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(I know this isn't your project, so I'm not expecting you to have the answers; these are questions that occur to me) |
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| Perhaps I expressed myself poorly. If that was the case, I apologize, and will clarify: When I mean that you could easily get 1 million heads in a row during a sequence of 1 billion tosses, I didn't mean that there wouldn't be on the whole an approximately even number of both heads and tails; I'm definitely agreeing with that notion. What I'm saying is something that Mr. Tsakiris seems to misconstrue, which is that over the 1 billion tosses, there's no reason to expect that it needs to alternate evenly. This is to say that his expectation seems to be a series like this: 0101010101010101010... (With 0 being tails, 1 being heads) But random isn't so uniform. There may indeed be stretches like the above in a random series of coin tosses, but it might look like this: 0110001010000011001110101000011111100011111 Over time, this would even out numerically, but you will have stretches of repetition. Given the vastness of 1,000,000,000, there are 1,000 places for a 1,000,000 chain of heads or tails to be contained. Considering that in such a tossing sequence there would be roughly 500,000,000 of each, a million here or there won't make much difference in terms of the distribution. The "order" that the GCP is looking for seems illusory, since randomness anticipates "nonrandom" sequences throughout. |
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I think I have enough statistics knowledge left to work this out with the help of a text book, but maybe somebody who is handier here can show us the way (please). |
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| All I know is that this reminds me of the opening to "Rosencrantz and Guildenstern are Dead." Rosencrantz & Guildenstern Are Dead (Heads) - Video - YouTube (you have to get about 35 seconds in to get to the salient portion) |
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The GCP people claim to have done so, and that the results they see are even less likely than the normally unlikely results one might expect. I think this point is often lost in the discussion. |
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I am no expert in statistics, but I have learned enough to be very confident that 1) professional experimental scientists (on both sides) are pros compared to you and me 2) statistics is probably slightly more friendly to amateurs than dentistry, but maybe not by much 3) you and I are amateurs or worse And that's really a big part of the problem with commenting on GCP. Maybe a little hyperbole there, but probably not that much. |
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In italics is another quote for proper context. Each of their RNGs does 200 "tosses"per second. Mathematically one expects 100 1s. Mind: "Expects" is mathematical jargon! It means that if you generate a few million (or more, the more the better) seconds of data then you will find that the average number of 1s per second approaches 100. However: In less than 6% of all seconds will the number be actually exactly 100. IE in any ordinary sense we don't expect exactly 100. It's still the case that any specific combination of 1s and 0s is equally likely. There simply are more combinations with 100 1s than, for example, 200 (for which there is only 1 combination). Basically when we analyse the data like that we lump together a lot of combinations simply based on how many 1s they have and that is why some outcomes end up more likely than others. In the above examples we have an equal number of 1s which means that this way of analyzing would not distinguish between the two. There are 2^200 (2 to the power of 200) patterns you can get from 200 "coin tosses". When you work out how many patterns there are that have exactly 100 1s you will get 5.63%. Another 5.58% will have 101 1s and and equal number will have 99 1s. And so on... This is a so-called binomial distribution. Anytime you have 2 seperate outcomes (hit/miss, win/lose) you will get a binomial distribution. The 2 possibilities need not have equal likelihood. Getting a 6 or not with a dice is also binomial. Here is a binomial calculator: Binomial Calculator n is the number of tries and p is the probability. Say you want to know what the likelihood is of getting a 6 at least 3 times in 10 dice throws then you enter: n=10 (for 10 dice throws) p=0.167 (=1 divided by 6; the probability of getting a 6 on a 6-sided dice) And the last setting should be obvious. Point here is that these curves result from the way we lump together outcomes. Getting a 6 is less likely than not because there are more ways in which the dice can show a "not-6". Getting many 6 is less likely than getting few because there are more ways in which one get few than many. The distribution derives simply from sorting and counting patterns. When you have an RNG that produces 1s and 0s like the GCP then you expect to find the same distribution in practice as we have just theoretically worked out. However, you "expect" this in the same way that you expect 100 1s per every 200 "throws". I don't have a suitable calculator at hand so I'm not giving numbers here. Suffice it to say that when you compare a real distribution to a theoretical and repeat this a lot of times then you'll get another distribution similar (but not identical!) to the binomial one. That's basically one of the tests they do at the GCP. Now what does statistical significance mean? Imagine a 20-sided dice. Role-players use those. An ordinary dice has 6 sides and a 1/6th probability for every side; a 20-sided dice (a d20) has 20 sides and a 1/20th probability for every side. A 1/20th is 5%. You throw the d20 100 times you "expect" five 20s. (Using the linked calculator you find exactly 5 happens about 18% of the time) When you do a significance test you basically translate your results to a d20 dice throw. The reasoning is basically: Throwing a 20 is kinda rare so if it happens then maybe the dice was biased. I guess it is obvious that one must be careful with significant results. Say you throw 100 fair d20s and 10 that are biased so that they will show a 20 50% of the time instead of only 5% of the time. From your total of 110 throws you get 10 that are "significant". Only alf of which come from biased dice. OTOH hand among the 100 "non-significant dice" there are still 5 biased dice. How do you do the translating? Please turn to the online calculator again. If you toy around with it a little you will find that the probability of getting more than 111 1s among 200 "throws" is 5.18% and of getting more than 112 is 3.84%. IOW getting 112 ones is less likely than getting a 20 on a d20. So the reasoning goes maybe your "throws" were biased. |
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Please answer DoctorAtlantis' question. |
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