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| ▲ | fainpul 15 hours ago | parent | next [-] | | Isn't 100 / 100 the most likely outcome for a fair coin? Sure it's unlikely that you hit exactly that result, but every single other result is even less likely. What I'm trying to say: if you get 100 / 100, that's not a sign of an unfair coin, it's the strongest sign for a fair coin you can get. | | |
| ▲ | shakow 15 hours ago | parent | next [-] | | > every single other result is even less likely. But the summed probability of the “not too far away results” is much higher, i.e. P([93, 107]\{100}) > P([100]). So if you only shoot 100/100 with your coin, that's definitely weird. | | |
| ▲ | tshaddox 15 hours ago | parent | next [-] | | Okay, but it doesn't make sense to arbitrarily group together some results and compare the probability of getting any 1 result in that group to getting 1 particular result outside of that group. You could just as easily say "you should be suspicious if you flip a coin 200 times and get exactly 93 heads, because it's far more likely to get between 99 and 187 heads." | | |
| ▲ | wat10000 14 hours ago | parent [-] | | It's suspicious when it lands on something that people might be biased towards. For example, you take the top five cards, and you get a royal flush of diamonds in ascending order. In theory, this sequence is no more or less probable than any other sequence being taken from a randomly shuffled deck. But given that this sequence has special significance to people, there's a very good reason to think that this indicates that the deck is not randomly shuffled. In theory terms, you can't just look at the probability of getting this result from a fair coin (or deck or whatever). You have to look at that probability, and the probability that the coin (deck etc.) is biased, and that a biased coin would produce the outcome you got. If you flip a coin that feels and appears perfectly ordinary and you get exactly 100 heads and 100 tails, you should still be pretty confident that it's unbiased. If you ask somebody else to flip a coin 200 times, and you can't actually see them, and you know they're lazy, and they come back and report exactly 100/100, that's a good indicator they didn't do the flips. | | |
| ▲ | tshaddox 10 hours ago | parent [-] | | > It's suspicious when it lands on something that people might be biased towards. Eh, this only makes sense if you're incorporating information about who set up the experiment in your statistical model. If you somehow knew that there's a 50% probability that you were given a fair coin and a 50% probability that you were given an unfair coin that lands on the opposite side of its previous flip 90% of the time, then yes, you could incorporate this sort of knowledge into your analysis of your single trial of 200 flips. | | |
| ▲ | wat10000 8 hours ago | parent [-] | | If you don’t have any notion of how likely the coin is to be biased or how it might be biased then you just can’t do the analysis at all. | | |
| ▲ | tshaddox 2 hours ago | parent [-] | | You can certainly do the frequentist analysis without any regard to the distribution of coins from which your coin was sampled. I’m not well studied on this stuff, but I believe the typical frequentist calculation would give the same results as the typical Bayesian analysis with a uniform prior distribution on “probability of each flip being heads.” |
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| ▲ | fainpul 15 hours ago | parent | prev | next [-] | | > So if you only shoot 100/100 with your coin, that's definitely weird. Not if you only try once. | | |
| ▲ | shakow 15 hours ago | parent | next [-] | | Even if you shoot only once, you still have a higher chance of hitting something slightly off the middle than the perfect 100/100. And this because that's one point-precise result (100/100) vs. a cumulated range of individually less-probable results, but more probable when taken as a whole. For a fair coin, hitting 100/100 is ~5%, vs. ~30% falling in [97; 103] \ {100}. You can simulate here: https://www.omnicalculator.com/statistics/coin-flip-probabil... | | |
| ▲ | tshaddox 14 hours ago | parent | next [-] | | > you still have a higher chance of hitting something slightly off the middle than the perfect 100/100 That's because "something slightly off the middle" is a large group of possible results. Of course you can assemble a group of possible results that has a higher likelihood than a single result (even the most likely single result!). But you could make the same argument for any single result, including one of the results in your "slightly off the middle" group. Did you get 97 heads? Well you'd have a higher likelihood of getting between 98 and 103 heads. In fact, for any result you get, it would have been more likely to get some other result! :D | | |
| ▲ | zdragnar 14 hours ago | parent [-] | | > But you could make the same argument for any single result Isn't that the point? The odds of getting the "most likely result" are lower than the odds of getting not the most likely result. Therefore, getting exactly 100/100 heads and tails would be unlikely! | | |
| ▲ | tshaddox 13 hours ago | parent | next [-] | | But as I said, getting any one specific result is less likely than getting another other possible result. And the disparity in likelihoods is greater for any one specific result other than the 50% split. | |
| ▲ | alwa 12 hours ago | parent | prev [-] | | I think the disagreement is about what that unlikeliness implies. "Aha! You got any result? Clearly you're lying!"... I'm not sure how far that gets you. There's probably a dorm-quality insight there about the supreme unlikeliness of being, though: out of all the possible universes, this one, etc... | | |
| ▲ | zdragnar 10 hours ago | parent [-] | | Let's look at the original quote: > "Remember, if you flip a coin 200 times and it comes heads up exactly 100 times, the chances are the coin is actually unfair. You should expect to see something like 93 or 107 instead". Inverting the statement makes it read something like this: You are more likely to not get 100/100 than you are to get exactly 100/100 ...which is exactly what I was saying. Nobody is arguing that there is a single value that might be more likely than 100/100. Rather, the argument is that a 100/100 result is suspiciously fair. |
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| ▲ | e12e 13 hours ago | parent | prev [-] | | Should that be 25% for 97..193 excluding 100? | | |
| ▲ | shakow 12 hours ago | parent [-] | | “[97; 103] \ {100}” means the interval [97; 103] without the set {100}; so no, still ~30%. |
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| ▲ | kalaksi 15 hours ago | parent | prev [-] | | I'm sorry, but try what once? 200 flips once? | | |
| ▲ | dredmorbius 6 hours ago | parent [-] | | In statistics, various examples (e.g., coin flips) often stand in for other activities which might prove expensive or infeasible to make repeated tries of. For "coin flips", read: human lives, financial investments, scientific observations, historical observations (how many distinct historical analogues are available to you), dating (see, e.g., "the secretary problem" or similar optimal stopping / search bounding problems). With sufficiently low-numbered trial phenomena, statistics gets weird. A classic example would be the anthropic principle: how is it that the Universe is so perfectly suited for human beings, a life-form which can contemplate why the Universe it so perfectly suited for it? Well, if the Universe were not so suited ... we wouldn't be here to ponder that question. The US judge Richard Posner made a similar observation in his book "Catastrophe: Risk and Response" tackles the common objection to doomsday predictions that all have so far proved false. But then, of all the worlds in which a mass extinction event has wiped out all life prior to the emergence of a technologically-advanced species, there would be no (indegenous) witnesses to the fact. We are only here to ponder that question because utter annihilation did not occur. As Posner writes: By definition, all but the last doomsday prediction is false. Yet it does not follow, as many seem to think, that all doomsday predictions must be false; what follow is only that all such predictions but one are false. -Richard A. Posner, Catastrophe: Risk and Response, p. 13. <https://archive.org/details/catastropheriskr00posn/page/13/m...> |
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| ▲ | grraaaaahhh 10 hours ago | parent | prev [-] | | >But the summed probability of the “not too far away results” is much higher, i.e. P([93, 107]\{100}) > P([100]). That's true of every result. If you're using this to conclude you have a weird coin then every coin is weird. |
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| ▲ | ptrl600 5 hours ago | parent | prev | next [-] | | Low kolmogorov complexity equals suspicious | |
| ▲ | voakbasda 15 hours ago | parent | prev | next [-] | | I would guess that a single trial of 200 flips can be treated as one event, so getting 100/100 is but one outcome. It may be the most likely individual outcome, but the odds of getting that exact result feel less likely than all of the other possible outcomes. The 100/100 case should be seen the most over repeated trials, but only marginally over other nearby results. Intuitively, this seems right to me, but sometime statistics do not follow intuition. | |
| ▲ | dooglius 15 hours ago | parent | prev | next [-] | | "fair coin" refers to both the probability of heads and tails being equal (which is still justified) as well as the trials being independent (unlikely with 100/200; more likely the "coin" is some imperfect PRNG in a loop) | | |
| ▲ | tshaddox 14 hours ago | parent [-] | | > more likely the "coin" is some imperfect PRNG in a loop "More likely"? How can you even estimate the likelihood of the coin being "an imperfect PRNG" based on a single trial of 200 flips? | | |
| ▲ | dooglius 14 hours ago | parent | next [-] | | You can use combinatorics to calculate the likelihood. If your PRNG is in a cycle of length N in its state space (assuming N>200), and half the state space corresponds to heads (vs tails), then the likelihood would be (N/2 choose 100)^2/(N choose 200) versus your baseline likelihood (for a truly random coin) of (200 choose 100)/2^200. Graphing here https://www.wolframalpha.com/input?i=graph+%28%28N%2F2+choos... and it does look like it's only a slight improvement in likelihood, so I did overstate the claim. A more interesting case would be to look at some self-correcting physical process. | |
| ▲ | ajuc 14 hours ago | parent | prev [-] | | Bayesian vs frequentist in a nutshell :) If a student was tasked with determining some physical constant with an experiment and they got it exactly right to 20th decimal place - I'll check their data twice or thrice. Just saying. You continue believing it was the most likely value ;) |
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| ▲ | buildsjets 10 hours ago | parent | prev | next [-] | | A truly fair coin would never land heads or tails. It would land standing on it's edge, every single time. | |
| ▲ | Joker_vD 15 hours ago | parent | prev [-] | | Well, yes. But the expected deviation from the mean is still ≈7.07. And the probability that the outcome will be either 93/107 or 107/93 is (slightly) higher than the outcome being exactly 100. | | |
| ▲ | fainpul 15 hours ago | parent | next [-] | | But those are 2 results. 100 / 100 is more likely than 93 / 107 (or any other specific result) is what I'm saying. | | |
| ▲ | paddleon 15 hours ago | parent | next [-] | | You can tell the difference between 93 heads, 107 tails and 93 tails, 107 heads but not between 100 heads, 100 tails and 100 tails, 100 heads. So while those are two results (93/107 and 107/93), they really only count as two separate outcomes if you pre-specify that the first number is heads. If instead you consider symmetries, where there are 2 ways to get 93/107 and only one wall to get 100/100, then you have more likelihood for the 93/107 outcome because you have two ways to get it. | | | |
| ▲ | kalaksi 15 hours ago | parent | prev | next [-] | | I don't think that makes 100 / 100 the most likely result if you flip a coin 200 times. It's not about 100 / 100 vs. another single possible result. It's about 100 / 100 vs. NOT 100 / 100, which includes all other possible results other than 100 / 100. | | |
| ▲ | andoando 11 hours ago | parent [-] | | If you flip a coin 200 times, and repeat that a billion times, 100/100 will be more likely than any other ratio. It will be a bell curve with 100/100 as the peak |
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| ▲ | Joker_vD 15 hours ago | parent | prev [-] | | Depends what you count as a result, I guess. "There is exactly N flips of a single kind" is also a viable definition, just as "The exact sequence of flips was x_0, x_1, ... x_199" is. |
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| ▲ | eviks 15 hours ago | parent | prev [-] | | Why not go one abstraction further and go expected deviation from deviation? Probably the word "expected" plays a mind trick? "Expected" doesn't mean the probability increases, the easiest way to understand it is just by looking at the probability distribution function chart for coin tosses - you'll immediately see that mean has the highest chance of happenning, so exactly 100 is the most likely outcome |
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| ▲ | munchbunny 15 hours ago | parent | prev | next [-] | | The chance of exactly 100 heads from 200 fair coin flips is approximately 5-6%. Qualitatively, that's not particularly strong evidence for an unfair coin if you did only one trial. You could also argue that 100 out of 200 on a fair coin is more likely than any other specific outcome, such as 93/200, so if the argument is that the coin is "too perfect", you then also have to consider the possibility that the coin is somehow biased to produce 93/200 much more often than anything else, vs. 100/200. | | |
| ▲ | nearbuy 2 hours ago | parent [-] | | There is also no way to weight a coin that makes it more likely to get 100/200 than a fair coin. |
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| ▲ | n1b0m 15 hours ago | parent | prev [-] | | In a real-world scenario, if you saw a result significantly far from 100 (like 150 heads), you might suspect the coin is unfair. However, seeing exactly 100 heads gives no reason to suspect the coin is unfair; it's the result most consistent with a fair coin. |
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