Throw a biased coin $N+1$ times vs $N$ times If I throw a biased coin ($P$ probability for heads) $N+1$ times and you throw it $N$ times, what is the probability that I get more heads than you?
What if I throw it $N+2$ times instead of $N+1$?
 A: Note: @ThomasAndrews gave a correct solution in the comments of the main post.  His solution involves a double summation, while mine below only involves one or two single summation(s), and also attempts to highlight the symmetry aspect.  However, I don't know how to turn any of them into a closed form.
First, the case when A throws $N+1$ coins vs B throws $N$ coins:
Suppose B goes first, then A goes, and A temporarily stops after $N$ coins.  Three things can happen:


*

*Event $E=$ they have the same number of Heads.  $P(E)$ can be calculated as a single summation, summing over them each getting $k$ Heads, for $0 \le k \le N$, i.e.


$$P(E) = \sum_{k=0}^N \bigl( {N \choose k} p^k (1-p)^{N-k} \bigr)^2 $$


*

*Event $F=$ A has more Heads

*Event $G=$ B has more Heads

*By symmetry, $P(F)=P(G) = {1 - P(E) \over 2}$
The overall answer is $P(F) + p \times P(E) = \frac12 + P(E) (p - \frac12)$.  I don't see a closed form for $P(E)$, but I'm not sure.

Generalizing to the $N+2$ case requires one more single summation, to keep track of event that B has exactly one more Head.
Again, suppose B goes first, and A temporarily stops after $N$ coins.  Five things can happen:


*

*Event $E=$ they have the same number of Heads.  $P(E)$ given above.


*

*In this event, A can win if he throws at least $1$ more Head, prob $(1 - (1-p)^2)$


*Event $F_1 = $ A has exactly $1$ more Head.

*Event $F_2 = $ A has at least $2$ more Heads (i.e. $2$ more, or $3$ more, or... compared to B).

*Event $G_1 = $ B has exactly $1$ more Head.


*

*In this event, A can win if he throws Heads in his last $2$ throws, prob $p^2$.


*Event $G_2 = $ B has at least $2$ more Heads.

*By symmetry, $P(F_1) = P(G_1)$, and by explicit counting:
$$P(F_1) = P(G_1) = \sum_{k=0}^{N-1} {N \choose k} p^k (1-p)^{N-k}  {N \choose k+1} p^{k+1} (1-p)^{N-(k+1)}$$


*

*By symmetry, $P(F_2)=P(G_2) = {1 - P(E) - P(F_1)-P(G_1) \over 2}$
The overall answer is 
$$P(F_2) + P(F_1) + P(E)(1 - (1-p)^2) + P(G_1) p^2 = \frac12 + P(E)(\frac12 - (1-p)^2) + P(G_1) p^2$$
