If $A$ flips $n+1$ and B flips n fair coins, show that the probability that $A$ gets more heads than $B$ is $\frac{1}{2}$ . Hint: Condition on which player has more heads after each has flipped n coins. (There are three possibilities.)

$H_A$ - the number of heads by $A$

$H_A^{\text{'}}$ - the number of heads by $A$ in the first $n$ flips

$H_B$- the number of heads by $B$(in $n$flips)

$H$- event that $A$flips head in the . flip

Event $H_A^{\text{'}}>H_B$is that after $n$flips, $A$has more heads.

Regarding the result after $n$flips there are three mutually exclusive events: $H_A^{\text{'}}>H_B,H_A^{\text{'}}=H_B$and $H_A^{\text{'}}<H_B$and those events make up the whole outcome space $\left(0\right)$.

The formula of total probability,

$P\left(H_A>H_B\right)=P\left(H_A>H_B\mid H_A^{\text{'}}>H_B\right)P\left(H_A^{\text{'}}>H_B\right)$

$+P\left(H_A>H_B\mid H_A^{\text{'}}<H_B\right)P\left(H_A^{\text{'}}<H_B\right)$

$+P\left(H_A>H_B\mid H_A^{\text{'}}=H_B\right)P\left(H_A^{\text{'}}=H_B\right)$

After both flipped $n$ times, the situation is symmetrical, the probability that $A$ has more heads is equal to the probability that $B$ has more heads, that is:

$P\left(H_A>H_B\mid H_A^{\text{'}}>H_B\right)=1$

$P\left(H_A>H_B\mid H_A^{\text{'}}=H_B\right)=P\left(H\right)=\frac{1}{2}$

$P\left(H_A>H_B\mid H_A^{\text{'}}<H_B\right)=0$

Equation becomes:

$=\frac{1}{2}\left(2P\left(H_A^{\text{'}}>H_B\right)+P\left(H_A^{\text{'}}=H_B\right)\right)$

$\stackrel{\left(1\right)}{=}\frac{1}{2}\left(P\left(H_A^{\text{'}}>H_B\right)+P\left(H_A^{\text{'}}<H_B\right)+P\left(H_A^{\text{'}}=H_B\right)\right)$

$\stackrel{\left(0\right)}{=}\frac{1}{2}$

The equality is proven,

The probability that $A$ has more heads is equal to the probability that $B$ has more tails up to $n$-th flip.