Q:

How does expected value work with multiple probabilities?

I'm familiar with how expected value works normally (EV = probability * value of event) but I'm not sure how to handle a situation with multiple probabilities.
As a simple example, let's say that I'm trying to calculate how much time I will take to park a car. I can either park at the end of the road, which takes only 1 minute or I can park in a big parking garage but that will take me at least 10 minutes of driving around in circles and riding the elevator.
Due to the time of day, there's an 80% chance that I'll snag a spot at the end of the road. The parking garage is huge so there's a 100% chance of getting a spot there.
How would this work?
Feel free to ask for additional information.

A:

Andre's answer is correct if it takes you exactly 10 minutes to get from the end of the road to the parking space in the lot and you flip a coin to decide which way to go.  But you said it takes at least ten minutes. If the is a continuous distribution starting at 10 minutes then it would change.  Also from the way you worded the problem i think you will automatically try the end of the road first and then move on to the parking lot if a space is not available (which you said happens with probability 0.2).  The answer to this problem is 1+(0.2)∫xf(x)dx where f is the density of times to go from the end of the road to the parking space in the lot. The limits of integration are 10 to ∞.  In practical terms there is probably a time T > 10 at which a parking spaced would have been found no matter how much circling you had to do.  So f is 0 beyond T and you could make the upper limit T.  Note that the result depends on what you select for the density f. This is the additional information needed that Robert referred to.

