The Expected Value

Question

A gambler, proposes the following game of chance. You roll a fair die. If you roll a 1, then you receive $40. If you roll a 2, you receive $20. If you roll a 3, you receive $15. If you roll a 4 or a 5, you receive $5, and if you roll a 6, you receive nothing. What is the break-even cost for this game ?

 

 

This is what I got so far:

 

0 = (1/20)(40) + (2/20)(20) + (3/20)(15) + _____________ ? = E.V ?

      Receive         Receive        Receive        Receive Nothing

 

 

But I don't even know if that's right. 

Answer 1

Hey there!

You are on the right track! Remember that the expected value is the sum of the outcomes multiplied by the probabilities of those outcomes. So, for example: If you roll a 1, you get $40. So that is the outcome. However, the probability of rolling a 1 is not 1/20 as you have it listed, but it is 1/6 (1 side of a 6 sided die). So the first part would be:

(1/6)(40). Then, do this for every number and add them. 

 

Note that the expected value will be greater than 0. This makes sense because there is no way for you to lose money, ie. paying $100 if you roll a 6. When the question asks for a "break-even cost", it is asking what value would you have to pay in order to, in the long run, end up gaining or losing 0 money. Remember that the expected value would represent the amount of money that you would win, on average, per play. This means that the "break-even cost" would be the same as the expected value. In other words, if you expected to win around $18 per game, you would have to be paying $18 per game to play in order to break even.