A player tosses 3 fair coins and wins Rs. 8 if three heads occur, Rs. 3 if 2 heads occur and Rs. 1 if one head occurs. If the game as to be fair (i.e., expected value is zero) how must should he lose, if no head occur ?

Answer

To solve the problem step by step, we will calculate the expected value of the game and determine how much the player should lose if no heads occur to make the game fair (i.e., the expected value is zero).

### Step 1: Determine the total number of outcomes when tossing 3 coins.
When tossing 3 coins, each coin has 2 possible outcomes (Head or Tail). Therefore, the total number of outcomes is given by:

\[ \text{Total Outcomes} = 2^3 = 8 \]

**Hint:** Remember that each coin has 2 outcomes, and when multiple coins are tossed, you multiply the outcomes.

### Step 2: List all possible outcomes.
The possible outcomes when tossing 3 coins are:
1. HHH (3 Heads)
2. HHT (2 Heads, 1 Tail)
3. HTH (2 Heads, 1 Tail)
4. THH (2 Heads, 1 Tail)
5. HTT (1 Head, 2 Tails)
6. THT (1 Head, 2 Tails)
7. TTH (1 Head, 2 Tails)
8. TTT (0 Heads)

**Hint:** Make sure to account for all combinations of Heads and Tails.

### Step 3: Calculate the probabilities for each outcome.
- Probability of getting 3 Heads (HHH):
\[ P(3H) = \frac{1}{8} \]
- Probability of getting 2 Heads (HHT, HTH, THH):
\[ P(2H) = \frac{3}{8} \]
- Probability of getting 1 Head (HTT, THT, TTH):
\[ P(1H) = \frac{3}{8} \]
- Probability of getting 0 Heads (TTT):
\[ P(0H) = \frac{1}{8} \]

**Hint:** The probability of an event is the number of favorable outcomes divided by the total number of outcomes.

### Step 4: Calculate the expected winnings for each outcome.
- Winnings for 3 Heads: Rs. 8
- Winnings for 2 Heads: Rs. 3
- Winnings for 1 Head: Rs. 1
- Winnings for 0 Heads: Let’s denote the loss as \( L \).

Now, calculate the expected value (EV):

\[ EV = P(3H) \cdot 8 + P(2H) \cdot 3 + P(1H) \cdot 1 + P(0H) \cdot (-L) \]

Substituting the probabilities:

\[ EV = \left(\frac{1}{8} \cdot 8\right) + \left(\frac{3}{8} \cdot 3\right) + \left(\frac{3}{8} \cdot 1\right) + \left(\frac{1}{8} \cdot (-L)\right) \]

**Hint:** The expected value is the sum of the products of each outcome's value and its probability.

### Step 5: Simplify the expected value expression.
Calculating each term:

\[ EV = 1 + \frac{9}{8} + \frac{3}{8} - \frac{L}{8} \]

Combine the terms:

\[ EV = 1 + \frac{12}{8} - \frac{L}{8} = 1 + 1.5 - \frac{L}{8} = 2.5 - \frac{L}{8} \]

**Hint:** Ensure you combine like terms carefully.

### Step 6: Set the expected value to zero for a fair game.
For the game to be fair, we set the expected value to zero:

\[ 2.5 - \frac{L}{8} = 0 \]

### Step 7: Solve for \( L \).
Rearranging gives:

\[ \frac{L}{8} = 2.5 \]

Multiplying both sides by 8:

\[ L = 2.5 \times 8 = 20 \]

### Conclusion:
The player should lose Rs. 20 if no heads occur to make the game fair.

**Final Answer:** The player should lose Rs. 20 if no heads occur.