# The coin flip conundrum

When the Wright brothers had to decide who would be the first to fly their new airplane off a sand dune, they flipped a coin. That was fair: we all know there’s an equal chance of getting heads and tails. But what if they had a more complicated contest?

What if they flipped coins repeatedly, so that Orville would win as soon as two heads showed up in a row on his coin, and Wilbur would win as soon as heads was immediately followed by tails on his? Would each brother still have had an equal chance to be the first in flight? At first, it may seem they’d still have the same chance of winning. There are four combinations for two consecutive flips.

And if you do flip a coin just twice, there’s an equal chance of each one — 25%. So your intuition might tell you that in any string of coin flips, each combination would have the same shot at appearing first. Unfortunately, you’d be wrong. Wilbur actually has a big advantage in this contest. Imagine our sequence of coin flips as a sort of board game, where every flip determines which path we take. The goal is to get from start to finish. The heads/tails board looks like this.

And this is the head/head board. There’s one critical difference. Heads/heads has a move that sends you all the way back to the start that heads/tails doesn’t have. That’s why heads/heads takes longer on average. So we can demonstrate that this is true using probability and algebra to calculate the average number of flips it would take to get each combination. Let’s start with the heads/tails board, and define x to be the average number of flips to advance one step. Focus only on the arrows.

It has two identical steps, each with a 50/50 chance of staying in place or moving forward. Option 1: If we stay in place by getting tails, we waste one flip. Since we’re back in the same place, on average we must flip x more times to advance one step. Together with that first flip, this gives an average of x + 1 total flips to advance. Option 2: If we get heads and move forward, then we have taken exactly one total flip to advance one step. We can now combine option 1 and option 2 with their probabilities to get this expression.

Solving that for x gives us an average of two moves to advance one step. Since each step is identical, we can multiply by two and arrive at four flips to advance two steps. For heads/heads, the picture isn’t as simple. This time, let y be the average number of flips to move from start to finish. There are two options for the first move, each with 50/50 odds. Option 1 is the same as before, getting tails sends us back to the start, giving an average of y+1 total flips to finish. In Option 2, there are two equally likely cases for the next flip.

With heads we’d be done after two flips. But tails would return us to the start. Since we’d return after two flips, we’d then need an average of y+2 flips in total to finish. So our full expression will be this. And solving this equation gives us six flips. So the math calculates that it takes an average of six flips to get heads/heads, and an average of four to get heads/tails.

And, in fact, that’s what you’d see if you tested it for yourself enough times. Of course, the Wright brothers didn’t need to work all this out; they only flipped the coin once, and Wilbur won. But it didn’t matter: Wilbur’s flight failed, and Orville made aviation history, instead. Tough luck, Wilbu

## Responses