Tossing a Biased Coin

When we talk about a coin toss, we think of it as unbiased: with probability one-half it comes up heads, and with probability one-half it comes up tails. An ideal unbiased coin might not correctly model a real coin, which could be biased slightly one way or another. After all, real life is rarely fair. This possibility leads us to an interesting mathematical and computational question. Is there some way we can use a biased coin to efficiently simulate an unbiased coin? Specifically, let us start with the following problem: Problem 1. Given a biased coin that comes up heads with some probability greater than one-half and less than one, can we use it to simulate an unbiased coin toss? A simple solution, attributed to von Neumann, makes use of symmetry. Let us flip the coin twice. If it comes up heads first and tails second, then we call it a 0. If it comes up tails first and heads second, then we call it a 1. If the two flips are the same, we flip twice again, and repeat the process until we have a unbiased toss. If we define a round to be a pair of flips, it is clear that we the probability of generating a 0 or a 1 is the same each round, so we correctly simulate an unbiased coin. For convenience, we will call the 0 or 1 produced by our simulated unbiased coin a bit, which is the appropriate term for a computer scientist. Interestingly enough, this solution works regardless of the probability that the coin lands heads up, even if this probability is unknown! This property seems highly advantageous, as we may not know the bias of a coin ahead of time. Now that we have a simulation, let us determine how efficient it is. Problem 2. Let the probability that the coin lands heads up be p and the probability that the coin lands tails up be q 1 p. On average, how many flips does it take to generate a bit using von Neumann’s method? Let us develop a general formula for this problem. If each round takes exactly f flips, and the probability of generating a bit each round is e, then the expected number of total flips t satisfies a simple equation. If we succeed in the first round, we use exactly f flips. If we do not, then we have flipped the coin f times, and because it is as though we have to start over from the beginning again, the expected remaining number of flips is still t. Hence t satisfies t e f 1 e f t