A fair coin is tossed until one of the two sides occurs twice in a row. The probability that the number of tosses required is even is

Options

A. `1//3`
B. `2//3`
C. `1//4`
D. `3//4`

Explanation

To solve the problem, we need to find the probability that the number of tosses required to get either heads or tails twice in a row is even. Let's break this down step by step.

### Step 1: Understanding the Problem
We are tossing a fair coin until we get either heads (H) or tails (T) twice in a row. We want to find the probability that the total number of tosses is even.

### Step 2: Identifying Possible Outcomes
The possible sequences of tosses that lead to two consecutive heads (HH) or tails (TT) can be represented as follows:
- HH (2 tosses)
- HTT (3 tosses)
- HHT (3 tosses)
- TTH (3 tosses)
- TT (2 tosses)
- THT (3 tosses)
- THH (3 tosses)
- HTH (4 tosses)
- HTHT (4 tosses)
- THTH (4 tosses)
- THTHT (5 tosses)
- HTHTH (5 tosses)
- ... and so on.

### Step 3: Finding the Probability of Even Tosses
To find the probability of getting an even number of tosses, we can analyze the sequences that result in an even number of tosses.

1. **Even Toss Outcomes**:
- The sequences that result in an even number of tosses are:
- HH (2 tosses)
- TT (2 tosses)
- HTHH (4 tosses)
- HTHT (4 tosses)
- THTH (4 tosses)
- TTHH (4 tosses)
- ... and so forth.

2. **Probability Calculation**:
- The probability of getting HH or TT in 2 tosses is:
\[ P(2) = P(HH) + P(TT) = \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]
- For 4 tosses, the sequences can be calculated similarly, and we can observe that the pattern continues.

3. **Generalizing the Probability**:
- The probability of getting an even number of tosses can be expressed as an infinite geometric series:
\[ P(\text{even}) = 2 \left( \frac{1}{2^2} + \frac{1}{2^4} + \frac{1}{2^6} + \ldots \right) \]
- This series can be summed up using the formula for the sum of an infinite geometric series:
\[ S = \frac{a}{1 - r} \]
where \( a = \frac{1}{4} \) (first term) and \( r = \frac{1}{4} \) (common ratio).

4. **Calculating the Sum**:
\[ S = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \]
Therefore, the probability of getting an even number of tosses is:
\[ P(\text{even}) = 2 \times \frac{1}{3} = \frac{2}{3} \]

### Final Answer
The probability that the number of tosses required is even is \( \frac{2}{3} \).

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