# Probability Help expected value

• Jan 28th 2007, 02:08 PM
air2slb
Probability Help expected value
Will someone help me step by step on this problem. Im sure its not as hard as I make it out to be.

When you bet \$6.00 in a common gamble, there is a probability of 133 that you will win \$6.00 and a probablilty of 138 that you will lose 271
What is the expected value?

Thanks!!!!!!!!
• Jan 28th 2007, 02:15 PM
CaptainBlack
Quote:

Originally Posted by air2slb
Will someone help me step by step on this problem. Im sure its not as hard as I make it out to be.

When you bet \$6.00 in a common gamble, there is a probability of 133 that you will win \$6.00 and a probablilty of 138 that you will lose 271
What is the expected value?

Thanks!!!!!!!!

are always between 0 and 1, and the probability of a favourable outcome
and that of all other outcomes sum to 1.

Here we have probabilities of 133 and 138, which in terms of probabilities
make no sense. Also when you lose what is the 271 you lose?

RonL
• Jan 28th 2007, 02:26 PM
air2slb
The problem posted funny. Its 133 over 271 and 138 over 271 133/271 and 138/271. Does that make more sense??? Thanks!!!
• Jan 28th 2007, 03:49 PM
chogo
ok so the expectation simply represents the average amount one "expects" as the outcome of the random trial when identical odds are repeated many times.

so if we have a random variable R that has values -> x1, x2
with probabilities -> p1,p2

The expected value of R

E(R) = p1x1 + p2x2

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So here we have two probabilities
p1 = 133/271 (win)
and
p2 = 138/271 (loose)

The values of x which go with these are

x1 = 6 (amount if u win)
x2 = -6 (amount if u loose)

thereore the expected value = (133/271)*6 + (138/271)*-6 = -0.110701107

Which means on average you loose 11 cents for every 6 dollar bet.

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