There is a dice game at the casino that costs $3 to play. A player rolls two dice. If the player gets a sum of 2 or 12 he or she gets back $ 28 (wins $25). the person gets a 7he or she gets back $5 (wins $2). If the player rolls anything other than a 2, 12 or 7 he or she gets nothing back (loses $3). Find the expected value for the player?

The expected value for the player is -$0.63.

Step-by-step explanation:

The dice game has the following rules:

The outcomes for a sum of 2 or 12 are: {(1, 1) and (6, 6)}

The probability of a sum of 2 or 12 is: 2/36 = 0.055.

The outcomes for a sum of 7 is: {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1)}

The probability of a sum of 7 is: 6/36 = 0.167.

The probability of a sum other than 2, 12 or 7 is: 1 - 8/36 = 0.778

The probability distribution is:

        Result          X         P (X)

Sum 2 or 12        $25       0.055

   Sum 7               $2        0.167

   Others             -$3       0.778

    Total                             1.000

Compute the expected value as follows:

Thus, the expected value for the player is -$0.63.

2x^2+5x+17

your welcome. : )

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