side molecule that makes them different.

Regards,
Lee

whiskey the hedonist

If I can connect to the beginning and the end than I can start with any sequence of two from anywhere in the sequence, so my odds of a good first draw are higher than 1/20, they are 13 in 20. The odds of draw 2 have to be calculated from a knowledge of draw one, so they have to be calculated 13 times, plus allowances for front/back connections, etc. Then, my odds of a good third have to be calculated 13x12 times for each 2 letter combination, plus allowances for front/back connections. But after all that calculating is done:

There are 20 amino acids. There are 20^32 ways to connect them in a set of 32, no matter where you start or how you do it, there are that many combinations. The odds of getting any given combination connected are 1 in 20^32.

In order for the odds to be different, there have to be more than 20^32 ways to arrange 32 amino acids, which there simply aren't.

Lets see what fun piece of weirdness Lee comes up with next!

whiskey the hedonist

Actually, given 20 amino acids and 3 draws, it is exactly 1/20^3. The reason is that there are 8000 ways to connect 3 amino acids, and that is one of them.

lee_merrill

Hi WTH,

That's what I will try next...

But this seems to be assuming that you have in your vat all the combinations of 3 amino acids, already pre-made, and you pick one of them! But that would mean that the 3-amino acid chain we need is already in the vat, in the right proportion to all the other 3-amino acid combinations.

So the way to proceed would be to first pick all the needed amino acids one by one in any order (let's assume for now that they are all different, and we need n of them), which would be a probability of n!/(20^n). Now we need to arrange them in the right order, we need to connect them, and the probability of doing that correctly is 1/(n!). So the overall probability is 1/(20^n), the n! on the top and bottom cancel out.

So you are right, and I am wrong, I missed that factor of n! on the top, and I would say it was Alf who was on the right track here. And the result is similar if there are duplicates as well, so the talk.orgins calculation seems correct.

Except! What stops the amino acid chain from adding another amino acid, once we have our needed molecule? Why are we so lucky, once we get our molecule by adding amino acids, that then the new molecule is whisked away to an environment where it can now self-replicate, instead of adding on more amino acids?

Regards,
Lee