Probability of getting more Heads between two gamblers I'm trying to solve this probability question I thought of. If two gamblers are playing a coin toss game and Gambler A has $(n+2)$ and B has $n$ fair coins. What is the probability that A will have more heads than B if both flip all their coins? 
I tried to solve it like this, using symmetry. If we compare the first $n$ coins of of $A$ and $B$:
$E_{1}$: Event that A and B have the same number of Heads
$E_{2}$: Event that A has 2 Heads less than B
$E_{3}$: Event that A has 1 Head less than B
$E_{4}$: Event that A has 3 Heads less than B
$E_{5}$: Event that A has 2 Heads more than B 
$E_{6}$: Event that A has 1 Head more than B 
Let $P(E_{1}) = y$, $P(E_{2}) = P(E_{5}) = x$, $P(E_{3}) = P(E_{6}) = z$, $P(E_{4}) = k$ which implies $y + x + x + z + z + k = 1$ and probability of A having more heads = $y*0.75 + x + z + z(0.5^{2})$. However, I don't know how to solve the question from here.
Thank You

You were right for using symmetry. Suppose A and B both have n coins. The probability of A winning would be 0.5. Now suppose A has $(n+2)$ coins and flips $n$ coins first then flips the other 2 coins. The probability is 0.5 for A to win with $n$ coins, and adding the 2 coins, the probability of at least one of them being heads is 3/4. So the probability of A winning is $3/4$.
