Can a coin with an unknown bias be treated as fair?
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 $\frac{1}{2}$.
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.
http://en.wikipedia.org/wiki/Probability_interpretations
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?