The coin flip conundrum PoShen Loh
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, Wilbur.