Question

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?

Answer 1

The expected value for the player is -$0.63.

Explanation:

The dice game has the following rules:

• The game 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).
• If the person gets a 7 he 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).

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:

`E(X)=\sum x\cdot P(x)\\\\`

`=(25* 0.055)+(2* 0.167)+(-3* 0.778)\\\\=1.375+0.334-2.334\\\\=-0.625\\\\\approx -0.63`

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