Compute probability of winning a game with coins, without using series I found the following question in a book with an answer.
Question: You have two kinds of coins. The number of coin $A$ you have is $n$. The number of coin $B$ you have is $n+1$. When you throw all coins at the same time, calculate the probability such that the number of the-obverse-side-$B$s is larger than the number of the -obverse-side-$A$s.  
Answer: $1/2$.
However, this book told us nothing about additional information except one sentence:
"There is a way to solve this question without using $\sum$".
I've tried to find it, but I'm facing difficulty. Then, here is my question.
Question: Could you show me a way to solve above question without using $\sum$ ?
 A: 
Look for some symmetry in the problem. If none is apparent, make some!

Say that you win if there are strictly more B heads than A heads. Throw all the coins except one coin B. Call $p$ the probability that you are now seeing the same number of A heads and B heads. Recall that you are observing as many coins A as coins B. Thus, by symmetry, the probabilities that you are seeing strictly more A heads than B heads or strictly more B heads than A heads are equal. This shows that both are $\frac12(1-p)$. Furthermore:


*

*If you are seeing strictly more A heads than B heads, the last B coin will not make you win hence you win with (conditional) probability $\color{red}{\mathbf 0}$. 

*If you are seeing the same number of A heads and B heads, the last B coin will make you win if and only if it shows heads, which happens with probability $\color{green}{\mathbf{\frac12}}$.

*If you are seeing strictly more B heads than A heads, the last B coin will not make you lose hence you win with (conditional) probability $\color{blue}{\mathbf 1}$. 
Hence the (absolute) probability to win is
$$
\tfrac12(1-p)\cdot\color{red}{\mathbf 0}+p\cdot\color{green}{\mathbf{\tfrac12}}+\tfrac12(1-p)\cdot\color{blue}{\mathbf 1}=\text{____}.
$$
