# Prob question on dice rolling and expected value

So let's say I roll 3 dice.

• If they show all the same number, I will EARN 10£
• If all numbers are different, I will LOSE 2£
• If they show two numbers equal and one different, I will EARN 5£

What would be my expected return per roll?

If we calculate the probabilities (if I am not wrong)

• 6/216 --> earn 10£
• 90/216 --> earn 5£
• 120/216 --> lose 2£

How would I proceed to calculate the expected value per roll?

EDIT

Would the EV be: (6/216)x10 + (90/216)x5 - (120/216)x2 = 1.25?

• Probability that all numbers are different $= \frac{6 \times 5 \times 4}{6 \times 6 \times 6} = \frac{120}{216}$. Probability that two are equal and one different is the remaining probability, that is, $= 1 - \frac{120}{216} - \frac{6}{216} = \frac{90}{216}$. Commented Dec 9, 2020 at 22:22
• Thank you!! how would I calculate the expected value per roll? Commented Dec 9, 2020 at 23:04
• I have edited the Q with my guess, is that right? Commented Dec 9, 2020 at 23:10
• Yes that's right. Commented Dec 9, 2020 at 23:10

By definition for discrete random variable, the expected value is the sum of the product of possible values of the random variable with it's probability. Say, X can be $$10$$ with probability $$\frac{1}{2}$$, X can be $$\left(-100\right)$$ with probability $$\frac{1}{4}$$, and X can be $$0$$ with probability $$\frac{1}{4}$$. Then the expected value is $$10 \cdot\frac{1}{2} + \left(-100\right)\cdot\frac{1}{4} + 0\cdot\frac{1}{4} = \left(-20\right)$$
So, the expected value is $$10 \cdot\frac{6}{216} + 5\cdot\frac{90}{216} + \left(-2\right)\cdot\frac{120}{216} = \frac{270}{216}$$