Q.3.21 - A First Course in Probability

Question

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.) 

Step-by-Step Solution

Answer

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

Step 1: Given Information

$H_A$ - 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

Step 2: Explanation

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,

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

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

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

Step 3: Explanation

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:

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

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

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

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

Step 4: Explanation

Equation becomes:

$P\left(H_A>H_B\right)=1·P\left(H_A^{\text{'}}>H_B\right)+0·P\left(H_A^{\text{'}}<H_B\right)+\frac{1}{2}·P\left(H_A^{\text{'}}=H_B\right)$

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

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

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

The equality is proven,

Step 5: Final Answer

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