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 of finding the probability that the number of tosses required to get two consecutive heads (HH) or tails (TT) is even, we can follow these steps:

### Step 1: Define the Problem
We are tossing a fair coin until we get either HH or TT. We need to find the probability that the total number of tosses is even.

### Step 2: Identify Possible Outcomes
The possible sequences of coin tosses that can lead to HH or TT can be represented as follows:
- For two tosses: HH, TT
- For four tosses: HTHH, HTTH, TTHH, THTT, HTHT, THTH, TTHH, THTT
- For six tosses: HTHHTT, HTHTHT, THTTHT, etc.

### Step 3: Calculate Probabilities for Even Tosses
We can express the probability of getting two consecutive heads or tails in terms of even tosses.

1. The probability of getting HH or TT in 2 tosses is:
- P(2) = P(HH) + P(TT) = (1/2 * 1/2) + (1/2 * 1/2) = 1/4 + 1/4 = 1/2

2. For 4 tosses, we can derive the sequences:
- The sequences that lead to HH or TT after 4 tosses can be calculated similarly. The probability of reaching HH or TT after 4 tosses can be calculated as:
- P(4) = (1/2)^4 * number of favorable sequences = (1/16) * 6 = 3/8

3. For 6 tosses, we can continue this pattern:
- P(6) = (1/2)^6 * number of favorable sequences = (1/64) * 12 = 3/16

### Step 4: Generalize the Pattern
From the above calculations, we can see a pattern forming. The probability of getting HH or TT in an even number of tosses can be represented as:
- P(even) = P(2) + P(4) + P(6) + ...

This can be expressed as:
- P(even) = 1/2 + 3/8 + 3/16 + ...

### Step 5: Use the Geometric Series Formula
The series can be summed up using the formula for an infinite geometric series:
- S = a / (1 - r)
Where:
- a = first term of the series
- r = common ratio

In our case:
- a = 1/2
- r = 1/4 (as each term is multiplied by 1/4)

Thus, we can calculate:
- S = (1/2) / (1 - 1/4) = (1/2) / (3/4) = (1/2) * (4/3) = 2/3

### Final Step: Conclusion
Therefore, the probability that the number of tosses required is even is:
- **P(even) = 2/3**

### Answer
The correct answer is option 2: **2/3**.

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