Question

Cal has a choice between two gambles. The first gamble offers a 50 percent chance of winning $20 and a 50 percent chance of losing $20. The second gamble offers a 20 percent chance of winning $100 and an 80% chance of losing $20. Which choice has the higher expected value?

Answer 1

Answer:

The second gamble has the higher expected value. EV = 4

Explanation:

In betting, expected value can be defined as (Amount won per bet * probability of winning) – (Amount lost per bet * probability of losing)

For the first gamble:

$EV=(0.5*20) - (0.5*20) = 0$

For the second gamble:

$EV= (0.2*100) - (0.8*20) = 4$

This means that Cal is expected to earn $4 for each $20 waged on the second gamble while he is expected to break even in the first gamble.

Therefore, the second gamble has the higher expected value.