A, B and C in order throw an unbiased coin. The first who throws head wins. Find the respective probabilities of winning

Answer

To find the respective probabilities of winning for A, B, and C when they throw an unbiased coin in order, we can follow these steps:

### Step 1: Calculate the Probability of A Winning
A can win in two scenarios:
1. A throws a head on the first toss.
2. A throws a tail, followed by B and C also throwing tails, and then A throws a head again.

The probability of A winning on the first toss is:
- P(A wins on first toss) = P(A throws head) = \( \frac{1}{2} \)

If A throws a tail, the probability of this happening is \( \frac{1}{2} \). Then, for B and C to also throw tails, the probability is:
- P(B throws tail) = \( \frac{1}{2} \)
- P(C throws tail) = \( \frac{1}{2} \)

So, the combined probability for A, B, and C all throwing tails is:
- P(A, B, C all tails) = \( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \)

Now, if they all throw tails, A gets another chance to throw a head. The probability of A winning in this case is:
- P(A wins after all tails) = \( \frac{1}{8} \times \frac{1}{2} = \frac{1}{16} \)

This scenario can repeat indefinitely, leading to an infinite series:
- Total probability of A winning = \( \frac{1}{2} + \frac{1}{8} \times \frac{1}{2} + \frac{1}{8^2} \times \frac{1}{2} + \ldots \)

This can be expressed as:
- Total probability of A winning = \( \frac{1}{2} + \frac{1}{16} + \frac{1}{64} + \ldots \)

This is a geometric series with:
- First term \( a = \frac{1}{2} \)
- Common ratio \( r = \frac{1}{8} \)

Using the formula for the sum of an infinite geometric series \( S = \frac{a}{1 - r} \):
- \( S_A = \frac{\frac{1}{2}}{1 - \frac{1}{8}} = \frac{\frac{1}{2}}{\frac{7}{8}} = \frac{4}{7} \)

### Step 2: Calculate the Probability of B Winning
For B to win, the following must happen:
1. A throws a tail.
2. B throws a head.

The probability of A throwing a tail is \( \frac{1}{2} \) and the probability of B throwing a head is \( \frac{1}{2} \):
- P(B wins on first toss) = \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)

If A throws a tail and B throws a tail, followed by C throwing a tail, the probability is:
- P(A, B, C all tails) = \( \frac{1}{8} \)

Then B gets another chance to throw a head:
- P(B wins after all tails) = \( \frac{1}{8} \times \frac{1}{2} = \frac{1}{16} \)

This leads to the infinite series:
- Total probability of B winning = \( \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \ldots \)

This series has:
- First term \( a = \frac{1}{4} \)
- Common ratio \( r = \frac{1}{4} \)

Using the formula for the sum of an infinite geometric series:
- \( S_B = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \)

### Step 3: Calculate the Probability of C Winning
For C to win, the following must happen:
1. A throws a tail.
2. B throws a tail.
3. C throws a head.

The probability of this happening is:
- P(C wins on first toss) = \( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \)

If A, B, and C all throw tails, the probability is:
- P(A, B, C all tails) = \( \frac{1}{8} \)

Then C gets another chance to throw a head:
- P(C wins after all tails) = \( \frac{1}{8} \times \frac{1}{2} = \frac{1}{16} \)

This leads to the infinite series:
- Total probability of C winning = \( \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \ldots \)

This series has:
- First term \( a = \frac{1}{8} \)
- Common ratio \( r = \frac{1}{4} \)

Using the formula for the sum of an infinite geometric series:
- \( S_C = \frac{\frac{1}{8}}{1 - \frac{1}{4}} = \frac{\frac{1}{8}}{\frac{3}{4}} = \frac{1}{6} \)

### Final Probabilities
- Probability of A winning = \( \frac{4}{7} \)
- Probability of B winning = \( \frac{2}{7} \)
- Probability of C winning = \( \frac{1}{7} \)