The Coin Toss Game

Probability Level 4

There are two people F and S who are playing a game. The game is a fair coin toss with one side called X and one side called Y. The following are the conditions set by the players that define how they are able to win.

Player F can win if and only if side Y appears two times in a row.
Player S can win if and only if side X appears three times in a row.

What is the probability that player F will win the game?

Give your answer in decimal form.

The answer is 0.7.

1 solution

Thomas Kim
Jun 10, 2014

This is one of the solutions that do not involve case-working. This also happens to be the solution I used.

Solution: Let the probability that F wins if the first toss is X be denoted as $p$ and let the probability that he wins if the first toss is Y be denoted as $q$ . Therefore, the probability that F wins is $\frac { 1 }{ 2 } p+\frac { 1 }{ 2 } q$ .

If the first toss is X, there is a $\frac{1}{4}$ probability that there will be two more X and F loses. Otherwise, the streak of X's will be broken. The probability that F wins after the Y that breaks the streak can be defined as $q$ . This gives $p = \frac{3}{4}q$ .

If the first toss is Y, there is a $\frac{1}{2}$ chance that the next toss will be Y and F wins. Otherwise, the next toss is X. The probability that F wins after that can be defined as $p$ . From this, $q = \frac{1}{2}+\frac{1}{2}p$ .

Solving this system gives $p = \frac{3}{5}$ and $q = \frac{4}{5}$ . The probability that F wins is $(\frac{1}{2} \times \frac{3}{5}) + (\frac{1}{2} \times \frac{4}{5}) = \frac{3}{10} + \frac{4}{10} = \frac{7}{10} = \boxed{0.7}$

Can anyone do this solution case by case; as it gives 0.6875

Shubham Maurya - 5 years, 10 months ago

The Coin Toss Game

The answer is 0.7.

1 solution

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