You have $100$ biased coins. The probabilities of seeing heads when you toss these coins are equal to $1/3$, $1/5$, $1/7$, $1/9$, and so on, up to $1/201$ for the last coin (in general, for the $k$-th coin, the probability is $\frac{1}{2k+1}$. Suppose that you toss all these coins at the same time. What is the probability that you will see an odd number of heads?
Can someone help me figure out where to start with this, please? Thank you!
Best Answer
Hint: Consider the generating function
$$ f(x) = \prod ( \frac{ 2k}{2k+1} + \frac{1}{2k+1} x)$$
The coefficient of $x^n$ gives you the probability that there are $n$ heads.
Hint: Let's find the sum of coefficient of even powers. This is equal to
$$ \frac{ f(1) + f(-1) } { 2}$$
Hint: $f(1) = 1$.
Hint: $f(-1) = \frac{ 1}{201} .$
Hence, conclude that the answer is $\frac{100}{201}$.