Biased Flips

Let

$P(x) = \prod_{i = 1}^{27} ( \frac{h}{2 \cdot i + 1} + \frac{2 \cdot i}{2 \cdot i + 1})$

Looking at the coefficient of the $h^{j}$ th term, we see that this corresponds exactly with the probability that $j$ heads are flipped. This is because in each term of the product, you are choosing whether or not to use the term with the $h$ , which has a coefficient equivalent to the probability. Functions like these are known as a generating functions .

If we plug in -1 for $h$ in $P$ , we see that all of the odd coefficients turn negative and the $h$ term disappears. Therefore, if we subtract $P(-1)$ from $P(1)$ , we simply get double the probability that there are an odd number of heads. So dividing by two will give us our answer!

Now, we wish to compute $\frac{P(1) - P(-1)}{2}$

We see that $P(1)$ is 1 immediately since all of the individual factors in the product turn to $1$ . On the other hand, $P(-1)$ becomes $\prod_{i = 1}^{27} \frac{2 \cdot i - 1}{2 \cdot i + 1}$ . This series is a telescoping series - product and simply becomes $\frac{1}{55}$ . Therefore, out solution is simply half of $1 - \frac{1}{55} = \frac{27}{55}$ .

Instead, suppose we had $n$ coins, and let the probability of flipping an odd number of heads be $p_n$ .

Flip the last coin. If it flips tails, then flipping the remaining coins should result in an odd number of heads. If it flips heads, then flipping the remaining coins should result in an even number of heads. In all

$p_n = \frac{2n}{2n+1}p_{n-1} + \frac{1}{2n+1}(1-p_{n-1})$ which simplifies to $(2n+1)p_n - (2n-1)p_{n-1} = 1$ Adding from $n=1$ to $n=m$ and using $p_0=0$ , we get

$p_m = \frac{m}{2m+1}$ In particular, the answer to this question is $\frac{27}{55}$ .