Can a coin with an unknown bias be treated as fair? This morning, I wanted to flip a coin to make a decision but only had an SD card:



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?
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 $\frac{1}{2}$.
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.