Calculating dice probability and risk-reward value

Question

Having trouble how to model this cost-risk question with Probability.

Assume that you are given 5 standard six-sided dice, what's the probability you get at least three 2s?

I have calculated this:

There are 6^5 = 7776 events possible.

Bad events include rolling exactly 0 2's, 1 2's, or 2 2's.

Good events are rolling exactly 3 2's, 4 2's, or 5 2's.

Thus the probability that I roll at least three 2's is the sum of the probabilities of the good events.

Answer: P = 276/7776.

But now here is a follow-up: A casino will pay 1000 dollars if I roll at least three 2's, but I have to pay the casino 200 dollars if I don't. Does it make sense to take the bet? Why or Why not?

I'm unsure of how to model the reward-risk of cash given the calculated probabilities?

Answer 1

For your final question:

What you're dealing with here is called Expected Value.

For some event $X$ with a set of possible outcomes $S$, we denote the Expected Value $E(X)$ as $$E(X) = \sum\limits_{s \in S} p(s) \cdot X(s)$$ where $p(s)$ is the probability of this outcome, and $X(s)$ is the value of this outcome.

Answer 2

On the follow-up, we can write the expected value as follows:

$E(x) = \sum x\cdot P(x) = 1000\cdot \frac{276}{7776} - 200\cdot \frac{7500}{7776} = 200 \cdot \frac{1380-7500}{7776} <0$. So this is a bad deal!