Probability and evolution

[mr friendly guy]

Various anti-evolutionary arguments exist using the probability argument. The flaw with these types of arguments are that the models they calculate the probability for, do not represent evolution hence they are attacking a strawman.

One of the ways I explain it to lay people is to take say 6 numbers, lets say 1,2,3,4,5,6 (although they can be in any other order). Using a six sided die the probability of getting that particular combination at once is (1/6) multiplied by itself 6 times, ie 1 in 46656. In other words, the probability something has ALL features coming about at the same time is low.

Using this example as an analogy to descent by modification, it would be if we rolled the die the first time, and then we get to "save" the first number if it turns out to be 1, and we get to "save" the second number if it turns out to be 2 and so on. The advantage is that we don't have to reroll the die the first time if it already lands on the correct number.

What I want to know, is there a way to calculate the probability of getting the combination of 1,2,3,4,5,6 if we are allowed to "save" the numbers as they come. And also please can someone explain how they derive the numbers instead of just quoting it.

Any help would be appreciated.

[Lord Zentei]

If you calculate the probability of a specific combination of numbers as rolled on some dice, you multiply the respective probabilities of each dice in turn (you always use this method for calculating composite events consisting of independant random events). However, if you can "save" the earlier numbers, it is given what the first numbers are, so the probability that the first dice is 1, for instance is 100% if you saved it after having rolled a 1.

Put differently, saying that "the first numbers are 1 and 2. What is the probability that the next four dice in turn are 3, 4, 5 and 6?" is equivalent to asking "I have 4 dice. What are the chances that I'll get 3, 4, 5 and 6 if I roll them each in turn?" The reason is that the probability of the outcome of the last four dice are completely unaffected by the outcome of the first two, or even of whether you rolled them at all.


So: chance of getting 1, 2, 3, 4, 5, 6 with six dice = (1/6)^6 = 1/46656.

Given that you have rolled 1, 2; chances that you get 3, 4, 5, 6 with the next four dice = 1*1*(1/6)^4 = 1/1296. (since the probility of getting 1 and 2 on the first two dice is trivially 100% and 100% respectively if you assume that you get 1 and 2 for the first two dice)

Chance of getting 3, 4, 5, 6 with four dice = (1/6)^4 = 1/1296.

The fact of having rolled the first two dice and what you got there is irrelevant, since these are independant random variables. You just proceed from the point where you "saved" the results.

[mr friendly guy]

So how would we apply this knowledge to answering how many people need to roll the die to get the correct combination.

For example with the first case of probability 1 in 46656, if we have 46656 people rolling a die simultaneously, we hope one person gets the right combination (ok this isn't strictly true, since "chance doesn't remember), but we do get a rough idea.

With the second case of probability, can we gauge how many people need to roll before the 1,2,3,4,5,6 combination comes up.

Ok, its not an exact science, but can we get an idea of how much likely is it to get the correct combination using the second method?

[nasor]

What I want to know, is there a way to calculate the probability of getting the combination of 1,2,3,4,5,6 if we are allowed to "save" the numbers as they come. And also please can someone explain how they derive the numbers instead of just quoting it.

The probability would be 6!/6, or 6/6 x 5/6 x 4/6 etc. The first number that you throw has to be 1-6, so it is guaranteed to be useful. When you throw the dice the second time there are now 5 numbers that you need and one number that you don't, so there is a 5/6 chance of getting a useful number. By the 6th throw there will only be one number left that you need to complete your 1-6 sequence, so you have a 1/6 chance of throwing it. The odds would be 1 in 120.

But that's not really a good way to explain it, since it ignores the fact that the chemical reactions involved in life aren't random; they are governed by physical laws that dictate how the chemical building blocks of life will behave. The analogy that I like to use is this:

Imagine that you place 5 magnets randomly in a large plastic bin. You then place the bin in the back of your car and drive around over bumpy roads all day, constantly jostling the bin. At the end of the day, when you look in the bin you will most likely find that the magnets are all stuck to each other. A creationist would look in the bin and say, "Look at that! These little magnets were being shaken all day, and somehow they all ended up in a final position where they are all touching each other! With all the ways that these five magnet could be randomly distributed in the box, what are the odds that they would happen to end up all touching! Clearly some intelligent force must have deliberately arranged them this way, because it's way too unlikely that it could happen by chance!"

What the creationist explanation overlooks, of course, is that the magnets weren't just randomly arranged in the box by the jostling of the car - their behavior was governed by physical laws that strongly influenced them toward aggregating rather than settling randomly.

[Lord Zentei]

GHETTO:

You may also wish to consider the fact that as far as evolution is concerned, there are multiple events taking place simultaneously. As an analogy, consider the chances against getting fist prize in a state lottery. Not very likely. But if millions of people play, the chances that someone will is a lot better. If you have a primordial soup, you have a lot of molecules to work with, and these are interacting independantly of one another, and simultaneously.

[Lord Zentei]

GHETTO 2:

The wording of the first ghetto edit is relevant to abiogenesis, of course; I'm not sure which you are after - abiogenesis or evolution, though the probability argument is more often used against abiogenesis.

[Lord Zentei]

So how would we apply this knowledge to answering how many people need to roll the die to get the correct combination.

For example with the first case of probability 1 in 46656, if we have 46656 people rolling a die simultaneously, we hope one person gets the right combination (ok this isn't strictly true, since "chance doesn't remember), but we do get a rough idea.

With the second case of probability, can we gauge how many people need to roll before the 1,2,3,4,5,6 combination comes up.

Ok, its not an exact science, but can we get an idea of how much likely is it to get the correct combination using the second method?

ROFL: I had not seen this post when I posted my GHETTO.

When you want to calculate that at least one person acheves something improbably you use the following formula:

P(at least one) = 1-(1-P)^N

Where P is the probability that one person in isolation gets lucky and N is the number of people. I.e. you take the probability that each person in turn does NOT get lucky, multiply the probabilities and that is the chance that NONE of them gets lucky. The chance that AT LEAST ONE gets lucky is 100% minus this probability.

So for instance, if you have 10 people rolling one die each, the chance of at least one getting a 6 is 1-(5/6)^10 = 1 - 0.1615 = 0.8385 or about 84%.

[drachefly]

As I understand it, the first die had to be a 1, the second die had to be a 2, and so forth. You get to save if the die is the right number for that slot, regardless of other dice being right.

If you're saving, it just becomes 6 'and'-ed 'have I gotten this digit yet' questions.

Mean number of tries for one die to get one particular digit: 6
Number of tries at which the probability of having gotten a particular digit is 5/6: at least 12 (probability 3/4), but not more than 18 (probability 7/8). Say, 15.

So, the number of tries is now around 15. Compared to 46656, that's not a lot.


Now, if you don't let the 2 lock in until the 1 has locked in, and so forth, it gets a little longer but it's actually easier.

#targets * mean tries = 36. Subtract off 5 because each time you get one you get to check the next one for 'free'.

Even in that more restrictive regime, it's a mere 31 tries.

[mr friendly guy]

What I want to know, is there a way to calculate the probability of getting the combination of 1,2,3,4,5,6 if we are allowed to "save" the numbers as they come. And also please can someone explain how they derive the numbers instead of just quoting it.

The probability would be 6!/6, or 6/6 x 5/6 x 4/6 etc. The first number that you throw has to be 1-6, so it is guaranteed to be useful. When you throw the dice the second time there are now 5 numbers that you need and one number that you don't, so there is a 5/6 chance of getting a useful number. By the 6th throw there will only be one number left that you need to complete your 1-6 sequence, so you have a 1/6 chance of throwing it. The odds would be 1 in 120.

No its not. Because the numbers have to be in that particular order, and not just have 1-6 in any order.

But that's not really a good way to explain it, since it ignores the fact that the chemical reactions involved in life aren't random; they are governed by physical laws that dictate how the chemical building blocks of life will behave.

While this is true, the point of my analogy was mainly to illustrate that if you have descent with modification, you need less number of tries to create the required feature, than if you reset each time you don't get it.

Its meant to be a counter to the "irreducible complexity" argument. Say there is structure with 6 parts, which I will call part 1, 2, 3 and so on. The structure has the parts connected in such an order as 1,2,3 and so on. It is much easier if you can "save" the first part when you get it, and then add the other parts on. The reason of course why you can "save" the first part, is that natural selection preserves it because each it allows the organism to survive to spread its genes. Richard Dawkins used a similar argument, but he used specific phrases, and I think he used a computer to keep on trying until it derived the required phrase. Since I don't have such a program handy, I thought it would be easier just to use a few numbers.

The analogy that I like to use is this:

<snip>

Great analogy. I will remember it for next time.

[Wyrm]

If you only want the probability of getting the right sequence after x number of times you have to roll dice (ie, you roll six dice the first round, adding the x_1 number of dice you have to roll the next round, adding the x_2 number of dice you have to roll the following round, and so on) before they're all correct, then there's a known distribution that gives the probability of success after x number of dice rolls.

If you were to roll one die until you get the right number, then this is a geometric distribution: P(x|p) = p(1-p)^{x-1}, where x is the number of times you have to roll, and p is the probability you succeed (here, 1/6). This situation is just the sum of r geometric distributions (r number of dice to get right), so the total number of dice rolls (y) has a negative binomial distribution: C(y-r,r-1) p^r (1-p)^{y-r}.

Turning this into a number of rounds will take a little more thought, but the number of rounds will not exceed the total number of dice rolls.

[Wyrm]

Okay, I have the exact calculation for the probability of getting all the dice having the correct number after Y trials.

First of all, let X_1, X_2, ..., X_n be the number of times the i-th die was rolled and did not come up with the correct number. These random variables are iid geometric random variables, with pmfs of

P(X_i = x_i) = p(1-p)^x_i

The probability of rolling the correct number on the i-th die after not more than y-1 failures for that die is

P(X_i ≤ y-1) = ∑^{y-1}_{k=0} p(1-p)^k = 1-(1-p)^y

The probability we want is the probability that ALL of the X_i's are less than y, so

P(all X_i ≤ y-1) = ∏_{all i} 1-(1-p)^y = [1-(1-p)^y]^n (X_i's iid)

But the condition "all X_i ≤ y-1" is simply the statement that we succeed fixing all the dice by at least the y-th roll. Therefore, P(all X_i ≤ y-1) = P(Y ≤ y). Fortunately, Y is a discrete random variable on the natural numbers, so the probability of succeeding by the Y-th roll is P(Y = y) = P(Y ≤ y) - P(Y ≤ y-1) = [1-(1-p)^y]^n - [1-(1-p)^{y-1}]^n