Probability for coins If A has n+1 coins and B has n coins then what is the probability that A has more heads than B
Cases I could make:
I. If A gets more heads than B when A and B toss n coins each independent of $(n+1)^{th}$ trial
II. If A and B get equal number if heads when each toss n coins and outcome of $(n+1)^{th}$ trial is head
I'm not able to carry out the calculations
 A: After $n$ tosses, by symmetry,
$A$ and $B$ will have equal probability of having more heads = $z,\;$say,
then P(tie) $= 1-2z$
With one more toss, $A$ is sure to win if leading, (Case I) and win with $Pr = \frac12$ if tied,  (Case II)
thus P(A wins) $= (1\cdot z) +\frac12(1-2z) = \frac12$
A: Intuitively it should depend on the extra coin that p has and the probability should be $\frac{1}{2}$. Answering based on two cases that you have come up with,
In first case when we ignore $(n+1)^\text{th}$ coin that person p has, p and q have the same probability of throwing more heads owing to symmetry but there is also a probability that they are tied. If $x$ is the probability of p having more heads and $y$ is the probability that they are tied after $2n$ tosses,
$2 x + y  = 1 \implies y = 1 - 2x$
Now the outcome of the $(2n+1)^\text{th}$ toss only affects the probability of p throwing more heads if they are tied after $2n$ tosses (just as a side note - if q was ahead by $1$ head after $2n$ tosses, they can get to a tie after the last toss). If $z$ is the probability of p throwing more heads after the last toss,
$z = x + \frac{y}{2} = x + \frac{1-2x}{2} = \frac{1}{2}$
A: Call a coin good if it is head and belong to A or it is tail and belong to B. If A has more head than B then more than half of the $2n+1$ coins are good. The probability is $\frac{1}{2}$
A: Let $X:$"number of head in $n+1$ trials" and $Y:$"number of head in $n$ trials", then $X\sim\text{Bin}(n+1,0.5)$ and $Y\sim\text{Bin}(n,0.5)$ independent. So, $\mathbf{P}(X>Y)=\mathbf{P}(X+n-Y>n)$ where $n-Y\sim\text{Bin}(n,0.5)$. This implies than $X+n-Y\sim\text{Bin}(2n+1,0.5)$ and then:
$$\mathbf{P}(X>Y)=\sum_{k=n+1}^{2n+1}\left(\begin{array}{c}2n+1\\k\end{array}\right)\frac{1}{2^{2n+1}}=\frac{2^{2n}}{2^{2n+1}}=\frac{1}{2}$$
I don't really understand your questions.
A: Assuming the coins are all independent and fair, you are asking
$$
\mathbb{P}\{X > Y\}
$$
where $X,Y$ are independent with $X\sim\operatorname{Bin}(n+1,1/2)$ and $Y\sim\operatorname{Bin}(n,1/2)$.
Here is an answer without any explicit computation.

*

*First, I encourage you to try the cases $n=0$ and $n=1$: they are simple, and give the same answer $1/2$. Based on this, it's natural to think that $1/2$ may always be the answer. Let's prove it!


*Write $X = Z+B$, where $Z\sim\operatorname{Bin}(n,1/2)$ and $B\sim\operatorname{Bern}(1/2)$ are independent. Then
$$\begin{align*}
\mathbb{P}\{X > Y\}
&= \mathbb{P}\{Z+B > Y\} \\
&= \mathbb{P}\{Z > Y \mid B=0\}\mathbb{P}\{B=0\} + \mathbb{P}\{Z \geq Y \mid B=1\}\mathbb{P}\{B=1\}\\
&=\mathbb{P}\{Z > Y\}\cdot\frac{1}{2} + \mathbb{P}\{Z \geq Y \}\cdot \frac{1}{2} \\
&=\frac{1}{2}\left(2\mathbb{P}\{Z > Y\}+\mathbb{P}\{Z = Y\}\right)
\end{align*}$$
where the third equality uses that $Z,B,Y$ independent and $B$ is $\operatorname{Bern}(1/2)$. Now, by symmetry (since $Z,Y$ are i.i.d.)
$$
\mathbb{P}\{Z > Y\} = \mathbb{P}\{Z < Y\} = \frac{1-\mathbb{P}\{Z = Y\}}{2}
$$
and so
$$\begin{align*}
\mathbb{P}\{X > Y\}
&=\frac{1}{2}\left(2\mathbb{P}\{Z > Y\}+\mathbb{P}\{Z = Y\}\right)
=\boxed{\frac{1}{2}}
\end{align*}$$
A: Since the probability of the outcome of head and tail are equal, we can ignore it. Let $a$ be the number of heads A has and $b$ be the number of heads B has. The set of all possible pairs $(a,b)$ is $$S = \{(a,b) | 0 \leq a \leq n +1, 0\leq b \leq n, a,b\in \mathbb{Z}\}.$$
Note that $S$ has $(n + 1)(n + 2)$ elements.
Now we only need to count when A has more heads then B, that is the number of element in the following subset T of $S,$ $$ T = \{(a,b\} \in S | a > b\}.$$
We can count the number of element of $T$ by considering a particular $a = 0,1,2,3,...,n + 1$ and get all $b$ possible. (Or other counting method you know)
Note that the element of $T$ is exactly $$ 1 + 2 + 3 + \dots + (n + 1) = \frac{1}{2}(n+1)(n + 2).$$
Hence the probability of $T$ is $\frac{|T|}{|S|} =\frac{1}{2}$.
