Simple Probability Distribution Question Alice flips a fair coin n times and Bob flips another fair coin n + 1 times, resulting in independent X $\sim$ Bin(n, 1/2) and Y $\sim$ Bin(n + 1, 1/2). 
Can someone please explain to me, in simple english, why this statement is true? 
$$P(X<Y)=P(n−X<n+1−Y)$$
The random variable Y is not inherently larger than X, so why is that statement true?
 A: Binomial coefficients have a symmetry that can be written
$$\binom{n}{k} = \binom{n}{n-k}$$
$Y$ is a random variable following a binomial distribution with parameters $n+1$ and $1/2$. When the second parameter is $1/2$, the symmetry in binomial coefficients extends to a symmetry in the probability mass function:
\begin{align}
P(Y=k) &= \binom{n+1}{k}\left(\frac{1}{2}\right)^k \left(1 - \frac{1}{2}\right)^{n+1-k} \\
&= \binom{n+1}{k} \left(\frac{1}{2}\right)^{n+1}\\\\
P(Y=n+1-k) &= \binom{n+1}{n+1- k }\left(\frac{1}{2}\right)^{n+1-k} \left(1 - \frac{1}{2}\right)^{n+1 - (n+1-k)}\\
 &=\binom{n+1}{k} \left(\frac{1}{2}\right)^{n+1}\\\\
\end{align}
We have, for all $k$, the symmetric relationship
$$P(Y=k) = P(Y = n + 1 - k) = \binom{n+1}{k} \left(\frac{1}{2}\right)^{n+1}$$
Similarly, $X$ has parameters $n$ and $1/2$, so we have, for all $h$, 
$$P(X = h) = P(X = n - h) =\binom{n}{h} \left(\frac{1}{2}\right)^{n}$$
A: As you do not specify, I will assume that $X$ and $Y$ count the number of tails for Alice and Bob.


*

*$X<Y$ means that Alice has less tails than Bob 

*$n-X$ is the Alice's number of heads (# of trials - # of tails) 

*$n+1-Y$ is Bob's number of heads (# of trials - # of tails)


therefore


*

*$n-X < n+1-Y$ means that Alice has less heads than Bob.


Now, if the coin is fair, counting heads leads to the same probability as counting tails (this is symmetry); therefore
$$
P(X<Y)=P(n-X < n+1-Y).
$$
