Question on expected value

Question

I think my brain is officially fried lol. Any help is greatly appreciated.

 

 A game at a fund-raiser carnival involves throwing a dart at a wall of balloons. If the balloon that is popped has money inside, the player wins the money. Suppose 70% of the balloons have no money inside, 15% have $2 inside, 8% have $5 inside, 4% have $10 inside, 2% have $20 and the remaining 1% have $100 bills inside. The player has to pay $1.50 per game. What is the expected value if a player choose to play this game? (Assume that each player pops exactly one balloon.)

Answer 1

OK, so, let's go step by step.

 

You have a 70% chance of losing $1.50.

 

0.7(-1.50) = -1.05

 

You have a 15% chance of gaining $0.50.

 

0.15(0.5) = 0.075

 

You have an 8% chance to gain $3.50.

 

0.08(3.5) = 0.28

 

Keep going, then add all your probabilities.

 

-1.05 + 0.075 + 0.28...

 

The sum, rounded to the nearest hundredth, is your answer.

Answer 2

To compute expected value, we take each event, multiply its probability by its net payment, and add the resulting numbers. When calculating the payment, remember that you pay $1.50 to play, so this subtracts from any winnings you earn.

 

There is a 70% chance of popping an empty, so you lose $1.50. So we have 0.7·(-1.5).

There is a 15% chance of popping a $2 balloon, which nets you $0.50 after the fee to play. So we have 0.15·(0.5). Note that this number is positive, since we earn money doing this.

 

Doing this for the remaining numbers gives a sum of 0.7·(-1.5)+0.15·0.5+0.08·3.5+0.04·8.5+0.02·18.5+0.01·98.5=1

 

So on average, you will make a profit of $1/per on this game.