Question

Suppose A has (n+1) and B has n fair coins which they flip. Let $p_k$ denote the probability that A obtains exactly $k(0\le k\le n+1)$ heads more than B. Value of

Answer 1

In total there are (2n+1) fair coins where P(heads)=P(Tails)=$\frac{1}{2}$
Total number of elements in the sample set will be
$=2^{2n+1}$
Now
$P_0$ implies both A, and B have equal heads.
This is the middle value,
i,e the middle term in the expansion of $(1+x{)}^{2n+1}{|}_{x=1}|$
Middle term will be
$=T_{n+1}$ and $T_{n+2}$ term.
Hence coefficient will be
$={}^{2n+1}{C}_{n+1}$
$={}^{2n+1}{C}_n$
Hence probability
$=\frac{{}^{2n+1}{C}_{n+1}}{2^{2n+1}}$
$=\frac{{}^{2n+1}{C}_n}{2^{2n+1}}$
Now consider B has n heads, Hence A must have n+2 heads
$P_2$
$=\frac{{}^{2n+1}{C}_{n+2}}{2^{2n+1}}$
$=\frac{{}^{2n+1}{C}_{n-1}}{2^{2n+1}}$
And
$P_n$ will be
$\frac{{}^{2n+1}{C}_{2n}}{2^{2n+1}}$ ... considering B has n heads.
$=\frac{2n+1}{2^{2n+1}}$
Also
${\sum }_{K\ge 1}\left(P_k\right)$
$=\frac{2^{2n+1-1}}{2^{2n+1}}$
$=\frac{2^{2n}}{2^{2n}.2}$
$=\frac{1}{2}$