That's An Odd Probability

The answer is $\boxed{\dfrac12}$ .

Claim : If I toss $2n$ fair coins, then the probability that I get an odd number of heads is the same as the probability that I get an even number of heads.

Proof : The value of the probability that $\mathsf{Pr} (m\text{ heads})$ can be computed using binomial distribution ,

$\mathsf{Pr} (m\text{ heads}) = \dbinom {2016}m p^m \cdot q^{2016-m} \; ,$

where $p$ and $q$ denotes the probability of obtaining a heads and a coin, respectively, and $p+q=1$ . Since we are given that it is a fair coin, then $p=q= \dfrac12$ .

Therefore, $\mathsf{Pr} (m\text{ heads}) =\dbinom {2016}m \left( \dfrac 12\right)^{2016}$ .

$\begin{aligned} \mathsf{Pr} (\text{odd number of heads}) &=& \mathsf{Pr} (1\text{ head}) + \mathsf{Pr} (3\text{ heads}) + \mathsf{Pr} (5\text{ heads}) + \cdots + \mathsf{Pr} (2015\text{ heads}) \\ &=& \dbinom {2016}1 \left( \dfrac 12\right)^{2016} + \dbinom {2016}3 \left( \dfrac 12\right)^{2016} + \dbinom {2016}5 \left( \dfrac 12\right)^{2016} + \cdots + \dbinom {2016}{2015} \left( \dfrac 12\right)^{2016} \\ &=& \left(\dfrac12\right)^{2016} \sum_{k\text{ odd}}^{2016} \dbinom{2016}k \\ &=& \left(\dfrac12\right)^{2016} \sum_{k\text{ even}}^{2016} \dbinom{2016}k , \qquad \qquad \text{ See } (\bigstar) \\ &=& \dbinom {2016}0 \left( \dfrac 12\right)^{2016} + \dbinom {2016}2 \left( \dfrac 12\right)^{2016} + \dbinom {2016}4 \left( \dfrac 12\right)^{2016} + \cdots + \dbinom {2016}{2016} \left( \dfrac 12\right)^{2016} \\ &=& \mathsf{Pr} (0\text{ head}) + \mathsf{Pr} (2\text{ heads}) + \mathsf{Pr} (4\text{ heads}) + \cdots + \mathsf{Pr} (2016\text{ heads}) \\ &=& \mathsf{Pr} (\text{even number of heads}) \qquad \blacksquare \end{aligned}$

Since the total number of heads can be either odd or even only, and we have proven that these two probability are equal, then the probability in question is $\dfrac12$ .

---

Note : The various proof for $(\bigstar)$ has been written up by here by the author himself!.

Imagine you've flipped 2015 coins and have 1 coin to go. There are 2 cases:

Case 1: You've flipped an odd number of heads, so the number of heads is even if the final flip is heads (which has a probability of 1/2)

Case 2: You've flipped an even number of heads, so the number of heads is even if the final flip is tails (which has a probability of 1/2)

So the probability of getting an even number of heads at the end is 1/2

How many coins, how much ever you flip, the possibility of getting head or tail is $\dfrac{1}{2}$