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A carnival game consists of three dice in a cage. a player can bet a dollar on any of the numbers 1 through 6. the cage is shaken, and the payoff is as follows. if the player's number doesn't appear on any of the dice, he loses his dollar. otherwise, if his number appears on exactly k of the three dice, for k = 1, 2, 3, he keeps his dollar and wins k more dollars. what is his expected gain from playing the carnival game once?

Answer :

Let's find what could happen to his dollar.
If the number doesn't appear on any dice, he would lose his dollar (-$1). The probability for this event is 5/6 * 5/6 * 5/6 = 125/216

If the number appears on one of the dice, he would earn $2. The probability for this event is 1C3 * 1/6 * 5/6 * 5/6 = 25/72

If the number appears on two of the dice, he would earn $3. The probability for this event is 2C3 * 1/6 * 1/6 * 5/6 = 5/72 

If the number appears on all those dice, he would earn $4. The probability for this event is 1/6 * 1/6 * 1/6 = 1/216

So the expected gain would be -1 * 125/216 + 2 * 25/72 + 3 * 5/72 + 4 * 1/216 = $0.34

Thus the answer is $0.34


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