Question 143

# 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?

Solution

$$\ \ \frac{\ 5\times\ 5\times\ 5}{6\times6\ \times6\ \ }=\ \frac{\ 125}{216}$$Let the player bet Re 1 on any particular number.

If exactly 1 die shows the number, the player will get Rs 2. (Operator has to lose Re 1 from his side). Probability of it is $$\ \frac{\ 1\times\ 5\times\ 5\times\ 3}{6\times\ 6\times\ 6}=\ \ \frac{\ 75}{216}$$

If exactly 2 die shows the number, the player will get Rs 3. (Operator has to lose Rs 2 from his side). Probability of it is $$\ \frac{\ 1\times\ 1\times\ 5\times\ 3}{6\times\ 6\times\ 6}=\ \frac{\ 15}{216}$$

If all die shows the number, the player will get Rs 4. (Operator has to lose Rs 3 from his side). Probability of it is $$\ \ \frac{\ 1\times\ 1\times\ 1}{6\times6\ \times6\ \ }=\ \frac{\ 1}{216}$$

If no die shows the number, the player will get Rs 0. (Operator will gain Re 1). Probability of it is $$\ \ \frac{\ 5\times\ 5\times\ 5}{6\times6\ \times6\ \ }=\ \frac{\ 125}{216}$$

Overall expected earning of Operator will be = $$\left(-1\right)\times\ \ \frac{\ 75}{216}+\left(-2\right)\times\ \ \frac{\ 15}{216}+\left(-3\right)\times\ \ \frac{\ 1}{216}+\left(1\right)\ \frac{\ 125}{216}$$

Overall Earning Per rupee is $$\ \frac{\ 17}{216}=\ 0.078$$

His earning per rupee is 7.87%

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