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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?

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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:

[tex]EV=(0.5*20) - (0.5*20) = 0[/tex]

For the second gamble:

[tex]EV= (0.2*100) - (0.8*20) = 4[/tex]

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.

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