# How many times do you have to flip a coin to ensure a heads **URL:** \ **Category:** Great Debates\ **Created:** [May 23, 2007, 11:06pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290 "2007-05-23T23:06:03Z")\ **Posts on this page:** 20\ **Page:** 4
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [May 24, 2007, 8:08am UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/61 "2007-05-24T08:08:50Z")
Indeed, we couldn’t even have hoped for pdfs, or anything based on pdfs, to work any better here. The values of pdfs are just derived from the values of the probability function, and are thus unable to give any more information. And, as I’ve been saying, the fundamental problem with trying to determine possibility/impossibility from the values of the probability function is that there’s just not enough information there; two variables can have the exact same probability distributions (in the sense that, for every set of values, both variables have the same probability of falling in that set), despite one variable being able to take on values that another never can. [For example, the X and Y defined in the second paragraph above have the exact same probability distribution, but X can be 0 while Y can’t and Y can be 3 while X can’t]. It’s perfectly well possible to talk about impossibility/possibility in mathematical terms, but you can’t base it on the values of probabilities alone. You might as well just take it to be a primitive notion in itself, connected slightly to probabilities (in that impossible events must have probability 0), but not determinable from them. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [May 24, 2007, 8:16am UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/62 "2007-05-24T08:16:43Z")
[QUOTE=Indistinguishable] pdfs aren’t uniquely determined by distributions (and, indeed, given any pdf, one can arbitrarily change its value at any one point to get another pdf for the same distribution) [/QUOTE] I suppose you may take the point of view that pdfs are defined as the derivative of the cumulative distribution function (rather than, as I was taking them to be, as any function whose integral gives the cumulative distribution function). Or you may have been restricting attention to continuous pdfs. In either of those cases, they are indeed unique (if they exist). Nonetheless, everything else I’ve said still holds. ---
**Author:** ![Polerius](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Polerius](https://boards.straightdope.com/u/Polerius)\ **Post date:** [May 24, 2007, 8:23am UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/63 "2007-05-24T08:23:33Z")
[QUOTE=Indistinguishable] For example, let X be a random variable uniformly distributed over (0, 1/2], let Y be either +1 or -1 with equal probabilities, and let Z be X \* Y. Like I said, Z will admit infinitely many pdfs [/QUOTE] I’m not sure what you mean by “will admit infinitely many pdfs”. Based on your definition of Z above, I’d say that Z is uniformly distributed in the set [-1/2, 0) U (0,1/2] So, f(Z) = 1 for Z in [-1/2, 0) U (0,1/2], and f(Z) = 0, otherwise. Which means that f(0) = 0. > [@](#): > > , but the most natural one assigns a value of 1 to the argument 0 [this one will fit your criterion that Pr(Z in [-D/2, D/2]) = D \* f(0)]. Yet, despite that, it is impossible for Z to equal 0. I disagree that f(0) = 1. As I mentioned above, f(0) = 0, which is consistent with the fact that it is impossible for Z to equal 0. > [@](#): > > The other direction doesn’t work either; let X be uniformly distributed over [-1, 1], and let Y be equal to X when X is nonzero but equal to 3 when X is 0. Again, there are infinitely many corresponding pdfs for Y, but the most natural one assigns a value of 0 to the argument 3 [this again being the one fitting your “probability of being close” criterion]. Yet, despite that, it is possible for Y to equal 3. Note that for Y, I would consider f(Y) = 1 when (Y in [-1,0) U (0,1]) or (Y = 3), and f(Y) = 0 otherwise. So, f(3) = 1, which means that Y = 3 is possible, though it has a probability of zero, which is consistent with what I said in the previous post. Overall, I agree that if we get into the details it can get hairy. To make things simpler, I should have noted that my example in the previous post was restricted to pdfs that are continuous functions. Otherwise, we’ll need to talk about pmf’s, Dirac Delta functions, etc (and, from what I know, mathematicians don’t consider Dirac Delta functions as “real functions”) I was just trying to set up an example where impossibility was denoted by a mathematical concept (i.e. the value of the pdf at a certain point). ---
**Author:** ![Polerius](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Polerius](https://boards.straightdope.com/u/Polerius)\ **Post date:** [May 24, 2007, 8:32am UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/64 "2007-05-24T08:32:21Z")
[QUOTE=Indistinguishable] I suppose you may take the point of view that pdfs are defined as the derivative of the cumulative distribution function. [/QUOTE] As an aside, it seems that cdfs behave much “nicer” that pdfs, and should be used in the more rigorous proofs. (e.g. when you have a continous density with a mass at some point, the pdf would be something like a continous function plus a Dirac Delta at the point of the probability mass. From what I know, this makes some purists nervous, and won’t consider such a pdf a ‘function’. However, the cdf is much nicer, having only a step where the probability mass is, so it behaves much more like a function) ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [May 24, 2007, 8:33am UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/65 "2007-05-24T08:33:28Z")
[QUOTE=Polerius] I’m not sure what you mean by “will admit infinitely many pdfs”. [/QUOTE] Yeah, I’ve now clarified my remarks on this. > [@](#): > > Based on your definition of Z above, I’d say that Z is uniformly distributed in the set [-1/2, 0) U (0,1/2] Correct. > [@](#): > > So, f(Z) = 1 for Z in [-1/2, 0) U (0,1/2], and f(Z) = 0, otherwise. > Which means that f(0) = 0. Why would f(0) = 0? You told us: > [@](#): > > Pr(x in [x0-D/2, x0+D/2]) = D\*f(x0) = D, when x0 in (0,1) . Keeping that in mind, we should take f(0) to be 1, as Pr(Z in [0 - D/2, 0 + D/2]) = D \* 1. > [@](#): > > I disagree that f(0) = 1. As I mentioned above, f(0) = 0, which is consistent with the fact that it is impossible for Z to equal 0. It’s not consistent with your “probability of being close” criterion for the values of the pdf. > [@](#): > > Note that for Y, I would consider > f(Y) = 1 when (Y in [-1,0) U (0,1]) or (Y = 3), and f(Y) = 0 otherwise. > > So, f(3) = 1, which means that Y = 3 is possible, though it has a probability of zero, which is consistent with what I said in the previous post. Again, though, not consistent with your “probability of being close” criterion for the values of the pdf. To be consistent with that, you need f(3) = 0. > [@](#): > > Overall, I agree that if we get into the details it can get hairy. To make things simpler, I should have noted that my example in the previous post was restricted to pdfs that are continuous functions. Otherwise, we’ll need to talk about pmf’s, Dirac Delta functions, etc (and, from what I know, mathematicians don’t consider Dirac Delta functions as “real functions”) Hairy indeed. But note that both the pdfs I proposed are continuous, while both the ones you propose in this post are discontinuous at the points of interest. > [@](#): > > I was just trying to set up an example where impossibility was denoted by a mathematical concept (i.e. the value of the pdf at a certain point). It was a good effort, but I’d say it failed to solve the problem of determining possibility from probability, because that problem can’t be solved. Possibility/impossibility is simply not determinable from probabilities, for all the reasons I’ve outlined. Again, consider the case of the X and Y which have the same probability distributions but different possibilities/impossibilities. Clearly, there isn’t enough information in the probability distributions from which to recover all the possibility/impossibility information. But as I’ve said before, this doesn’t mean possibility is not a mathematical concept; just that it’s not a purely probabilistic concept, as such. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [May 24, 2007, 8:37am UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/66 "2007-05-24T08:37:18Z")
[QUOTE=Polerius] As an aside, it seems that cdfs behave much “nicer” that pdfs, and should be used in the more rigorous proofs. (e.g. when you have a continous density with a mass at some point, the pdf would be something like a continous function plus a Dirac Delta at the point of the probability mass. From what I know, this makes some purists nervous, and won’t consider such a pdf a ‘function’. However, the cdf is much nicer, having only a step where the probability mass is, so it behaves much more like a function) [/QUOTE] Yeah, I agree, cdfs are usually “nicer” to work with, or feel “cleaner” or vaguely aesthetically preferable in some way, even though I’m not terribly opposed to working with generalized functions like the Dirac delta in this context. ---
**Author:** ![RTFirefly](https://avatars.discourse-cdn.com/v4/letter/r/c77e96/32.png) [@RTFirefly](https://boards.straightdope.com/u/RTFirefly)\ **Post date:** [May 24, 2007, 1:16pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/67 "2007-05-24T13:16:53Z")
[QUOTE=Kimstu] It’s like picking up a bridge hand of 13 cards dealt randomly from a pack of 52. For **any particular hand** , the odds of being dealt **exactly that hand** at random are about 1 in 400000000000000000000. (Yes, 20 zeroes. :eek: ) [/quote] While it doesn’t undermine the point you were making, I feel compelled to note that the number of distinct bridge hands is much, much smaller than that: C(52,13) = 52!/(13!39!) = 635,013,559,600 to be precise. ---
**Author:** ![Sapo](https://avatars.discourse-cdn.com/v4/letter/s/ec9cab/32.png) [@Sapo](https://boards.straightdope.com/u/Sapo)\ **Post date:** [May 24, 2007, 1:37pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/68 "2007-05-24T13:37:02Z")
[QUOTE=Indistinguishable] [QUOTE=Indistinguishable] pdfs aren’t uniquely determined by distributions (and, indeed, given any pdf, one can arbitrarily change its value at any one point to get another pdf for the same distribution) [/QUOTE] I suppose you may take the point of view that pdfs are defined as the derivative of the cumulative distribution function (rather than, as I was taking them to be, as any function whose integral gives the cumulative distribution function). Or you may have been restricting attention to continuous pdfs. In either of those cases, they are indeed unique (if they exist). Nonetheless, everything else I’ve said still holds. [/QUOTE] 😕 how many Indistinguishables are there? How can we distinguish them? ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [May 24, 2007, 1:42pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/69 "2007-05-24T13:42:50Z")
[QUOTE=Shamozzle] Any surface has an infinite number of points. [/QUOTE] Yes, that’s true for surfaces in 3-dimensional Euclidean geometry. However, it’s not true of surfaces in the real world, because of the finite size of atoms and sub-atomic particles, and because of quantum effects. Reality is only an approximation of Euclidean geometry: a theoretical sphere has an infinite number of points on its surface, but a real sphere (or even an approximate sphere (like a football or the Earth) has only a finite number of points on its surface. ---
**Author:** ![Apos](https://avatars.discourse-cdn.com/v4/letter/a/838e76/32.png) [@Apos](https://boards.straightdope.com/u/Apos)\ **Post date:** [May 24, 2007, 2:38pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/70 "2007-05-24T14:38:20Z")
I don’t get the problem in the OP. If you have an infinite number of coins, this is a very different situation than the one in which you are talking about the probability of one particular coin. The latter statement is poorly phrased: it is a statistical statement about probability _approaching_ 50% as more and more coin flips are made. That has no bearing on the fact that you’ve previously stipulated that there are infinite number of coins and thus are looking at the few that never flip heads. The vanishingly improbable occurance is being DEFINED as being inevitable with an infinite number of coins, and coins approaching a perfect 50% is an aggregate statistical conclusion, not a fact for what all coins must do over a very very long finite period or even an infinite one. ---
**Author:** ![Bill\_Door](https://avatars.discourse-cdn.com/v4/letter/b/50afbb/32.png) [@Bill\_Door](https://boards.straightdope.com/u/Bill_Door)\ **Post date:** [May 24, 2007, 4:19pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/71 "2007-05-24T16:19:29Z")
[QUOTE=RTFirefly] While it doesn’t undermine the point you were making, I feel compelled to note that the number of distinct bridge hands is much, much smaller than that: C(52,13) = 52!/(13!39!) = 635,013,559,600 to be precise. [/QUOTE] You only get the larger number, 3.95424\*10^21 if you take the order you receive the cards as being significant. In this model, A-K would be different than K-A. From a bridge point of view it’s invalid, but looking at the probability of being dealt a specific hand it’s not. ---
**Author:** ![Kimstu](https://avatars.discourse-cdn.com/v4/letter/k/ecd19e/32.png) [@Kimstu](https://boards.straightdope.com/u/Kimstu)\ **Post date:** [May 24, 2007, 4:23pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/72 "2007-05-24T16:23:57Z")
> [@RTFirefly](#): > > While it doesn’t undermine the point you were making, I feel compelled to note that the number of distinct bridge hands is much, much smaller than that: C(52,13) = 52!/(13!39!) = 635,013,559,600 to be precise. Yer absolutely right **Rufus** , sorry about that. I had a brain fart and calculated the cumulative probability of getting each card **given** that the previous ones had already been dealt, which isn’t the same thing. Thanks for the correction. So okay, if you were randomly dealing one bridge hand per second from the beginning of the universe to the present moment, you **would** almost certainly get any possible hand at some point. But not if you were dealing at the same rate for, say, a hundred years. Which in practical terms still makes it extremely improbable that any **particular** bridge hand should ever get dealt in the course of ordinary play—and yet, any particular hand that you encounter obviously **has** been dealt. So my analogy, although less gee-whizzical numerically than it originally appeared, basically still stands. ---
**Author:** ![gazpacho](https://avatars.discourse-cdn.com/v4/letter/g/6f9a4e/32.png) [@gazpacho](https://boards.straightdope.com/u/gazpacho)\ **Post date:** [May 24, 2007, 4:31pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/73 "2007-05-24T16:31:36Z")
[QUOTE=Giles] Yes, that’s true for surfaces in 3-dimensional Euclidean geometry. However, it’s not true of surfaces in the real world, because of the finite size of atoms and sub-atomic particles, and because of quantum effects. Reality is only an approximation of Euclidean geometry: a theoretical sphere has an infinite number of points on its surface, but a real sphere (or even an approximate sphere (like a football or the Earth) has only a finite number of points on its surface. [/QUOTE] Why do you equate points to sub-atomic particles? ---
**Author:** ![Lemur866](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lemur866/32/434_2.png) [@Lemur866](https://boards.straightdope.com/u/Lemur866)\ **Post date:** [May 24, 2007, 5:03pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/74 "2007-05-24T17:03:09Z")
[QUOTE=gazpacho] Why do you equate points to sub-atomic particles? [/QUOTE] It shouldn’t be sub-atomic particles, it should be Planck length. In “the real world” anything smaller than Planck length is meaningless. > **[Planck units](https://en.wikipedia.org/wiki/Planck_length)** > > In particle physics and physical cosmology, Planck units are a system of units of measurement defined exclusively in terms of four universal physical constants: c, G, ħ, and kB. Expressing one of these physical constants in terms of Planck units yields a numerical value of 1. They are a system of natural units, defined using fundamental properties of nature (specifically, properties of free space) rather than properties of a chosen prototype object. Originally proposed in 1899 by German physicis... ---
**Author:** ![gazpacho](https://avatars.discourse-cdn.com/v4/letter/g/6f9a4e/32.png) [@gazpacho](https://boards.straightdope.com/u/gazpacho)\ **Post date:** [May 24, 2007, 5:07pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/75 "2007-05-24T17:07:30Z")
[QUOTE=Lemur866] It shouldn’t be sub-atomic particles, it should be Planck length. In “the real world” anything smaller than Planck length is meaningless. > **[Planck units](https://en.wikipedia.org/wiki/Planck_length)** > > In particle physics and physical cosmology, Planck units are a system of units of measurement defined exclusively in terms of four universal physical constants: c, G, ħ, and kB. Expressing one of these physical constants in terms of Planck units yields a numerical value of 1. They are a system of natural units, defined using fundamental properties of nature (specifically, properties of free space) rather than properties of a chosen prototype object. Originally proposed in 1899 by German physicis... [/QUOTE] I don’t think that is exactly true. It is not like the universe is a three dimensional grid where everything moves one plank length at a time. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [May 24, 2007, 5:08pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/76 "2007-05-24T17:08:50Z")
Leaving aside coins, the premise of the OP seems to be that God exists because the universe is too improbable to exist unless God intentionally created it. But the flaw with that argument is that it presumes its own answer. Otherwise you have to address the issue of the probability of God’s existence and which is more improbable: a self-created universe or a universal creator? ---
**Author:** ![begbert2](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/begbert2/32/3357_2.png) [@begbert2](https://boards.straightdope.com/u/begbert2)\ **Post date:** [May 24, 2007, 5:19pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/77 "2007-05-24T17:19:27Z")
[QUOTE=Lemur866] It shouldn’t be sub-atomic particles, it should be Planck length. In “the real world” anything smaller than Planck length is meaningless. > **[Planck units](https://en.wikipedia.org/wiki/Planck_length)** > > In particle physics and physical cosmology, Planck units are a system of units of measurement defined exclusively in terms of four universal physical constants: c, G, ħ, and kB. Expressing one of these physical constants in terms of Planck units yields a numerical value of 1. They are a system of natural units, defined using fundamental properties of nature (specifically, properties of free space) rather than properties of a chosen prototype object. Originally proposed in 1899 by German physicis... [/QUOTE] My read of that cite is that the Planck length is simply a limit on measurement, not on distance. After all, we’re not proposing that any particles are that close together, just that there’s infinite variation on how the object could have landed. It’s not as if the sphere lands on a neat tripod of the three lowest atoms, after all. It’s landing position is determined by the interaction of the various energy fields and forces at that level - which are continuous, I believe. So it could have landed a little less than a Planck length to the left of where it did, balanced on those same fields in a slightly different position. > [@Little Nemo](#): > > Leaving aside coins, the premise of the OP seems to be that God exists because the universe is too improbable to exist unless God intentionally created it. But the flaw with that argument is that it presumes its own answer. Otherwise you have to address the issue of the probability of God’s existence and which is more improbable: a self-created universe or a universal creator? Leaving aside coins, the OP _presumes_ the creation of the universe, life, sentience, etc to be infinitely improbable, without any actual reason to. I mean, sure, it doesn’t seem to happen every day on every street corner, but that doesn’t make it impossible or anything. Just rare. (At least around here.) This is a thought discussion about a strawman, essentially. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [May 24, 2007, 5:23pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/78 "2007-05-24T17:23:04Z")
Yeah, the God connection is not particularly well-conceived. I say we all just drop it and keep the discussion focused on the issues of probability in themselves. ---
**Author:** ![RTFirefly](https://avatars.discourse-cdn.com/v4/letter/r/c77e96/32.png) [@RTFirefly](https://boards.straightdope.com/u/RTFirefly)\ **Post date:** [May 24, 2007, 6:49pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/79 "2007-05-24T18:49:16Z")
[QUOTE=Kimstu] Yer absolutely right **Rufus** , sorry about that. I had a brain fart and calculated the cumulative probability of getting each card **given** that the previous ones had already been dealt, which isn’t the same thing. Thanks for the correction. So okay, if you were randomly dealing one bridge hand per second from the beginning of the universe to the present moment, you **would** almost certainly get any possible hand at some point. But not if you were dealing at the same rate for, say, a hundred years. Which in practical terms still makes it extremely improbable that any **particular** bridge hand should ever get dealt in the course of ordinary play—and yet, any particular hand that you encounter obviously **has** been dealt. So my analogy, although less gee-whizzical numerically than it originally appeared, basically still stands. [/QUOTE] Absolutely. If I got dealt a nondescript hand like the following: S: J 7 5 4 H: K 6 D: Q 10 3 C: Q J 9 7 As you say, I could exclaim, “look at this hand I just got dealt - did you know that the odds against getting this hand were 635 billion to 1 against?!” and I’d be absolutely correct. Everything that happens is extremely improbable at some level, but (with the possible exception of the Universe’s big-banging into existence in the first place) _something_’s gotta happen. So just that it’s extremely, extremely improbable for X to happen, doesn’t mean that a Designer had to be behind it. ---
**Author:** ![Taber](https://avatars.discourse-cdn.com/v4/letter/t/ac91a4/32.png) [@Taber](https://boards.straightdope.com/u/Taber)\ **Post date:** [May 24, 2007, 8:06pm UTC](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290/80 "2007-05-24T20:06:00Z")
[QUOTE=Shodan] You can’t flip a coin an infinite number of times. Regards, Shodan [/QUOTE] But could a thousand monkeys with an infinite number of coins in a slower growing infinite amount of time flip a coin an infinite number of times while typing hamlet an infinite number of times on three typewriters while humming the national anthem of Kazakhastan? [Previous page](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290.md?page=3) [Next page](https://boards.straightdope.com/t/how-many-times-do-you-have-to-flip-a-coin-to-ensure-a-heads/405290.md?page=5)