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
 A: The probability that $A$ wins when it has two extra throws is related to the probabilities involving them both having $n$ throws ...
$$P( A \text{ wins in }n+2 \text{ throws})
 \\=P( A \text{ wins in }n \text{ throws})\\
 +\frac 34 P( A \text{ ties in }n\text{ throws})\\
 +\frac 14P( A \text{ down by one  in }n \text{ throws})$$
Consider the following problems ...
Assume $A$ and $B$ both throw $n$ fair coins
(a) What is the probability they toss the same number of heads ?
$$P(\text{ same }) = \sum_{k=0}^n  \binom nk^2 (0.5)^{2n}\\  = (0.5)^{2n}\sum_{k=0}^n  \binom nk^2 \\
= (0.5)^{2n}\binom {2n}{n} $$
Where the last equality comes from the normalization of the hypergeometric distribution
$$\sum_{k=0}^{X}\binom nk \binom {N-n}{X-k}=\binom N X$$
so 
$$\sum_{k=0}^n  \binom nk^2 =\sum_{k=0}^n  \binom nk \binom {n}{n-k}\\ =\sum_{k=0}^{n}\binom nk \binom {2n-n}{n-k}=\binom {2n}{n}$$
So 
$$ P( A \text{ ties in }n\text{ throws}) = (0.5)^{2n} \binom {2n}{n}$$
and by symmetry 
$$ P( A \text{ wins in }n\text{ throws}) = \frac 12 \left( 1 - (0.5)^{2n} \binom {2n}{n} \right )$$
(b) What is the probability that $B$ tosses one more head than $A$  when they both throw $n$ fair coins ?
$$P(\text{ down by one  }) = \sum_{k=1}^{n}  \binom nk \binom{n}{k-1} (0.5)^{2n}
 \\  = \sum_{k=1}^{n}\binom nk \binom{2n-n}{n-k+1} (0.5)^{2n} 
 \\=  \binom{2n}{n+1} (0.5)^{2n} 
$$
So putting it all together
$$ P( A \text{ wins in } n+2 \text{ throws})
\\ = \frac 12 \left( 1 - (0.5)^{2n} \binom {2n}{n} \right ) + \frac 34 (0.5)^{2n} \binom {2n}{n}+\frac 14 (0.5)^{2n} \binom {2n}{n-1}
\\=\frac 12 +(0.5)^{2n+2}\left[ \binom {2n}{n} +\binom {2n}{n-1} \right ]
$$
A: For the probability that A wins after $n+2$ tosses, i.e. more heads than B, I got:
$(1/2)^{2n+2}\sum\limits_{j=0}^{n+2} \sum\limits_{k=0}^{j-1}\binom{n+2}{j}\binom{n}{k}$
I'd really appreciate if someone could confirm my result, I tested the case $n=1$ by bruteforce and it seems to be correct.
A: 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 $3/4$.
