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.

• If your number appears on one die only, you get your money back plus the same amount.
• If two dice show your number, you get your money back plus twice the amount you placed on the square.
• If your number appears on all three dice, you get your money back plus three times the amount.
• 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.

Hint:

Do not just add the three \(\frac{1}{6}\) chances together. Work out the exact probability of matching on 0, 1, 2, or 3 dice using \(6^3=216\) total outcomes, then weigh each payout against those probabilities.

Answer 1

The Correct Option is C

Step 1: Work out the correct probability of each outcome.
With three independent dice, there are \(6^3 = 216\) equally likely outcomes in total, not a simple sum of three separate \(\frac{1}{6}\) chances.
Probability your number appears on exactly one die: choose which one of the three dice shows it (3 ways), that die matches (probability \(\frac{1}{6}\)), and the other two dice must miss it (probability \(\frac{5}{6}\) each):
\[ P(\text{exactly 1}) = 3 \times \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{75}{216} \]
Probability your number appears on exactly two dice:
\[ P(\text{exactly 2}) = 3 \times \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} = \frac{15}{216} \]
Probability your number appears on all three dice:
\[ P(\text{exactly 3}) = \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216} \]
Probability your number appears on none of the dice:
\[ P(\text{none}) = \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{125}{216} \]
These add to \(75+15+1+125=216\), which checks out.

Step 2: Work out the operator's net gain or loss for a 1 Rupee bet, outcome by outcome.
If the number appears on none of the dice, the operator keeps the player's 1 Rupee: operator gains 1.
If the number appears on exactly one die, the operator pays 1 extra Rupee: operator loses 1.
If the number appears on exactly two dice, the operator pays 2 extra Rupees: operator loses 2.
If the number appears on all three dice, the operator pays 3 extra Rupees: operator loses 3.

Step 3: Combine these into the operator's expected earning per Rupee staked.
\[ E = \left(\frac{125}{216}\right)(1) + \left(\frac{75}{216}\right)(-1) + \left(\frac{15}{216}\right)(-2) + \left(\frac{1}{216}\right)(-3) \]
\[ E = \frac{125 - 75 - 30 - 3}{216} = \frac{17}{216} \]

Step 4: Convert to a percentage.
\[ \frac{17}{216} \approx 0.0787 = 7.87\% \approx 7.8\% \]

Final Answer:
The operator earns a small but real profit of about 7.8% on every Rupee bet. The naive claim that three separate \(\frac{1}{6}\) chances simply add up to a fair \(\frac{1}{2}\) ignores that the payout only grows 1, 2, or 3 times the stake, while matching your number twice or three times is far rarer than matching it once, so the odds favor the house.
\[ \boxed{\text{Operator profit} \approx 7.8\%} \]