Unbiasing a Biased Coin

Flip the coin twice. If it lands heads-tails, you win. If tails-heads, your friend wins. If both flips match, discard the result and flip again.

Let $p$p be the probability of the coin landing heads, and $1-p$1−p the probability of tails. When we flip the coin twice, there are four possible outcomes:

$\text{HH}$HH	$p\times p=p^2$p×p=p2
$\text{HT}$HT	$p\times (1-p)$p×(1−p)
$\text{TH}$TH	$(1-p)\times p$(1−p)×p
$\text{TT}$TT	$(1-p)\times (1-p)=(1-p{)}^2$(1−p)×(1−p)=(1−p)2

Notice that $P\left(\text{HT}\right)=p(1-p)$P(HT)=p(1−p) and $P\left(\text{TH}\right)=(1-p)p$P(TH)=(1−p)p. These are identical — heads-then-tails and tails-then-heads are equally likely no matter what the bias is. So we assign one outcome to each player and ignore the rest.

If we get $\text{HH}$HH or $\text{TT}$TT, we simply discard that round and flip again. Since $\text{HT}$HT and $\text{TH}$TH are equally likely whenever they do occur, each player has exactly a 50% chance of winning.

This works for any bias, so long as 0<p<1 (that is, neither heads nor tails is impossible).