Question

Two persons $A$ and $B$ have respectively $n+1$ and $n$ coins, which they toss simultaneously. Then probability $P$ that $A$ will have more heads than $B$ is:

• A. $P>\frac{1}{2}$
• B. $P=\frac{1}{2}$
• C. $\frac{1}{4}<P<\frac{1}{2}$
• D. $0<P<\frac{1}{4}$

Answer 1

The correct option is B $P=\frac{1}{2}$

Let the number of heads and tails thrown by A be $a$ and $a^{\prime }$ respectively and the number of heads and tails thrown by B be $b$ and $b^{\prime }$ respectively. Then $a+a^{\prime }=n+1$ and $b+b^{\prime }=n$

The required probability P is the probability of the inequality, $a>b$

The probability $1-P$ of the opposite event $a\le b$ is at same time, the probability of the inequality $a^{\prime }>b^{\prime }$ i.e., $1-P$ is the probability that A will throw more tails than B. For

$a\le b⟹n+1-a^{\prime }\le n-b^{\prime }⟹1-a^{\prime }\le -b^{\prime }⟹a^{\prime }\ge b^{\prime }+1⟹a^{\prime }>b^{\prime }$

But the two events that A will throw more heads than B and A will throw more tails than B are equally likely,

$\therefore P=1-P$ or $2P=1$ or $P=\frac{1}{2}$