For Inquisitor, et al: What is Evolution?

Since so many people have tried to make probability calculations, I thought I'd proffer an actual question, just to see if people understand:Suppose I have n darts, each with a 1/n chance of striking the target. What is the probability of hitting the target at least once given 10 darts? 20 darts? An infinite number of darts? How many darts must you have to be within 99% of the value of an infinite number of darts?Hint: Don't try to calculate the number of ways to hit the target. Instead, notice that the question is equivalent to asking for the probability of not hitting the target at all and then subtracting that from one since there is only one way of not hitting the target at all: Every dart misses.------------------
Rrhain
WWJD? JWRTFM!

buzsaw writes:

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I've read the arguments both of Morris and of his critics on this. He calls it "hydraulic" sorting, similar to what I alluded to.

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But hydraulic sorting would sort things by density, not by morphology.Have you ever bought a box of cereal or a bag of chips? Have you noticed that all of the big flakes/chips are on top while all of the crumbs and dust are at the bottom?Do you think the manufacturers deliberately did that?So why is it that if you were to turn the box/bag upside down and gently shake it, you find that it would re-sort itself so that all the big ones are on top and all the little ones are on the bottom?If the flood were to do things like Morris suggests, then we should see all the gigantic fossils on top and all of the tiny fossils on bottom.But we don't. They're sorted not by density but by morphology.Ergo, Morris is incorrect in his hypothesis.

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And again you folks have this huge problem of things sitting around to be fossilized somehow magically.

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What problem is there for a dead organism not to move?------------------
Rrhain
WWJD? JWRTFM!

crashfrog responds to me:

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Suppose I have n darts, each with a 1/n chance of striking the target.

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This part I don't understand. Your use of n twice confuses me. Do you mean that if I have 2 darts, I have a 1/2 chance of hitting the target? 3 darts, 1/3 chance? By this logic if I only have one dart (1/1 chance of hitting) it hits every time.

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You should be more careful picking your variable names if the above isn't what you meant.

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No, you got it exactly right: If I have 4 darts, each has a probability of only 1/4 of hitting...10 darts means each has only a 1/10 probability of hitting...an infinite number of darts means each has an infinitesimal probability of hitting.And, indeed, a single dart means there is a guaranteed chance of hitting. But, the question put before us is to determine a solution for all possible cases.That's why I made the hint that I did: Rather than try to determine directly the odds, calculate the opposite problem: Completely missing. Subtract that from 1 and you've got your answer. All you need to do, then, is plug in n and you get your answer.------------------
Rrhain
WWJD? JWRTFM!

crashfrog responds to me:

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No, you got it exactly right: If I have 4 darts, each has a probability of only 1/4 of hitting...10 darts means each has only a 1/10 probability of hitting...an infinite number of darts means each has an infinitesimal probability of hitting.

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And, indeed, a single dart means there is a guaranteed chance of hitting. But, the question put before us is to determine a solution for all possible cases.

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What I don't understand is - are these hypothetical darts whose accuracy decreases as the number of them increases, or do you believe this to be fundamental property of real-life darts?

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Hypothetical darts. It's a thought experiment designed to see how well people who make arguments of probability can calculate a simple probability.I don't think you could come up with an actual infinite number of darts. But then again, that was the last part of the question:How many darts must you have to be within 99% of the value of an infinite number of darts?While we can't get an infinite number of darts, we can get arbitrarily close to that value with a finite number of darts.

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I realize this is nit-picking, but I'm curious.

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S'aright. You'd want to be sure you understand the questiion before answering it.

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Given n darts, each with a constant chance of hitting the target,

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Not just a constant chance, but a specific chance.

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the chances of all of them not hitting the target decreases as n increases. Is that what you're after?

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That's a general result, yes, but I'm looking for the specific answer with all work shown. Does the chance of them all not hitting decrease without limit or not?That's why I asked the questions that I did:Suppose I have n darts, each with a 1/n chance of striking the target. What is the probability of hitting the target at least once given 10 darts? 20 darts? An infinite number of darts? How many darts must you have to be within 99% of the value of an infinite number of darts?------------------
Rrhain
WWJD? JWRTFM!

Vunderkind responds to me:

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Suppose I have n darts, each with a 1/n chance of striking the target. What is the probability of hitting the target at least once given 10 darts? 20 darts? An infinite number of darts? How many darts must you have to be within 99% of the value of an infinite number of darts?

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Why don't you ask a Statistics Professor?

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Because the average statistics prof isn't making claims that evolution is impossible because of a probability calculation.As I said, the reason I am asking the question is to see if those who are making arguments based upon probability are capable of calculating a very simple probability.

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Sheesh.

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So you don't know? I gave an amazingly good hint:Don't try to calculate the number of ways in which you might hit the target. Instead, calculate the number of ways in which you miss the target completely. There's only one way to do that, so the final answer is that result subtracted from 1.

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Crashfrog is no stats professor; he's a moron, and clearly not an individual with more than a High School diploma.

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Now that you have the ad hominem commentary out of your way....

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But anyway, please post the answer, as I want to take it next door to my neighbor who happens to be a Stats Prof. at a major accredited university boasting of over 47,000 FULL-TIME enrollment!!!

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And what will that prove? I have my training in Applied Mathematics, too.This is an extremely simple question. You can find it in almost any introductory textbook on probability.

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And not once in this thread have I seen any argument come close to containing the language consisent with that of Statistical Analysis I heard in my stats classes.

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And that's usually because most people who are making arguments based upon probability don't actually understand probability.

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What's more puzzling, is that the same arguments are too independantly simple to be of any advanced merit within the stats' field.

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That's part of the reason why I asked the question I did. In order to understand the more advanced problems, one will have to be able to comprehend the simpler ones. I have a gut reaction that the people who are making arguments from probability are simply parroting the words of others without understand what they are saying. Thus, I ask a simple question just to feel out their abilities.If they truly don't understand probability, it will be very difficult to explain why the argument from probability is flawed.

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Nevertheless, I can't wait to print this thread out and show it to my neighbor. Y'all just might spontaneously create a creationist by the utter incoherent foolishness of your arguments. But how one is to weave such a hopeful monster into the ToE is the question...

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Um, I think you misunderstand.I am not the one making an argument from improbability. I am not saying that we can calculate the probability of any species showing up from a common ancestor. Instead, it is the creationists who are saying that evolution can't happen because it is "too improbable."Therefore, let's have a simple test: For those who are making an argument of improbability, answer this simple question:Suppose I have n darts, each with a 1/n chance of striking the target. What is the probability of hitting the target at least once given 10 darts? 20 darts? An infinite number of darts? How many darts must you have to be within 99% of the value of an infinite number of darts?Hint: Don't try to calculate the number of ways in which you might hit the target. Instead, calculate the number of ways in which you miss the target completely. There's only one way to do that, so the final answer is that result subtracted from 1.------------------
Rrhain
WWJD? JWRTFM!

Percipient responds to me:

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Suppose I have n darts, each with a 1/n chance of striking the target. What is the probability of hitting the target at least once given 10 darts? 20 darts? An infinite number of darts? How many darts must you have to be within 99% of the value of an infinite number of darts?

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Just noticed this statistics problem for the first time, I'll give it a shot.

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Well, to be anal about it, it's probability, not statistics, but that's really just a nit.

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Here's the general solution for n, just do the math to find the answers for 10, 20 and ∞ darts:

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P_n = 1 - (1 - 1/n)ⁿ

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Yes, that is the correct formula, but you neglected to say why. You need to show your work.

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To find how many darts you must have to be within 99% of the value for ∞ darts, simply solve this equation for n:

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P_n - P_∞ = .01

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(1 - (1 - 1/n)ⁿ) - (1 - (1 - 1/∞)^∞) = .01

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The term for ∞ goes to 0, so this becomes simply:

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(1 - (1 - 1/n)n) = .01

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That's the answer, but I don't know how to solve the equation.

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Incorrect. That is not the answer. The term for ∞ does not go to 0.That's part of the reason that I asked you to calculate it for 10 darts and for 20 darts...to let you see where the trend is going.You have the right idea, though. You need to solve:1 - (1 - 1/n)ⁿ = 0.01 + [1 - (1 - 1/∞)^∞]for n.Hint: Logarithm.------------------
Rrhain
WWJD? JWRTFM!

Percipient responds to me:

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Here's the general solution for n, just do the math to find the answers for 10, 20 and darts:

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P_n = 1 - (1 - 1/n)ⁿ

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You need to show your work.

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This particular part is too elemenary for that, I just wrote it down off the top of my head. I could break it down for you, but why don't we let someone who's figuring it out for the first time do that? This isn't the interesting part of the problem anyway.

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Well, I'd say that for someone just figuring out the problem, this is very interesting. It opens up ways of thinking about problems. Before you forge ahead, it is a good thing to stop and really look at the question and what it is entailing. This is a very good example of something that I've been talking about in another thread: Working the opposite problem. That is, some problems are much more easily solved by looking at the opposite problem rather than by trying to tackle it directly.The interesting part is that the process makes you think about what it is you're trying to do. Thinking about thinking, and all that.

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Yes, that is the correct formula, but you neglected to say why.

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The original hint you provided explains why.

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Yes, but not directly. That is, where did you get the terms? I was looking for something like, "The chance of hitting the target is blah, this means that the chance of not hitting the target is blah."Sorry...my math professors were very strict about detail in proofs. You can't just say, "Do this an infinite number of times." You have to construct the process as a limit function and show the result as the terms go to infinity.

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For example, I don't know how to find log(1-1/n). Just by inspection it looks like n is less than 20, though.

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Well, there's the brute force method, and it works just as well even if it isn't very elegant: Just plug in numbers until you hit the desired tolerance.

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You originally introduced the problem in a reply to Daddy's Message 38 where he was talking about zero chances in a kajilion. Can you tie your probability problem back into the misconception Daddy was experiencing? I think that would be pretty helpful.

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It is a common claim of creationists that evolution is so unbelievably improbable that it would be ridiculous to suggest it as an actual answer to how life diversified. This is often done by showing how a bunch of events are somewhat unlikely and then treating them as independent events, thus multiplying them together, and coming up with an extremely small probability.Ignoring the claim that these probabilities are independent (that's another question), I'm going on the question of multiple trials. That is, while a single improbable event is unlikely to occur given only one chance, it becomes more probable when you get more than one chance to have it happen.There's a "scam" I have always thought about for winning the lottery: The typical 6/49 lottery (where you have 49 balls of which you need to choose 6 without regard to order) has about 14 million ways of coming up. Thus, borrow $14 million from the bank and then purchase a ticket for each possible outcome.You are guaranteed to win and not just the big prize but all of the smaller prizes for gettting 5 of 6 right, etc. That's because while the odds of winning the lottery with one ticket are quite small, the odds of winning the lottery with every ticket are quite large. Multiple trials increase your odds.Now, you will have to wait until the jackpot is larger than the amount you're going to borrow. And there is the danger of multiple winners reducing the pot. And then there is the problem of convincing the bank to go along with this scheme....------------------
Rrhain
WWJD? JWRTFM!

Percipient responds to me:

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Your problem showed up a bug in MathCad, which believes the expression for P_∞ goes to 0.

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Yeah, alas, computers trying to work with infinity is hard. As soon as you hit underflows, the term becomes an actual zero even when it isn't.It is interesting that it makes the term go to 0 and not to 1, though. The program is doing something right.That is, 1 - 1/n can be calculated a couple of ways: You can plug n in immediately and then subtract the quotient from 1. This term as n goes to infinity goes to 1. If you then raise this to infinity as an exponent, the term would be 1.The other way is to convert the fraction to (n - 1)/n. Again, this fraction also goes to 1 as n goes to infinity, but it is more apparent that the fraction is never 1. Since a common shortcut is that a number between 1 and 0 raised to the infinite power goes to 0, you'd hit 0.The thing is, neither answer is correct. This is one of the freaky things about infinity. There is no stock method of behaviour. You have to look at each case and calculate it individually.------------------
Rrhain
WWJD? JWRTFM!

OK, since nobody has actually come up with the answers, here they are.The original questions were:Suppose I have n darts, each with a 1/n chance of striking the target. What is the probability of hitting the target at least once given 10 darts? 20 darts? An infinite number of darts? How many darts must you have to be within 99% of the value of an infinite number of darts?In order to solve this question, it is much easier to solve the opposite problem. That is, rather than try to calculate the number of ways in which one dart can hit and then add the number of ways two darts can hit, etc., solve the problem of how many ways no darts can hit and then subtract that from the total number of possibilities. There's only one way to have no darts hit, so you only have to perform one calculation.Thus, if we have a dart that has a 1/n chance of hitting, then it has a 1 - 1/n chance of not hitting. Thus, if we have n darts, then we have a probability of:(1 - 1/n)ⁿfor absolutely no darts to hit.Thus, the probability of getting at least one to hit is:1 - (1 - 1/n)ⁿNow, all we need to do is pop in the appropriate number:What is the probability of hitting the target at least once given 10 darts?1 - (1 - 1/10)¹⁰ = 1 - (9/10)¹⁰ = 1 - .349 = .651Notice that this is a nearly two-thirds chance of hitting the target at least once.20 darts?1 - (1 - 1/20)²⁰ = 1 - (19/20)²⁰ = 1 - .358 = .642It's getting smaller, but not much.An infinite number of darts?This is where some fancy-schmancy math comes in. It turns out that (1 - 1/n)ⁿ goes to 1/e as n goes to infinity. e is Euler's number, equal to approximately 2.71828182845904523536... (some people memorize pi...I memorize e.)Thus, we have:1 - 1/e = 1 - .368 = .632Again, notice that this is nearly a two-thirds chance...even though each individual dart has an infinitesimal chance of hitting.How many darts must you have to be within 99% of the value of an infinite number of darts?For this, we solve the equation:[1 - (1 - 1/n)ⁿ] - (1 - 1/e) < .01
1 - (1 - 1/n)ⁿ < .01 + (1 - 1/e)
(1 - 1/n)ⁿ > 1 - (.01 + 1 - 1/e) (which equals 1/e - .01 which equal .358)A little plugging and chugging and we find that the answer is, indeed, 10 darts. All you need are 10 darts and you are within 99% of the infinite answer.Now, the main point I am making with this is to show that even with an extremely low probability of getting a hit on any individual outcome, the probability of getting at least one hit is actually fairly high if you have enough attempts.And, of course, if you have even more attempts, the probability goes even higher. If each dart has a 1/10 chance and you have 20 chances, you have a probability of 1 - (1/10)²⁰ = .878. Thus, it's a less than one-eighth chance of not hitting at all.------------------
Rrhain
WWJD? JWRTFM!