- Expected Value is the predicted loss or gain from an action over the long run.
- Expected value is usually associated with an event that will succeed or fail, and the costs associated with success or failure.
- Example 1, Problem 12, page 684
P(Success) = 60%, Profit on success = $80,000
P(Failure) = 40% Loss on failure = $20,000
E = P(success) * profit + P(failure) *( -loss)
E = .6 * $80,000 - .4 * $20,000 =
= $48,000 - $8,000
= $40,000
- Example 2
What is the expected value of a $1 bet on a single number on a two 0 roulette
wheel that pays 35 to 1.
P(success) = 1/38 = .026 Profit = $35
P(failure = 37/38 = .974 Loss= $1
E = .026 * $35 + 1 * .974
= .927 - .974
= -$0.05, or a 5 cent loss
- Notice the importance of "over the long run".
- Will you lose 5 cents on any one bet?
- If there are multiple possible payoffs, you must consider each.
- Example 3
What is the expected value of a dice game that where you toss two dice
for a $2 bet. You win as follows
$1 if you roll a 7, $2 if you roll a 2 or 12 and you lose otherwise.
P(success1) = 6/36 = .167 Profit1 = $1
P(success2) = 2/36 = .056 Profit2 = $2
P(failure) = 28/36 = .778 Loss = -2
E = .167 * 1 + .056 * 2 - .778*2
= .17 + .11 -1.56
= -$1.28
- The fair price = expected value + price to play
- Example 4, problem 26
P(success) = 1/1000 Profit = $800
P(failure = 999/1000 Loss = 1
E = 800 * 1/1000 - 999/1000
= -.099
= -$0.10
Fair price = expected value + price to play
= -.10 + 1.00
= $0.90
- Example 5, problem 28
P(E1) = 1/10,000 profit = $10,000
P(E2) = 1/10,000 profit = $5,000
P(E2) = 2/10,000 profit = $1,000
P(loss) = 9,996/10,000 loss = $5.00
E = 10,000/10,000 + 5,000/10,000 + 2,000/10,000 - (5 * 9996)/10,000
= (17,000 - 49,980) / 10,000
= -3.2
Fair price = 5.00 - $3.2
= $1.80