## Flip Better

After posting this, I realized I made an error that seemed small but messed up everything.
You can see the correction here.

Depending on the internet circles you run in, a friend may have linked you to this quotation recently:

When faced with 2 choices, simply toss a coin. It works not because it settles the question for you, but because in that brief moment, when the coin is in the air, you suddenly know what you are hoping for.

Or, as another pointed out to me, this poem by Piet Hien:

Whenever you’re called on to make up your mind,
and you’re hampered by not having any,
the best way to solve the dilemma, you’ll find,
is simply by spinning a penny.
No – not so that chance shall decide the affair
while you’re passively standing there moping;
but the moment the penny is up in the air,
you suddenly know what you’re hoping.

I recently had a binary choice to make, so I gave this tactic a shot. I report that it doesn’t work if you’re counting on it. That is, it might work by serendipity, but if you toss the coin specifically so that you’ll understand your subliminal desires, the effect is ruined. I want my money back. (I didn’t pay anything for the advice, but I dropped three coins down a gutter before I finally managed to catch one I had flipped.)

This reminds of a paper saying that if you flip a coin vigorously, it’ll come up the same way it started about 51% of the time.

Here, I’ll ask if you can substantially avert this bias by flipping best-two-of-three. (Another way to avoid the 51% problem is not to flip the coin at all. Take it out of your pocket, look at it, and go with that.)

Let’s say that when you flip a coin, its chance to land the same way it started is $p$, regardless of whether it started heads or tails. Then its probability to flip to the other state is $1-p$, and let’s agree to call that $q$ for short. In an ideal world, $p = q = 0.5$, but Diaconis et. al. suggest this may not be true.

Suppose the coin starts with Heads up ($H$). Then what is the probability that heads wins two out of three tosses? Is it closer or further from $0.5$ than $p$, and by how much? That is, is two-out-of-three more fair than a single coin flip?

There are four ways that heads can win 2 of 3 tosses.

$HHH$

$HHT$

$HTH$

$THH$

To find the probability of each, we need to know not whether each toss is heads or tails, but when the coin is changing state. For example, take the sequence

$HTH$.

The coin starts out $H$, and then we make our first flip and get another $H$. The coin didn’t change state, so this has probability $p$. We flip again and get $T$, a switch. This has probability $q$. The last flip we go back to heads, another switch, and another probability $q$. The final probability for this sequence is $p*q*q = pq^2$.

Here is a mildly-relevant image from the internet. It is the best thing that pops up when google "sexy quarter." This is the middle of the blog post, which is roughly where you are supposed to forcibly insert semi-relevant images without explanation, if I've learned anything from reading blogs.

The probabilities for each of the four sequences are:

$P(HHH) = p^3$

$P(HHT) = p^2q$

$P(HTH) = pq^2$

$P(THH) = q^2p$

GAAH! More $p$‘s than a kindergartener’s underpants. Rather than calculate, let’s simplify using our knowledge that coin tossing is not horrendously unfair, so that $p \approx 0.5$. Specifically, let’s set a small number $\epsilon$ such that

$p = \frac{1}{2} + \epsilon$.

This will let us simplify the calculations because we can ignore anything that has an $\epsilon^2$ in it as a second-order correction. Unless the first-order correction comes out to be zero, this is a good approximation.

We know that

$q = \frac{1}{2} - \epsilon$,

which lets us simplify a bit. First, lay out the ground rules

$\begin{array}{rcl}p^2 &=& \left(\frac{1}{2} + \epsilon\right)^2 \\ &=& \frac{1}{4} + 2*\frac{1}{2}\epsilon + \epsilon^2 \\ &\approx& \frac{1}{4} + \epsilon \end{array}$

This says that when you take a number a little bit bigger than one half and square it, you get something the same little bit bigger than one fourth. Let’s apply the same rule for $q$:

$q^2 \approx \frac{1}{4} - \epsilon$

and when multiplying $p$ and $q$, the over-compensation and under-compensation cancel out to first order, and we have

$p*q \approx \frac{1}{4}$.

The nice thing is that the we can use the binomial theorem to propagate out to higher powers of $p$ and $q$. If we have $p^mq^n$, for example, that is roughly

$p^mq^n \approx \frac{1}{2^{m+n}} + \frac{m-n}{2^{m+n}}*\epsilon$,

which of course is left as an exercise to the reader. It’s not the kind of exercise that will give you washboard abs, but it is the kind that will get you girls.

Returning to the probability for two heads out of three, we can fill in the probabilities like so:

$P(HHH) = p^3 = \frac{1}{8} + \frac{3}{8} \epsilon$

$P(HHT) = p^2q = \frac{1}{8} + \frac{1}{8} \epsilon$

$P(HTH) = pq^2 = \frac{1}{8} - \frac{1}{8} \epsilon$

$P(THH) = pq^2 = \frac{1}{8} - \frac{1}{8} \epsilon$

the sum is

$P(H wins) = \frac{1}{2} + \frac{\epsilon}{4}$

If it’s a little unfair to flip a coin, flipping for best two out of three cuts the bias by a factor of four. Or, in less coherent words, it quarters the quarter’s recording of its core uncordial proclivity.

There’s more interesting stuff here. How much better is it to play three-of-five, or four-of-seven, or $x$ of $2x-1$? Exercises to the very sexy reader.