Tossing unfair coins.

We take a break from blog posts on REU’s, career advice and conference announcements and ruminate a bit on mathematics. Here is a question that probably comes up in a probability class (ha ha, yes pun intended).


You are given a random coin and you have to come up with a way to determine two outcomes with equal probability. How do you do that?

“Well, OU Math Blog”, you might say, “just flip the coin!” Trouble is, most coins are biased. As a mathematician one would like to know a way to achieve a (near) perfect 50-50 outcome. Can this be done? Well, yes, and if you want to think about it, then Spoiler Alert! Stop reading now!

Let us say that whatever coin you are given has some (biased) probability of returning Heads; let us call this  α. Then assuming that the coin won’t land on its edge (a fairly safe assumption given the nature of gravity) the probability of getting Tails must be 1 – α. The following protocol will now generate a theoretically 50-50 outcome no matter the bias of the coin. Toss the coin twice and select the following two outcomes as “Heads” and “Tails”: let’s say the outcome HT is “Heads” and the outcome TH is “Tails”. If the outcome is HH or TT, then start over (i.e., toss twice more). By the independence of Bernoulli trials, the probabilities of the event HT is α(1 – α) while the probability of the event TH is (1 – α)α i.e., they are equal! The only drawback with this method is that this is not guaranteed to end in finite time. One could theoretically keep tossing HH or TT indefinitely and thus have to keep going. But, it provides an elegant solution to the problem of tossing a coin when one does not know the bias of the coin.

Question: The above solution is near perfect in that we had to discard the possibility of the coin landing on its edge. Is there a way out of this dilemma?