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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?

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    $\begingroup$ 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}$. $\endgroup$
    – sudeep5221
    Commented Dec 9, 2020 at 22:22
  • $\begingroup$ Thank you!! how would I calculate the expected value per roll? $\endgroup$
    – Tim Solor
    Commented Dec 9, 2020 at 23:04
  • $\begingroup$ I have edited the Q with my guess, is that right? $\endgroup$
    – Tim Solor
    Commented Dec 9, 2020 at 23:10
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    $\begingroup$ Yes that's right. $\endgroup$ Commented Dec 9, 2020 at 23:10

1 Answer 1

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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 $10 \cdot\frac{6}{216} + 5\cdot\frac{90}{216} + \left(-2\right)\cdot\frac{120}{216} = \frac{270}{216}$

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