Dryhad

As you should have. To back up your denial of examples put forward by skeptics

Because it wouldn't be Babylon! It would be an exercise in archeological recreationalism. If you call that Babylon, then my whole arguement that you're requiring too much falls appart.

That sounds really easy (comparitively). I apologise for wasting your time, I have no quarrel with you.

whiskey the hedonist

So apparently, if there are x possible combinations, the odds--all things being equal--of getting one particular one are not 1/x! Lee, this is a really, really, really exciting moment in probability calculations! How does this work!?! Oh... wait! I get it! Lee thinks that in the area where the protein is forming, there are only 32 amino acids, total! Oh... wow...

Now that is funny! And a little sad.

No, lee, amino acids are quite plentiful, as Musgrave demonstrates in his article. So drawing an 'R' doesn't deplete the supply or 'R's' in a statistically meaningful way. Besides, even if they weren't, and the pot only consisted of 32 acids total, the actual number of possible combinations goes DOWN to 2.65x10^32. (32 possible first draws, 31 possible second draws...) However, the assumption in the calculation is that you have all 20 acids in more or less plentiful supply.

Look, it IS easy to show that your critique of Musgrave is no good, Lee. In case you missed it, read my explanations again. What is difficult, clearly, is getting you to admit that you made a mistake in your critique. I'm bored with this, its like kicking a puppy.

So lets talk Tyre for a moment, Lee. I maintain that you are of the opinion that there is no possible chain of events that would lead to the Tyre prophecy NOT being true. You maintain I am misinterpreting you. What could possibly have happened to make the Tyre prophecy untrue? Is there any city on earth to which the Tyre prophecy would not apply? I mean, if the city had existed then, and I made that prophecy about that city, what city would it not be true for, and why?

whiskey the hedonist

Just to make sure that Lee, and everyone else, understands the nature of the problem.

Lets say I have about a million of each of 4 letters in a bag. I'm going to draw an order from the bag of 5 letters.

The letters are a,b,c,d

The possible orders are aaaaa to ddddd. Any combination of 5 of these 4 letters that you can think of is in that range, abdca, adcab, cccdd, and onward. How many unique combinations of letters and orders are there? What I mean by "unique" is that 'aaaab' is not the same combination as 'baaaa', each one is distinct, and has to be counted as such, a combination is a selection of particular letters, in a particular order.

Well, to answer this, we just count how many ways each selection can go. Our first draw can be any letter, so there are 4 possible draws. Our second draw can be any letter, so there are 4 possible draws. For two draws, then, there are 16 different combinations! There is aa, ab, ac, ad, ba, bb, bc, bd, ca,cb,cc,cd,da,db,dc,dd. The chances of drawing any one is 1/16 (actually, the chances of 5 a's is slighlty lower, something like 1/16.000004)

One thing to notice, here, is that there are 2 ways to draw 'a' and 'b', 'ab' and 'ba'. The odds of getting either one are 1/8, not 1/16, and the odds of getting 'ab' on a draw from a pool of 2 is 1/2, so the actual odds are still 1/8*1/2 = 1/16. The order is part of the original equation, and each combination is unique.

For our chain of five, there are 4x4x4x4x4 (4^5), or 1024 combinations. So the odds of drawing any particular combination is 1/1024.

In the protein example, there are 20 letters (20 amino acids). For a chain of 32, there are 20^32 combinations. We draw 1/20 for position 1, 1/20 for position 2, and so forth. So if I want a specific combination of amino acids, that is, particular acids in a particular order, I have a 1/20^32 chance of making that draw. That is what Musgrave's calculation actually says.

Lee wants us to believe that this is somehow not true, that this formula "misses the order." But in order to answer why this is, Lee must be able to do the same thing with 2 letter combinations of 4 letters, that is, he must also be able to show that the odds of 'ab' are not 1/16.

So, lee: why are the odds of drawing 'ab' in my hypothetical not 1/16?

lee_merrill

Thanks for your kind demeanor, Dryhad! Much appreciated...

If Neb didn't attack it, that would make it untrue...

Yes, the probability is 1/x, I agree.

Agreed again.

Yes.

It is indeed 1/16. The problem arises when you have to both draw them, and put them together, just drawing an "a" and a "b" is certainly necessary, and the probability of drawing an "a" and then a "b" is 1/(16*16), yet this is not the probability of drawing them and combining them properly! When we try and combine them, we can get "ab" or "ba", yet we require "ab", so I am saying that this extra step reduces the probability of getting the right protein even further, to 1/(16*16*2).

We can't just pick all the right proteins from the soup, and then assume that they will then be (lucky for us!) connected correctly together.

Regards,
Lee

whiskey the hedonist

We'll get back to Tyre, promise. But you really are completely wrong about this, and I think if you read my previous post a little more carefully you would see that I anticipated the above response and explained your error, so I'll try explaining again.

The odds of FIRST drawing an 'a', and THEN drawing a 'b' are 1/16 (1/4*1/4). But if we don't connect b to a based on the order of the draw, than it can't matter what order we drew them in, two draws (a then b, or b then a) result in an identical, unordered collection that contains an 'a' and a 'b'. We can build the probability as follows:

1. With a two step process, the order of drawing isn't important, just that after I've drawn twice, I have an 'a' and a 'b', which I then must combine.

2. So for my first draw, I am willing to take either an 'a' or a 'b'. I don't care which I draw, becuause it isn't connected to anything yet. So the odds of that draw are 1/2, because half the letters are either an 'a' or a 'b'. Lee may want to make this number 1/4, but those are the odds of drawing an 'a'. The odds of drawing a 'b' are also 1/4. I don't care which of these events occurs, so I add these odds together, and I get 1/2.

3. For my second draw, I do care whether I draw an 'a' or a 'b'. If I drew an 'a', I want a 'b', and vice versa. So I must draw 1 specific letter out of the four letters so my odds of drawing that letter are 1/4. Lee probably wants to multiply 1/4 times itself to make 1/16, but there are a maximum of 4 ways that any draw can go. I'm either going to draw an 'a' 'b' 'c' or 'd'. The odds of getting any specific letter on any specific draw can never exceed 1/4, no matter what. You are going to draw one of those four letters. There is nothing else to draw!

4. I calculate 1/2*1/4 is 1/8. So my odds of drawing an 'a' and a 'b' are 1/8. To verify this, count all the ways you can possibly draw two letters. There are 16 of them. Count all the ones that contain exactly 1 'a' and 1 'b'. There are 2 of them. 2 of the 16 draws are acceptable, or 1/8.

Having drawn my letters, I must put them together.

5. There are two ways to put them together, so when I combine them, the chances that they will be in the correct order are 1/2. 1/2 * 1/8 equals 1/16.

The odds of connecting a to b in that order, given 4 letters and 2 draws, are 1 in 16, no matter how you calculate it. My first calculation (x = all possible unique orders, 1 out of x is the odds of any one particular order) is the easy way. What I just did is the hard way. But they will always get the same result.

Lee, you are just flat out wrong about this, and because this is math, I've actually proven it. This is not a question of interpretation. Just admit that you were mistaken so we can move on.

Boro Nut

He's right you know. You've only got one fairy picking chemicals out of soup. You're forgetting the fairy who has to stitch all those chemicals together. Otherwise you're left relying on the fundamental laws of nature, and then where would we be? There has to be two fairies in any chemical reaction. He's only got one pair of hands after all, and his wings are not strictly speaking opposable. That's why chemicals don't always react in exactly the same way every time under exactly the same conditions. Fairies are only human after all.

Boro Nut

Alf

and this is exactly where you are wrong. The "putting together" is already in the 20^32 formula. To draw exactly one "a" and one "b" without regard for order with 4 possible letters abcd and a huge number of each is:

Chance of picking a single "a" is 1/4
Chance of picking a single "b" is 1/4
Chance of picking a single "a' and then a single "b" is 1/16
Chance of picking a single "b" and then a single "a" is 1/16
Chance of picking a single "a" and a single "b" ignoring order is 1/8.

The 4^2 formula is the ones incluiding the order. The ones ignoring order is a different formula. In general it is:

PIck n "a" and m "b" from a set of three types "a", "b" and "c" with we have probabilities for each and we have so huge number of each that the probabilies do not change noticably (i.e. it is the same as if we had put the one we draw back again before drawing a second time) Then we have the following situation:

Chance of picking "a" is P(a) is some number in the range 0 <= P(a) <= 1.
Chance of picking "b" is P(b) is some number in the range 0 <= P(b) <= 1
Chance of picking n a and m b is some number P(n a, m b) in the same range.

Now, chance of picking exactly n a and m b can occur in the following manner:

Either I pick an "a' - chance of that is P(a), then my next drawings must have exactly n-1 a and m b. so on condition I pick P(a) first the remaining have a chance P(n-1 a, m b)

Or, I pick a "b" and the next drawings must have exactly n a and m-1 b.
P(b) * P(n a, m-1 b)

The total sum of these two mutually exclusive cases is the chance for P(na, mb)

P(n a, m b) = P(a) * P( n - 1 a, m b) + P(b) * P( n a, m - 1 b)

This gives you a recursive formula to compute P(n a, m b) for the general case.

It turns out that binomials come to our rescue here so:

P(1 a 0 b) = P(a)
P( 0 a, 1 b) = P(b)
P(1 a, 1 b) = P(a) * P(b) + P(b) * P(a) = 2P(a)P(b) = binom(2,1) P(a) P(b)
P(2 a 0 b) = P(a)^2
P(0 a 2 b) = P(b)^2
P(n a 0 b) = P(a)^n
P(0 a m b) = P(b)^m
P(n a, m b) = binom( n + m, n) P(a)^n P(b)^m

This is the formula when you ignore the order. When we do take order into account we get the formula P(a)^nP(b)^m or if the chance for P(a) = P(b) we get P(a)^n+m or for a string of 32 amino acids and 20 possible acids in ample supply the formula - when considering order - is 20^32. If you do not consider order the formula is this value divided by all permutations of the 32 acids which is 32! so it is 20^32/32! and the chance is 32!/20^32 a much higher value I agree but that is besides the point.

The point is that your "picking" does not produce the formula you gave. The formula you gave was for "picking and connecting".

Alf

whiskey the hedonist

Lee is going to seize on this and say "Exactly! You have to pick, and then you have to connect!"

So to be clear: the formula he gave for the 2 draws of 4 choices problem (1/16*1/16*1/2) was actually for FIRST picking two letters in the desired order of connection, then, picking two letters in the desired order AGAIN, and THEN connecting two letters in the desired order. In other words, he wants to calculate the order three times, instead of once.

Alf

Err.. don't think it can be misunderstood this way. the "pick and connect" formula IS the 1/20^32. The "pick" where you ignore the order and thus have to put it in again is the 32!/20^32 formula. Consequently, when he give the formula of 20^32 you cannot say "wait, we did not consider the order", it is exactly what we do, if you did not consider the order, the formula would be different.

Alf

John A. Broussard

The best way to guarantee the fulfilment of a prophecy is to make it after the event occurred.

In fact, I absolutely guarantee that that prophecy will be be fulfilled.

There's no charge for this information.