In: Statistics and Probability
In the carnival game chuck-a-luck, three dice are rolled. You make a bet on a particular number (1, 2, 3, 4, 5, 6) showing up. The payout is 1 to 1 if that number shows on (exactly) one die, 2 to 1 if it shows on two dice, and 3 to 1 if it shows up on all three. (You lose your initial stake if your number does not show on any of the dice.) If you make a $1 bet on the number three, what is your expected net winnings? (answer to 3 decimal places)
Total number of combinations for 3 dice throw is computed here as:
= 6*6*6 = 216
Total number of combinations with exactly one of the dice having the number three
= Number of ways to select the dice with number three * Number of ways that the chosen dice will have a three * Number of ways that the other two dice dont have a three
= 3*1*5*5
= 75
Total number of combinations with exactly two of the dice having the number three
= Number of ways to select two dice with number three * Number of ways that the chosen two dice will have a three * Number of ways that the other dice dont have a three
= 3*1*1*5
= 15
And finally there is only 1 way that all three dice have the number three
Therefore the expected payout amount is computed here as:
P(X = 1) = 75/216
P(X = 2) = 15/216
P(X = 3) = 1/216
Therefore the expected winning amount here is computed as:
= Expected Payout - 1
= (1*75 + 2*15 + 3*1) / 216 - 1
= 108 / 216 - 1
= 0.5 - 1 = -0.5
Therefore the expcted payout is $0.5, but the expected loss from the game is $0.5 because of the fee of $1.