What is the expected outcome of a game of chance with varying probabilities?

[aprilryan]

I have one more probability problem for you. I just don't know where to begin with this one.

"In a game of chance you have a 30% chance of winning $\$5$, a 50% chance of losing $\$4$ and a 20% chance of breaking even. What is your expected profit or loss?"

 

[Ackbach]

So you need to take each outcome (positive = winning, negative = losing), multiply each one by its probability, and add it all up.

 

[aprilryan]

Let me get this straight:

I multiply 5 times 30 percent or 0.3
I also multiply 4 times 50% or 0.5
Then I add the sums of these two correct?

What do I do with the 20%?

 

[Siron]

Break even means no profit nor loss. Let $X$ be your profit, then $X$ is a random variable defined as (watch the signs!):$$ X = \left \{ \begin{array}{lll} 5, \quad & p = 0.3 \\ -4, \quad & p = 0.5 \\ 0, \quad & p = 0.2 \end{array} \right. $$Then the expected profit is given by $\mathbb{E}[X] = 5(0.3)+(-4)(0.5) = - 0.5$. This means that you expect a negative profit or hence a loss of $0.5 \$ $.

On the other hand, it's completely similar to say, let $Y$ be the loss, then $Y$ is a random variable defined as:$$ Y = \left \{ \begin{array}{lll} -5, \quad & p = 0.3 \\ 4, \quad & p = 0.5 \\ 0, \quad & p = 0.2 \end{array} \right. $$The expected loss is given by $\mathbb{E}[Y] = -5(0.3)+4(0.5) = 0.5$. Hence your expected loss is $0.5 \$$ which is exactly the same answer as above. It does not matter how you define the r.v as a loss of profit as long as you understand that a negative profit is a loss and vice versa.

 

[aprilryan]

Thanks Siron, this really helped break down everything for me! Thank you Ackbach for your assistance as well!