Carnival Game Expected Value

Question

Imagine a carnival game where you get to randomly pull numbered balls from a bag. The bag contains 100 balls uniquely numbered from 1-100. The game is $1 to try. If you pull any number 1-99, nothing happens. If you pull the ball labeled 100, you win $10,001.

 

Now imagine a second version of this game using the same bag with the same balls. Again, the game is $1 to try. This time if you pull any number 1-99, you win $101. If you pull the ball labeled “100”, nothing happens.

Answer 1

The expected value is calculated by finding the sum of each value multiplied by the chances of that value happening.

 

For the firth game, there are two possibilities. Either you lose $1, or win $10,000.

The chances of losing is 99/100, and the chance of winning is 1/100

 

The expected value for the first game would be -1*99/100 + 10,000*1/100 = $99.01

 

In the second version of the game, there are also two possibilities. You can lose $1, or win $100.

 

The chances of losing is now 1/100, and the chance of winning is 99/100.

The expected value for the second game is -1*1/100 + 100*99/100 = $98.99.

 

When playing the first game, you can "expect" to win $99.01. When playing the second game, you can "expect" to win $98.99. Surprisingly, you would be better off playing the first game than the second.