Players a and b toss a coin. The first player wins if heads…
Players $a$ and $b$ toss a coin. The first player wins if heads appears on the first, second or third toss; otherwise the second player wins. Determine the probability that the second player wins.
Explanation of Method
The second player wins exactly when the first three tosses are all tails. The tosses are independent, each tails with probability $\frac12$, so the probability is $\left(\frac12\right)^3$. The first player's chance follows by the addition rule over the first toss that shows heads.
Step 1: When does $b$ win?
$b$ wins iff none of the first three tosses shows heads, i.e. tails, tails, tails.
Step 2: Independence
\[ \boxed{P(b \text{ wins})=\left(\tfrac12\right)^3=\tfrac18=0.125.} \]
Step 3: Check through player $a$
$a$ wins if the first heads appears on toss $1$, $2$ or $3$ (mutually exclusive):
\[ P(a \text{ wins})=\tfrac12+\tfrac14+\tfrac18=\tfrac78, \qquad \tfrac78+\tfrac18=1 \ \checkmark. \]
Step 4: A fair game?
The game is heavily in favour of $a$. To make it fair with this rule, $b$ should receive $7$ times more when winning than $a$ does.
Result
| Answer | |
|---|---|
| $P(b \text{ wins})$ | $(1/2)^3=1/8$ |
| $P(a \text{ wins})$ | $7/8$ |