A coin is tossed until a head appears. What is the expectation of the number of tosses required ?

Answer

To find the expectation of the number of tosses required until a head appears when tossing a coin, we can follow these steps:

### Step 1: Define the Random Variable
Let \( X \) be the random variable representing the number of tosses required to get the first head.

### Step 2: Identify the Probabilities
When tossing a fair coin:
- The probability of getting a head (H) on any toss is \( P(H) = \frac{1}{2} \).
- The probability of getting a tail (T) is \( P(T) = 1 - P(H) = \frac{1}{2} \).

### Step 3: Calculate the Expected Value
The expected value \( E(X) \) can be calculated using the formula:
\[ E(X) = \sum_{n=1}^{\infty} n \cdot P(X = n) \]
Where \( P(X = n) \) is the probability that the first head appears on the \( n \)-th toss.

### Step 4: Determine \( P(X = n) \)
For the first head to appear on the \( n \)-th toss:
- The first \( n-1 \) tosses must be tails, and the \( n \)-th toss must be a head.
- Therefore, the probability is:
\[ P(X = n) = \left(\frac{1}{2}\right)^{n-1} \cdot \frac{1}{2} = \frac{1}{2^n} \]

### Step 5: Substitute into the Expected Value Formula
Now we substitute \( P(X = n) \) into the expected value formula:
\[ E(X) = \sum_{n=1}^{\infty} n \cdot \frac{1}{2^n} \]

### Step 6: Use the Formula for the Sum of a Series
To compute this sum, we can use the formula for the expected value of a geometric distribution:
\[ E(X) = \frac{1}{p} \]
where \( p \) is the probability of success (getting a head). Here, \( p = \frac{1}{2} \), so:
\[ E(X) = \frac{1}{\frac{1}{2}} = 2 \]

### Conclusion
The expected number of tosses required to get the first head is \( 2 \).

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