Question

On the counter are six squares marked 1, 2, 3, 4, 5, 6. Players are invited to place as much money as they wish on any one square. Three dice are then thrown.
1 If your number appears on one die only, you get your money back plus the same amount.
1 If two dice show your number, you get your money back plus twice the amount you placed on the square.
1 If your number appears on all three dice, you get your money back plus three times the amount.
1 If the number is not on any of the dice, the operator gets your money.
For example, suppose that you bet one Rupee on square No. 6. If one die shows a 6, you get your Rupee back plus another Rupee. If two dice show 6, you get back your Rupee plus two Rupees. If three dice show 6, you get your Rupee back plus three Rupees.
From a player’s point of view, the chance of his number showing on one die is \(\frac{1}{6}\) , but since there are three dice, the chances must be \(\frac{3}{6}\) or \(\frac{1}{2}\) , therefore the game is a fair one. Of course this is the way the operator of the game wants everyone to reason, for it is quite fallacious. What is the probable story ?

• A. Operator gets a profit of 6% on each Rupee bet.
• B. Operator suffers a loss of 7.8% on each Rupee bet.
• C. Operator gets a profit of 7.8% on each Rupee bet.
• D. The player suffers a loss of 6% on each Rupee bet.

Answer 1

The Correct Option is C

To solve this problem, we need to calculate the expected value for the player and compare it to the amount bet, which will reveal the operator's profit or loss percentage.
Given that we are dealing with dice, let's consider the various scenarios:

• The probability of a specific number appearing on one die is \( \frac{1}{6} \).
• The probability of a specific number not appearing on one die is \( \frac{5}{6} \).

Now, considering that there are three dice, the events are independent, leading to:

• Probability of the number appearing on exactly one die:
  \( C(3,1) \times \left(\frac{1}{6}\right)^1 \times \left(\frac{5}{6}\right)^2 = 3 \times \frac{1}{6} \times \frac{25}{36} = \frac{75}{216} \)
• Probability of the number appearing on exactly two dice:
  \( C(3,2) \times \left(\frac{1}{6}\right)^2 \times \left(\frac{5}{6}\right)^1 = 3 \times \frac{1}{36} \times \frac{5}{6} = \frac{15}{216} \)
• Probability of the number appearing on all three dice:
  \( C(3,3) \times \left(\frac{1}{6}\right)^3 = \frac{1}{216} \)
• Probability of the number not appearing on any of the dice:
  \(\left(\frac{5}{6}\right)^3 = \frac{125}{216} \)

Let's calculate the expected return for the player per 1 Rupee bet:

• Return from one die match:
  \( 2 \times \frac{75}{216} = \frac{150}{216} \)
• Return from two die match:
  \( 3 \times \frac{15}{216} = \frac{45}{216} \)
• Return from three die match:
  \( 4 \times \frac{1}{216} = \frac{4}{216} \)

The expected value, \(E\), for the player is:
\(E = \frac{150}{216} + \frac{45}{216} + \frac{4}{216} = \frac{199}{216} \approx 0.9213\)
The operator's profit per Rupee bet is thus:
\(1 - 0.9213 = 0.0787\) or approximately 7.8%
Therefore, the correct conclusion is: Operator gets a profit of 7.8% on each Rupee bet.