A Chuck-a-luck variant

Question

You're thinking of trying a variant of Chuck-a-luck. In this variant, you choose one number in the range $1$ to $6$ and five dice are rolled. These are the payouts:

Number of dice           
equal to your  Your         
number         payout    
---------------------
None          - 5 USD 
   1            0 USD        
   2          + 5 USD   
   3          +10 USD 
   4          +15 USD 
   5          +20 USD 

So for example if none of the dice show your number, you pay $\$5$, while if exactly 1 die shows your number, you don't pay or receive anything.

You suspect the odds are against you but you want to know exactly how much. You've also had a few too many, so you don't want to have to resort to inclusion-exclusion or anything complicated.

What is a quick and simple way to calculate the expected value of this game?

Answer 1

The expected value is

$-5/6$ USD

The payoffs in the table follow the formula

   payoff =   5 * (# dice equal to your number    minus 1)

Since there are six possible outcomes per dice, for a single die the expected number of dice equal to your number is $1/6$. By linearity of expectation the expected number of dice equal to your number is $5/6$, and $5\cdot(5/6-1)=-5/6$.