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

A cleaner way to see the house edge is to compute the total amount the operator expects to hand back to the player per Rupee staked, and compare that to the Rupee actually staked, instead of computing the operator's gains and losses outcome by outcome.

1. List what the player receives in each case: nothing back if the number misses all three dice, 2 Rupees back (the stake plus 1) if it hits exactly one die, 3 Rupees back if it hits exactly two dice, and 4 Rupees back if it hits all three dice.
2. Weight each payout by its true probability, using $P(\text{none}) = 125/216$, $P(\text{one}) = 75/216$, $P(\text{two}) = 15/216$, and $P(\text{three}) = 1/216$.
3. Compute the expected payout: $E[\text{payout}] = \dfrac{125}{216}(0) + \dfrac{75}{216}(2) + \dfrac{15}{216}(3) + \dfrac{1}{216}(4) = \dfrac{0 + 150 + 45 + 4}{216} = \dfrac{199}{216}$.
4. Compare to the stake: the player risked a full Rupee, worth $216/216$ in the same units, but only expects $199/216$ back on average.

The shortfall is $\dfrac{216 - 199}{216} = \dfrac{17}{216} \approx 0.0787$, or about 7.8% of every Rupee staked. That 7.8% is exactly the operator's expected profit margin, since whatever the player expects to lose, the operator expects to gain.

Let's summarize:

• The naive claim that three $1/6$ chances add to a fair $1/2$ ignores that the reward only scales 1x, 2x, or 3x while the chance of a double or triple hit is tiny.
• Weighting every payout by its real probability shows the player recovers only about 92.2% of the stake on average.

So the true story is that the operator earns a profit of about 7.8% on every Rupee bet, not that the game is fair.