This morning, I wanted to flip a coin to make a decision but only had an SD card:

enter image description here

Given that I don't know the bias of this SD card, would flipping it be considered a "fair toss"?

I thought if I'm just as likely to assign an outcome to one side as to the other, then it must be a fair. But this also seems like a recasting of the original question; instead of asking whether the unknowing of the SD card's construction defines fairness, I'm asking if the unknowing of my own psychology (e.g. which side I'd choose for which outcome) defines fairness. Either way, I think I'm asking: What's the exact relationship between not knowing and "fairness"?

Additional thought: An SD card might be "fair" to me, but not at all fair to, say, a design engineer looking at the SD card's blueprint, who immediately sees that the chip is off-center from the flat plane. So it seems fairness even depends on the subjects to whom fairness matters. In a football game then, does an SD card remain "fair" as long as no design engineer is there to discern the object being tossed?

Andrew Cheong
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    I've got to be honest, this sounds more philosophy than maths, but I also reckon if you tried to 'prove' that this was fair, you'd probably somewhere along the way decide that you'd assigned the outcomes with fifty fifty probability to each side of the SD card, so you'd really flipped a coin in your head before assigning the sides so yeah, I guess it would be fair. – CameronJWhitehead Apr 02 '14 at 14:31
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    While fairness in probability is a mathematical concept, our idea of fairness is philosophical. – RghtHndSd Apr 02 '14 at 14:31
  • @CameronJWhitehead, well, there's a tag for it [here](http://math.stackexchange.com/tags/philosophy). And I doubt they would accept it in the respective SE, people seem rather "picky" on edge-case questions on smaller sites. – JMCF125 Apr 02 '14 at 19:39
  • I don't understand the problem. Flip it 1000 times. If you get roughly 50% of each outcome, it's probably about fair. Where does subjectivity and knowledge come into it? – Jack M Apr 02 '14 at 21:48
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    @JackM: The question is what you know _before_ flipping it. What probabilities should you assign, based on your (no) knowledge, _before_ you have any information from flipping it? The OP has already discounted the case where you already have some knowledge (e.g. from its design), let alone from flipping it 1000 times. – ShreevatsaR Apr 03 '14 at 02:21
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    Isn't probability directly related to our knowledge of a system anyway? Like, if you knew exactly how biased a coin was, exactly how it would be flipped, exactly where it would land, and all the forces that would act on it the entire way through the air, you would be able to predict with 100% accuracy what it would land as, even if it stops on its side. It becomes random because we don't know enough... right? – AlbeyAmakiir Apr 03 '14 at 04:05
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    It's definitely not fair after the first flip. – user2357112 Apr 03 '14 at 04:45
  • I think it depends entirely on how you assign the outcomes to the sides. If you are really just as likely to assign either outcome to either side, no need to flip at all - just take the one you've assigned to the side in the image, and both are equally likely. But how to achieve choosing without bias? Well, take a fair coin... – RemcoGerlich Apr 03 '14 at 10:31
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    @JackM Let me amend what you said there to fit the question. Take 1000 *different* people, each with their own coin that has some bias. None of them know the bias of their own coin, but are asked to call it and flip. In this case, you will get similar results as to if they all flipped a fair coin. Flipping the same coin 1000 times doesn't work, because now you're gaining knowledge, which is eliminating a key factor of the question. – Cruncher Apr 03 '14 at 13:08
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    @ShreevatsaR But the OP's concern seems to be that "fairness" isn't well-defined, and it is. I don't understand where *"If I'm just as likely to assign an outcome to one side of this SD card as to the other, then it must be a fair."* comes from at all. – Jack M Apr 03 '14 at 13:10
  • @AlbeyAmakiir, you forget about Quantum Mechanics. `:)` There, randomness appears not because we don't know enough, but because information is not yet determined, does not exist, until a measurement is taken. – JMCF125 Apr 03 '14 at 18:17
  • @JMCF125, Well, fair point. :P But the effect of quantum mechanics on the flip of a coin or SD card is minimal compared to the other factors I mentioned. I was referring to things of a higher level. – AlbeyAmakiir Apr 03 '14 at 21:24
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    Instead of asking if a possibly weighted sd-card can be used as a one-time fair coin, you should ask if you have a 50% chance of successfully guessing which side is the heavy side? They're almost the same question. – Mooing Duck Apr 03 '14 at 22:42
  • I posit that if you are "just as likely to decide to assign an outcome to one side of this SD card as to the other" then you don't need a coin. In order for that to be true, your brain must be capable of generating the equivalent of fair coin tosses. – Tim Seguine Apr 04 '14 at 13:19
  • The answer depends on what YOU (to me...) expect. If you expect 50/50 and the SD card is not weighted properly then NO. Otherwise YES. – capikaw Apr 04 '14 at 21:03
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    See http://lesswrong.com/lw/oj/probability_is_in_the_mind/ – ike Apr 07 '14 at 02:44
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    @JackM: (1) See the answer by josinalvo below. (2) "If I'm just as likely to assign an outcome to one side of this SD card as to the other, then [the process is fair]" is clearly correct, though in that case the fairness comes from the just-as-likely assignment of outcomes to the two sides. – ShreevatsaR Apr 07 '14 at 11:39
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    What kind of world we live in... Someone wants to flip a coin, but they only find SD cards... – qqbenq Apr 07 '14 at 14:04
  • I'm surprised no one has been tempted enough to flip an SD card 100 times and report the results. – Witness Protection ID 44583292 Apr 08 '14 at 05:05
  • @qqbenq Well, a spun penny is [biased towards tails](http://web.archive.org/web/20080314023237/http://www.sciencenews.org/articles/20040228/fob2.asp). – Cole Tobin Apr 08 '14 at 05:18
  • Mathematically it should work out to be like a fair coin toss. Whether or not this is actually true physically is another question. – user60887 Apr 09 '14 at 03:16
  • @user2357112 Well, if the sd card is actually fair, then it still is fair after the first flip. But you can't determine fairness based on lack of knowledge anymore. – Cruncher Aug 13 '14 at 17:27
  • Even if the SD card always fell on the same side, the chance of you picking that side is still $\frac 1 2$ because you don't know which side that is. XORing non-random data with random data (in this case 0 or 1 depending on the side landed on/picked) will produce random data. –  Jun 22 '17 at 14:53
  • There many models where probabilities are intrumentalist or what one projects onto the world (projectivist bayesian) which is just the agents opinion. There are other accounts where are more objectivists which nonetheless have an stranger interaction with knowledge where chance vlaues pre-exist and so does reality and where your knowledge or perception literally causally interacts with the external objective world making or changing the outcome or chance in an objective ontic fashion Much like von neummans approach in the end to quantum mechanics – William Balthes Jul 07 '17 at 07:27

25 Answers25


Here's a pragmatic answer from an engineer. You can always get a fair 50/50 outcome with any "coin" (or SD card, or what have you), without having to know whether it is biased, or how biased it is:

  • Flip the coin twice.
  • If you get $HH$ or $TT$, discard the trial and repeat.
  • If you get $HT$, decide $H$.
  • If you get $TH$, decide $T$.

The only conditions are that (i) the coin is not completely biased (i.e., $\Pr(H)\neq 0, \Pr(T)\neq 0$), and (ii) the bias does not change from trial to trial.

The procedure works because whatever the bias is (say $\Pr(H)=p$, $\Pr(T)=1-p$), the probabilties of getting $HT$ and $TH$ are the same: $p(1-p)$. Since the other outcomes are discarded, $HT$ and $TH$ each occur with probability $\frac{1}{2}$.

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    What you call the engineer's approach is an example explained in tons of "theoretical" textbooks. – Did Apr 02 '14 at 15:18
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    He wanted to flip a fair coin, and MGA gave him a way to do it, whether the SD card is fair or not. That's +1 in my book. Note the method is from John von Neumann. – soakley Apr 02 '14 at 17:57
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    @soakley no, he wanted to *know* if what he is doing is fair. Not how to do it fairly. At least by my reckoning. – AncientSwordRage Apr 02 '14 at 18:25
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    @soakley: I think he originally wanted to flip a fair coin, but he must have figured it out long ago (possibly by getting an actual coin) — the question actually asked here is about how we can know whether something is fair. – ShreevatsaR Apr 02 '14 at 18:38
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    Would someone elaborate on why this approach works? – Vilhelm Gray Apr 02 '14 at 18:48
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    @VilhelmGray If H has probability $p$ and T has probability $p-1$, then $P(HT) = p(p-1)$ and $P(TH) = (p-1)p$. These are not 50%, but they are equal to eachother. Since you reflip everything else, it scales their probability up to 50% each. – Cruncher Apr 02 '14 at 19:28
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    @Cruncher I believe you meant that T has probability $1-p$? Same conclusion, of course. – Aaron Dufour Apr 02 '14 at 19:35
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    @AaronDufour Oh dear, how did I not see that. Too late to edit my comment. – Cruncher Apr 02 '14 at 19:40
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    This is one of my favorite questions from Cover and Joy's info theory text book :) – bright-star Apr 03 '14 at 01:41
  • This works as long as 0>p>1. If p somehow ends up to be 1 or 0, it no longer works.. – Rob Audenaerde Apr 04 '14 at 10:52
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    @RobAu Hence the "not completely biased" condition. – MGA Apr 04 '14 at 10:53
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    Actually this is good even if you do have a coin, since as far as I know most coins are not fair coins. Just out of curiosity is there a version that would work for dice? – Tim Seguine Apr 04 '14 at 12:41
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    Technically, yes, this can be generalized to an $N$-state discrete random variable, but note that as $N$ increases, you might grow old before making a decision. The idea is to assign the decisions to the sequences of length $N$ that contain each of the possible outcomes exactly once, and discard all the other sequences. For a 3-sided die, you could map the outcomes $ABC$ and $ACB$ onto Decision 1, $BAC$ and $BCA$ onto Decision 2 and $CAB$ and $CBA$ onto Decision 3. The conditions of no zero probabilities and of non-changing probabilities still apply. – MGA Apr 04 '14 at 13:01
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    @MGA I just did a back of the envelope calculation on that. In order to have a 95% probability of successfully generating just one guaranteed fair roll of a 20 sided die, you would have to roll it over two and a half billion times. – Tim Seguine Apr 04 '14 at 13:14
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    This is pragmatic, but not an answer. I an very curious in the answer to the OP's actual question. – jnm2 Apr 04 '14 at 17:08
  • @TimSeguine I think that the detailed analysis of this general process deserves its own question. Shall I post one so that you can supply your answer? – MGA Apr 04 '14 at 18:27
  • @MGA if I decide that I feel like writing it up, I will do an "answer your own question" post on it linking to your answer here. I think the expected number of rolls would be interesting to see as well, but I didn't do that calculation. – Tim Seguine Apr 05 '14 at 15:13
  • @Pureferret He wanted to know if what he was doing was fair, and this answer says "maybe, maybe not, but here's how to be sure:". – OJFord Apr 05 '14 at 23:17
  • +1,even if this is not a complete answer. This is very practical, and I prefer your engineer's approach in comparison to @Did's. – Ganesh Apr 07 '14 at 18:02
  • Wait, this is seriously a famous solution? I made this up on the spot in an interview once :D – Davor Apr 08 '14 at 10:15
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    Famous != difficult to come up with – Andrew C Apr 08 '14 at 22:49
  • This doesn't work for all biases(fails with bias 100% and 0%). And all biases close to that, make this impractical. The fun part is that the practical cases aren't likely to fall into these categories. – Cruncher Jul 14 '14 at 15:13
  • Can someone explain why condition (ii) needs to be met? – Stephen Dec 23 '14 at 23:11
  • @Lucas Trivial case, if the bias changes then it could be 99.99% probability of heads first time and %00.01 probability of heads second time. Then the probability would be at least 99.98% that we choose heads as our result, which isn't very fair. – gmatht Apr 17 '17 at 15:14

That is a very good question!

There are (at least) two different ways to define probability: as a measure of frequencies, and as a measure of (subjective) knowlegde of the result.

The frequentist definition would be: the probability of the sd card landing "heads" is the proportion of times it lands "heads", if you toss it many times (details ommited partially because of ignorance: what do we mean by 'many'?)

The "knowledge" approach (usually called bayesian) is harder to define. It asks how likely you (given your information) think an outcome is. As you have no information about the construction of the sd card, you might think both sides are equally likely to appear.

In more concrete terms, say I offer you a bet: I give you one dollar if 'heads', and you give me one if 'tails'. If we both are ignorant about the sd card, then, for us both, the bet sounds neither good nor bad. In a sense, it is a fair bet.

Notice that the bayesian approach defines more probabilities that the frequentist. I can, say, talk about the probability that black holes exist. Well, either they do, or they don't, but that does not mean there are no bets I would consider advantageous on the matter: If you offer me a million dollars versus one dollar, saying that they exist, I might take that bet (and that would 'imply' that I consider the probability that they don't exist to be bigger than 1 millionth).

Now, the question of fairness: if no one knows anything about the sd card, I would call your sd card toss fair. In a very meaningfull way: neither of the teams, given a side, would have reason to prefer the other side. However, obviously, it has practical drawbacks: a team might figure something out latter on, and come to complain about it. (that is: back when they chose a side, their knowledge did not allow them to distinguish the sides. Now, it does)

It the end: there is not one definition of probability that is 100% accepted. Hence, there is no definition of fair that is 100% accepted.


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  • (not very knowleable on the subject matter. Imprecisions may be present =P) – josinalvo Apr 02 '14 at 18:27
  • Is this answering the OP's question? – Did Apr 02 '14 at 19:23
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    Given that the OP wrote 'In either case, the core of my question appears to be: Does not knowing contribute to "fairness"?' and this is the only question that mentions the word "Bayesian", this probably (pun intended) is the only answer that answers the question. – JiK Apr 02 '14 at 22:31
  • "that would 'imply' that I consider the probability that they don't exist to be bigger than 1 millionth" -- the existence of lotteries proves that the utility of a bet is not equal to the expected cash prize (cash for winning multiplied by the probability of winning). So the quotes around 'imply' are certainly in order even if we we talking frequentist probabilities :-) – Steve Jessop Apr 03 '14 at 09:20
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    this is the answer that should be accepted! It is the only that actually answer the question. :-) – Ant Apr 06 '14 at 13:47
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    @JiK and josinalvo, a tiny quibble: I believe that what josinalvo calls "the 'knowledge' approach" to probablility in this answer is generally called the *subjectivist* approach (see, for example, http://plato.stanford.edu/entries/probability-interpret/). I'm not so clear on how the word "Bayesian" is typically used, but to me it means using Bayes's law to update probabilities in the face of new evidence. You can be a subjectivist without doing Bayesian updating (see http://en.wikipedia.org/wiki/Bayesian_inference#Bayesian_updating for a few references), and also perhaps vice versa. – Vectornaut Aug 18 '14 at 07:20
  • After months of monitoring answers and comments, I feel that this post offered the most direct answer to my question (as opposed to related issues, like _how_ to achieve a fair toss with a biased coin), and pointed me to further reading that was very relevant and enlightening. Still, I'm regretful that I must accept only one answer, when so many have contributed to shaping my understanding. My gratitude goes out to all who pondered and posted. – Andrew Cheong Sep 09 '14 at 22:17
  • I'd also like to recognize that \@CameronJWhitehead, \@RghtHndSd, @ike (especially), \@AlbeyAmakiir, and others did comment on the philosophical / frequentist vs. Bayesian nature of this problem. Had they expounded on their comments in an answer I might very well have accepted one of their posts, but alas. – Andrew Cheong Sep 09 '14 at 22:25

The SD card is biased (or not), that is, it falls on face A with probability $p$ and on face B with probability $1-p$, for some unknown $p$. If it is "just as likely to assign an outcome to one side of this SD card as to the other", this means that, either one is faithful to the SD card, that is, one answers the result produced by the SD card, or one is unfaithful to the SD card, that is, one answers the result opposite to the result produced by the SD card.

If faithfulness or unfaithfulness are decided independently of the result of the SD card and with equal probabilities, then the final result is fair, irrespectively of the value of $p$.

Note that, if the goal is to answer A or B with equal probabilities, one could as well forget the SD card altogether and simply rely on the perfect source of randomness that was used to choose between faithfulness and unfaithfulness, that is, to encode the result produced by the SD card. Note also that such perfect sources of randomness do not exist in real life (but that very good approximations do).

Is this mathematical factlet the source of your puzzlement?

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    I think for this to be complete you need to comment on whether or not, him not knowing and arbitrarily picking **is** a true source of randomness. And as I argued, I think it is. – Cruncher Apr 03 '14 at 13:05
  • @Cruncher Thanks for your comment. The answer **is** complete. – Did Apr 03 '14 at 15:37
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    I like this answer, but... it seems like, psychologically speaking, "faithfulness" is implemented as, "Do I map the logo-side of the SD card to Heads?" Generally I doubt that this decision will be made anything like randomly. The same issue would apply to any object found sitting on a table, with a presumed bias to "face-up" being heads. – kyle Apr 07 '14 at 21:22
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    (in theory, quantum mechanics should give us perfect randomness.. nitpicking of course :) .) – hadsed Apr 15 '14 at 00:20
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    @hadsed, you are not nitpicking, you are rightly pointing out that the statement "perfect sources of randomness do not exist in real life" is false. – Garrett Jun 12 '14 at 19:44
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    @Garrett Ha. Well... – Did Jun 12 '14 at 19:52
  • @Garrett Now that things have had the time to cool down, would you care to explain your comment above? If you are aware of any physical source of randomness proven to be pure, I am interested. – Did Jul 27 '18 at 16:23
  • @Did, I was referring to one of the most interesting phenomena that I learned in my undergrad physics: in quantum, probability is a part of reality, rather than just being a formalization of ignorance ([more](https://en.wikipedia.org/wiki/Quantum_indeterminacy)). Rereading my comment with some extra maturity, however, I can see it might look argumentative, but that wasn't my intention. I find myself bringing up this point whenever I can because I love sharing / discussing it with people / seeing their reaction. Just that now, I'm slightly older and slightly more diplomatic. – Garrett Jul 27 '18 at 20:21
  • @Garrett You seem to be referring to a (very nice) theoretical principle rather than to some actual practical device producing sequences as long as one wants of (demonstrably) purely random bits. – Did Jul 27 '18 at 21:28
  • @Did, ah, if you are asking for some kind of mass-market, affordable device for generating "true randomness", I don't know of any. – Garrett Jul 28 '18 at 17:55
  • @Garrett Empty comment ("mass-market" and "affordable" are of course quite offtopic here). – Did Jul 28 '18 at 18:23
  • @Did, Do you know if normal numbers would suffice as an object producing pure randomness? I would think so, but I am not in a position to make an argument for or against the case. I thought there were proofs that certain numbers are normal but I am not aware if they were constructive proofs (and so maybe only approximations exit). –  Oct 28 '19 at 22:23

You could say that there are two unknowns when you flip something: the probability to get side $A$ and the actual result.

So, say you have $3$ coins and $p$ is the probability you get heads:
Coin $A$ is fair ($p=0.5$)
Coin $B$ always gives heads ($p=1$)
Coin $C$ always gives tails ($p=0$)

Obviously, the only fair coin is $A$. But if you don't know which coin you have, say you pick one randomly, then the result is also fair:
$\frac13$ chance to get $A$, then $\frac12$ chance to get heads
$\frac13$ chance to get $B$, you always get heads
$\frac13$ chance to get $C$, you'll never get heads

Therefore, the chance to get heads is $$p=\frac13\times\frac12 + \frac13\times1 + \frac13\times0 = 0.5$$

So, if we assume that the distribution of the chance to get heads of the set of objects you might pick to flip is even, then you can say that the outcome of the SD card is fair as long as you do not, consciously or unconsciously, know or estimate the probability of the card before you select it.

In other words, you rely not on the randomness of the flip but on the randomness of the selection of the object.

Of course, this method won't work for repeated flips and it generates more trust issues compared to a plain coin when more people are involved.

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  • While I like the ideas this answer brings (+1), I cannot see it generalizing well to any pseudo-coins and reality... – JMCF125 Apr 02 '14 at 19:54
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    @JMCF125 hm, I wonder if it could be improved by considering the randomness of the assignment; if everything landed on one side but you didn't know which, your selection of which side corresponds to each choice would create a fair coin *for one flip* (thanks for the formating btw!) – Thanos Tintinidis Apr 03 '14 at 10:30
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    That equation could really use some brackets – Level River St Apr 04 '14 at 23:20

My look at it is this. You are moving probability from one place to the other.

Assume side 1 has probability $p$, and side 2 $1-p$.

If you were to use a fair coin to determine which side of the SD card you get, then you end up with $0.5p + 0.5(1-p)$ of flipping the side you picked(dictated by the coin). This happens to be exactly $0.5$.
EDIT: As this seems to have caused a little bit of confusion, what I mean is if you flip tails on the fair coin, then you'll call tails on the SD card, and the same for heads. This is the probability of flipping the same thing on the coin and SD card.

So the question at the end of the day is, how likely is your choice of side to be as fair as a fair coin toss? Assuming you know NO information, it should be an arbitrary decision, and therefore fair.

The problem is when you try to convince the person you're flipping against that you didn't know which side is better.

In a football game, then, does an SD card become "fair" if none of its viewers can know or imply anything about the construction of the SD card (whether in the stadium or on TV, however improbable)?

In this case actually, I would argue that even if everyone else knew how unfair the coin is, as long as the person calling it has no information about it(you have to assume here that no one hinted it and that he really doesn't know) then it is fair. Again, the same problem as I stated before though. It becomes difficult when you try to convince everyone that he really didn't know.

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  • No, if you only use the fair coin, then you have `0.5*1 + 0.5*1` (it's like flipping the SD card in a very unfair way that it determines the outcome with certainty). In contrast, `0.5p + 0.5(1 - p)` is by definition the probability that, flipping both the fair coin and the SD card, you get `(T on the fair coin and T on the SD card) or (H on the fair coin and H on the SD card)`. – ignis Apr 03 '14 at 07:31
  • @ignis I don't understand what you're saying in your first statement. In your second statement, you just stated what I said in my answer in a different way. Could you elaborate your objection please? – Cruncher Apr 03 '14 at 12:39
  • @ignis I think you misunderstood what I meant. When I said "If you were to use a fair coin to determine which side of the SD card you get" this is essentially saying the coin flip is calling the flip for the SD card. As is, if I get tails on the coin, I'll call tails on the SD card. Which is of course, exactly what you said in your comment. – Cruncher Apr 03 '14 at 12:42
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    Ok, I misunderstood your answer. Thanks for clarifying, +1 – ignis Apr 04 '14 at 06:28
  • Even if the SD card comes up as "heads" 2/3rds of the time, you only have a 50-50 chance of picking the favorable side, assuming you have no data to nudge you toward the correct choice. I think this answer helps answer the O.P.'s question: *Does not knowing contribute to "fairness"?* – J.R. Apr 07 '14 at 01:31

Imagine taking into account a "fair" coin's initial angular momentum, its initial velocity, initial orientation, the nearby air movements, etc. Assume we can disregard quantum effects and solve for the rest position of the coin in terms of purely classical physics. Assume you'll immediately find and correct any mistakes in your computation.

You should expect to determine perfectly the final state of the coin. It seems that the fairness of a "fair" coin is largely dependent on your state of partial information before the flip. In fact, attempting to ascribe fairness to a coin seems to be a case of mind-projection fallacy.

To be at least a little formal, you can visualize this as the error of defining your fairness function as


when instead it should be

fairness(coin, mind)

So, I recommend getting comfortable with the idea of relative fairness. I'm not sure I understand your objection to this idea based on fairness in hindsight. The fairness(coin, mind0) need not equal fairness(coin, mind1), where mind1 is the future state of the mind and thus has the advantage of hindsight.

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    Your first paragraph may be *theoretically* true, but it still seems to be possible to define the fairness of a coin objectively, by considering ensembles. There is a configuration space describing the initial conditions of the problem. I can define 'local fairness' in a region as the fraction of configuration space volumes leading to T and to F. Like most dynamical systems, a coin will have a large enough region where initial conditions leading to T/F outcomes are strongly mixed, and 'local fairness' will be approximately constant. The closer it is to 1, the fairer the coin (by definition). – Anton Tykhyy Apr 02 '14 at 20:24

Yes, it's fair for the first time you use it, then it won't be fair anymore.

If you don't know the possible outcome of something, and randomly map the outcomes of the thing to some result, the first time you use it it'll be fair (since the outcome was unknown by you). After the first time you use it, you've learnt something from it, and that can bias your output.

Another example: take a deck of cards, without looking to the face of any card, shuffle it, and consider that red will be head, black will be tail. You don't need to know if the deck was complete (all 52 cards on it), if it only had black cards, and so on.

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The whole idea of a biased coin is basically a convenient fiction useful for making probability examples. It's not really possible to make an object with a coin like aspect ratio significantly biased when tossed. It can be biased when spun (EDIT: or I guess if allowed to bounce). So yes it's actually very close to fair, not just due to your ignorance of it's properties.

Reference: here

Kai Sikorski
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    Not true. Take two coins of the same size and differing weights and glue them together. Their centre of gravity will be displaced off the division line. – AncientSwordRage Apr 02 '14 at 16:23
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    @Pureferret http://www.stat.columbia.edu/~gelman/research/published/diceRev2.pdf – Kai Sikorski Apr 02 '14 at 16:26
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    @Pureferret "Jaynes (1996) explained why weighting the coin has no effect here (unless, of course, the coin is so light that it floats like a feather): a lopsided coin spins around an axis that passes through its center of gravity, and although the axis does not go through the geometrical center of the coin, there is no difference in the way the biased and symmetric coins spin about their axes." You should further justify your objection or remove your negative feedback.. – Kai Sikorski Apr 02 '14 at 16:34
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    I cant remove the downvotes until you edit the answer. Regardless this does not answer the question, it merely circumvents it. – AncientSwordRage Apr 02 '14 at 16:46
  • @Pureferret That's a pretty pedantic criticism. I think my answer provides a pertinent and interesting piece of information that obviously many people are unaware of. – Kai Sikorski Apr 02 '14 at 16:49
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    Pertinent and interesting does not an answer make. – AncientSwordRage Apr 02 '14 at 16:52
  • @Pureferret ...? do what you want. I'll just go back to reading and learning about interesting things. – Kai Sikorski Apr 02 '14 at 16:52
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    That's fine. The question as I'm lead to believe however is 'is something fair if i don't know it's not' your answer is 'it's fair'. That doesn't answer the question, just your interpretation. – AncientSwordRage Apr 02 '14 at 16:58
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    @Pureferret question is based on a false premise; why wouldn't one point that out. There are already multiple answers dealing with the other aspect of the question. If you feel that the fact I'm circumventing the question is the so critical to your feedback maybe you should have raised the issue in your original comment instead of attacking my answer based on an intuitive and wrong understanding of the physics of coin flipping? Feel free to get the last word in; I'm done. – Kai Sikorski Apr 02 '14 at 17:05
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    Sure - last word: the critical part about answering questions is spotting which parts are crucial to answering the questioning and which parts aren't. In this case, coin could be replaced with dice, PRNG or pulling socks from a drawer. The medium of randomness (coin) is not a critical part of the question, especially considering the question was about SD cards. – AncientSwordRage Apr 02 '14 at 17:10
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    I didn't downvote this, and I thank you for this information and the reference (I've upvoted your comments), but this is not an _answer_ to the question; it's only (as you admit) "a pertinent and interesting piece of information that obviously many people are unaware of". The answer area is for actual answers to the question; you could post this as a comment on the question. – ShreevatsaR Apr 02 '14 at 18:00
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    Interesting information, but by reference to your source only useful if an SD card is more like a coin than it is like a pickle-jar lid. That is, does it have any aerodynamic features that differ substantially from a flat disc? The fact that it's a *disk* is not alone sufficient ;-) – Steve Jessop Apr 03 '14 at 08:53
  • @ShreevatsaR: At least on other SE network sites it's accepted to use the answer area for comments which are either long, formatted, or contain images. Provided, of course, that one begins by "I would have made this a comment but couldn't because _____" – Ben Voigt Apr 03 '14 at 22:49
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    @BenVoigt: The comment must at least tangentially attempt to address the question. The question here is about philosophy / interpretation of probability -- the OP explicitly discounts the effect of further information, like a design engineer knowing the physics of the card. On top of that, this "answer" is a factoid about coins, and it does not appear from the paper that it applies to an object like an SD card. Even if it did, this is like answering, say, a combinatorics question about counting seating arrangements with some trivia that sidesteps the question. :-) – ShreevatsaR Apr 04 '14 at 06:43
  • I'm sure you're all right in regards to the proper placement and i will abide by it in the future. Just found the other commenters, down vote, factually wrong attack on my comment and subsequent disregard for his mistake rude so i got a little carried away with being defensive. – Kai Sikorski Apr 04 '14 at 06:47

I would say that the SD card itself is not a 'fair coin', since a reasonable requirement of a 'fair coin' would be that the probability of landing on each side is essentially 0.5 each.

However, I think the entire process as a whole of assigning an outcome to each side of the card and then flipping the card is a 'fair coin':

The reason why we need coins to make trivial decisions for us is that we can't just cycle through the 2 options in our head, stop at a random point, and choose the option we were thinking of at that point. Not only is this hard to pull off, but you can't be sure that you didn't subconsciously/consciously want to stop on a particular option.

Since the card decides the final result and we don't know the bias of the card beforehand, there is no way that your conscious/subconscious desires would have any way of manifesting themselves when you label each side of the card, and so you label each side essentially randomly, resulting in a 50/50 probability of your two options.

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If the conceptual goal of a fair coin is to provide equal chances of two outcomes, then the problem you face is twofold:

  1. Is the SD biased?

  2. Can you fairly assign the real-world outcomes to the sides of the SD card?

My hunch is that (1) doesn't matter as long as you can achieve (2), but (2) essentially requires a "fair coin", which you don't have.

Rob Starling
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"If I'm just as likely to assign an outcome to one side of this SD card as to the other, then it must be a fair."

I think you're right with the reasoning. However, the SD card has a definite "front/back", which might be associated with "right/wrong" or "good/bad" in your head. I don't know how strong your bias is, but I would tend to assign the thing I want to the "front" because it's the "good" side. Then if the card is biased, which seems likely, your bias will bleed through to the outcome. I don't think it matters which direction the card is biased, because somebody else who has knowledge of this would recognize the flip as unfair, so it must be unfair on some level.

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There are various models of probability that mathematicians may use, such as epistemic, frequentist, or the trivial deterministic model where we assume the future is fixed. From a pure maths point of view picking any of those models is fine, so long as you don't mix models.

For example, I really don't know what the 99th digit of $\pi$ is. So I can say that my epistemic probability that it is 1 is $\frac{1}{10}$. For either a frequentist or deterministic perspective it is clearly 1 or 0, so if we assume that these probabilities are the same we can easily prove that $\frac{1}{10}$ equals either 0 or 1 and then it is easy to show 0=1.

As to which model is "correct", that depends on what you are trying to achieve. If I am trying to make a decision without prejudice I'd be content to pick digits of $\pi$ that I do not yet know. Once I learn the actual digit and my epistemic probability collapses to 0 or 1 it doesn't matter as I have already committed to my decision and moved on. On the other hand, it may be unwise to choose passwords based on digits of $\pi$, even if no-one has yet computed those particular digits. Likewise when Debian chose a random key generator that no-one knew was biased let alone how it was biased it was a disaster for security, despite all possible keys having the same epistemic probability at the time.

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As an aside, related to the question but not actually answering it: even if you assign sides fairly and randomly, and keep the assigned sides after the first toss, you still can't do multiple flips and still get the same outcome as a fair coin:

Suppose the probability of a "head" is $p$, and the probability of assigning event $A$ to heads is $1/2$. Then:

$P(HH) = p^2$

$P(TT) = (1-p)^2$

so $P(AA) = P(HH)/2 + P(TT)/2 = (1 - 2p + 2p^2)/2 = 1/4 + (p-1/2)^2$

If the "coin" is fair, then the probability of getting $A$ twice is $1/4$. But if the "coin" is not fair then this probability increases, up to $1/2$ if the "coin" only ever lands one way up.

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The card must be fair in both cases, or in no one, because its behavior does not depend on the observer's knowledge:

  • Case 1: an engineer knows whether the card is fair, flips it and tells you the result,
  • Case 2: you do not know whether the card is fair, and flip it.

Therefore, do not assume that the card is fair in case 2. You need a proof that it is fair.

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I'll give it another try:

Assume that the SD card falls on the back with probability $p$ and on the front with probability $1-p$. Further assume that you decide for option $A$ if the SD card falls on its back with probability $q$, and for $A$ with probability $1-q$ if the card falls on its front, and for option $B$ in the other cases. Then $$ P(A) = pq + (1-p)(1-q) = 2pq - p - q + 1.$$

Now if $q=.5$, i.e. you assign option $A$ fairly to one of the card's sides, we get $$ P(A) = p-p-.5+1 = .5, $$

a "fair coin". Does this make sense or answer your question?

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TL;DR: carry a true random number generator with known distribution on your person at all times[*].

If there's no correlation between your assignment of outcomes to faces of the SD card, and the SD card's bias, then yes, it doesn't matter whether the SD card is biassed. Saying "no correlation" means it's as if the outcomes were assigned to the faces of the SD card by an unbiassed random choice. So there's your randomness, regardless of what the SD card does. It could be 100% biassed as long as you assign the outcomes to faces randomly. This isn't disturbing :-)

Naively one can argue that if we're capable of assigning outcomes to faces "as if randomly" then we could just choose an outcome "as if randomly". Not so, of course, the point is that any knowledge we have of which outcome will actually happen affects our assignment. For an unbiassed coin flip we cannot have any such knowledge. For a biassed coin flip we're concerned that we might unknowingly act on any information or prejudice we do have. Note that even if we're in some sense trying to get a particular outcome and our information/prejudice is wrong, we still create an unfair choice. It's unfair in the opposite direction to the one we want, but still unfair. That's enough to lose at rock-paper-scissors, and surely the ultimate goal of all probability and game theory is to break even at rock-paper-scissors?

So, you could argue against the existence of that correlation in two ways:

  1. Experiment. Observe whether you in particular, or people in general, seem to create any bias of outcomes.
  2. Theory. State that since you have no knowledge of which side of the card is favoured, you cannot possibly correlate that to the choice of outcomes.

Experiment 1 seems overkill for this particular case, since the actual outcomes matter -- there might be some outcomes that your brain is capable of assigning in an unbiassed way and others which it is not. A more general such study might be good. Besides, if you could set that up you could probably just fetch a coin.

Theory 2 disappears as soon as you have used the SD card once. At that point you have some information (admittedly not a lot to start with) about its possible bias, so you can no longer claim to be ignorant.

In fact I think argument 2 is dubious anyway. Aside from anything else it's possible that you are subconsciously able to observe from the SD card some feature that (without you realising it) contributes to its bias when flipped. Furthermore that could affect your assignment of outcomes.

If you couldn't see the SD card then you'd be on firmer ground. Suppose that a third party coloured the faces of the SD card "red" and "blue", and you assigned the outcomes to "red" and "blue" before seeing the card. Then my proposed connection between flip bias and outcome bias needs two correlations:

  • you need to assign the outcomes to "red" and "blue" based on some intuition of which is more likely
  • the third party needs to assign "red" and "blue" to the faces based on some intuition of which face is more likely.

If either of those fails, then the choice of outcome becomes fair.

It starts to sound implausible both those things would hold, but the fact I'm even considering it is because I consider the original scenario plausible, i.e. your original process is not certainly free from bias.

For the football game where the coin is later discovered to be biassed, then provided everybody accepts that there is no way that bias could have affected the call or the person making the toss, I don't think there's a problem. But probably in practice everyone would accept that on somewhat shaky grounds. Good enough for football isn't necessarily good enough for mathematical theory.

[*] von Neumann's construction of a known distribution from an unknown distribution is fine, provided that you can't somehow (intentionally or unintentionally) influence the outcome of the second toss based on the result of the first. You can achieve that in the first round by using a camera, and only examining the result of the first toss after making the second. However if you get a pair and go to a subsequent round then you risk having some information about how to effect that result, which could alter your flipping action. Basically when you don't trust your own subconscious influences you're in deep trouble.

Steve Jessop
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If you toss the SD card, and hide the result.

Then, not knowing anything (nor the result, nor the probability of each side), you pick one side. We can then assume you had a 50% chance of picking a side — as you can not distinguish which side has greater probability.

You then uncover the SD card.

This way, the probability of each side is not taken into account, so if this manner of doing (first tossing, then picking unknowingly) is the same as first picking, then tossing, can we say the toss was fair ?

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you have a chance of $p_1$ that you choose a certain side, when you flip it there is a chance $p_2$ that that side comes up

this means that you have a chance of $p_1*p_2+(1-p_1)*(1-p_2) = 2*p_1*p_2-p_1-p_2+1$ to win and $p_1*(1-p_2)+(1-p_1)*p_2=p_1+p_2-2*p_1*p_2$ that you lose.

if neither is a perfect 50% (you are biased to picking a side and the card is biased to land on a certain side) then there will be a bias in the outcome.

ratchet freak
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It COULD be a fair toss. Prior knowledge of anything but adequate testing is of no help to you. You just cannot establish fairness without testing. It MAY be determined to be fair in the future - AFTER testing. Not knowing doesnt make it fair. Any deviation from determining fairness by testing is pointless. You could only convince someone who didnt understand probablity, and that would only be through trickery.

However, you originally state that you wanted to make a decision, then got distracted by the fairness of the method you chose. That's just procrastination. a fair item for a "coin toss" is not required to make a decision, because the whole point of tossing in that situation is to be aware of your desired outcome before the item shows you the result. In your situation, making a decision and determining the fairness of tossing an SD card are conflicting desires,


If there was no correlation between your (conscious and unconscious) knowledge of the SD card (which includes knowledge of any previous flips) and your choice, then it would be fair. However, this will not be the case.

The unknowing of your own psychology is not enough to eliminate the bias, no matter if the bias is negative or positive.

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I feel there are some disturbing implications if the above were true, but I'm unable to convince myself it definitely isn't true.

Why? Is it because you think that the SD card is obviously not balanced, so it cannot possibly be fair? But you said:

Given that I don't know the bias of this SD card, would flipping it be considered a "fair toss"?

If you don't know anything about the bias, you don't know. If you think it might not be balanced because of the asymmetrical structure, then you do know something. You don't necessarily know how it's biased, but you know that it might be biased - and fail to yield a 50/50 distribution.

Now we have really moved away from mathematical analysis of an ideal fair coin, and moved on to questioning whether a given real coin (in this case the SD card is taken to be a sort of "coin") is going to behave like the ideal coin.

This problem can be solved by applying hypothesis testing. Your null hypothesis is:

$H_0$: The coin is unbiased, and when flipped a sufficiently large number of times, will come up with one side up 50% of the time, and the other side up 50% of the time.

Then you perform an experiment. Let's say you flip it 52 times, and it comes up with the front up 36 times (69%) and the back up 16 times (31%). Now you must decide, does the null hypothesis still stand? Is it conceivable for such a difference to be generated by chance?

More cogently, you want to ask - what is the probability of this result if the null hypothesis were true? In this case, you can obtain the probability (called a p-value) from the binomial distribution. If the probability satisfies you, then you have no reason to suspect bias in your coin. If it doesn't satisfy you, then it is clearly biased.

You must make a judgement call on what probability is acceptable. Most people use 0.05 (1 in 20 chance).

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Why have we over-complicated this? Assuming neither of you knows the bias that the SD card has as to which side it will most likely land on, then you both have an equal chance to pick either "heads or tails," therefore making the toss fair...

Joe Harper
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As has been pointed out, there are 2 events involved: the decision as to which side of the SD card to associate with which course of action and the flipping event which determines which side lands upwards.

It's normal for those two events to occur in that order. But not necessary: an umpire may toss a coin and conceal the result while the opposing captains call heads or tails. In that event, the outcome of the coin toss is certain at the point at which the heads/tails call is made, but the toss is still considered fair.

Now suppose that the SD card is in fact certain to come down on one particular side (doesn't matter which). If you don't know this in advance, then by analogy with the above, the test is still fair as you don't know that in advance. If it is fair in this case, then it is fair for all values of p


The reality is that few coins are actually truly 50/50. If the center of gravity of the coin does not lie perfectly between the landing surfaces of the two sides then there will be some slight bias toward one side. Differences in the images imprinted on the two sides of a traditional coin normally cause the center of gravity to be closer to one side or the other.

That said, as long as there is no way to discern with the eye which side is more likely to land facing up, then the first flip will always be fair if one person is 'calling' a side for the win. It's the same as if the winner was determined by one person writing either a 1 or a 0 on a piece of paper and folding it up and the other person guessing if the number is a 1 or a 0. There is 100% bias toward the 1 or the 0 after it has been written down but until that bias is known by the person choosing the number the odds are 50/50. The SD card may land with a certain side up 100% of the time but until that bias is known the odds are 50/50.

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By definition, an unbiased coin is fair. So, I suspect the answer is "don't know".

Let $F=${"the coin is fair"} and $B=${"the coin is biased"} and $P\left ( B\cup \bar{B} \right )=P\left ( B \right ) + P\left ( \bar{B} \right ) = 1$ (obvious). Then $$P\left ( F | B \cup \bar{B} \right ) = \frac{P\left ( F\cap \left ( B \cup \bar{B} \right ) \right )}{P\left ( B\cup \bar{B} \right )}=P\left ( F\cap B \right ) + P\left ( F \cap \bar{B} \right )$$ But, total probability says: $$P\left ( F \right )=P\left ( F| B \right )\cdot P(B) + P(F|\bar{B})\cdot P(\bar{B}) = P(F\cap B) + P(F\cap \bar{B})$$ So $$P\left ( F | B \cup \bar{B} \right ) = P(F)$$ and $B \cup \bar{B}=${"the coin is either biased or not"} (sort of "don't know"). Or, as other people suggested, the fact that we don't know if the coin is biased or not, has no value for determining if the coin is fair or not.

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