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

Given that

I don’t knowthe 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 constructiondefines fairness, I’m asking if theunknowing 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 betweennot knowingand “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 seemsfairnesseven depends on the subjects to whom fairnessmatters. 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?

**Answer**

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), 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}.

**Attribution***Source : Link , Question Author : Andrew Cheong , Answer Author : MGA*