Probability and expected value

Question

You are in business and you are given two opportunities. The first you will have to invest $5000 but you could earn $25,000 less your initial investment. The second you only have to invest $2000 but then you will only earn $10,000 less your investment. The probability of getting $25,000 opportunity is 3/10 while the probability of getting $10,000 is 2/5. Which should you choose based on expected values?

Answer 1

The expected value is not simply the potential gain, but also the possibility of loss. You have to look at both in order to determine the expected value of the investment.

If you have 3/10 probability of gain, you also have a 7/10 probability of loss.

Your gain in option 1 is $20,000 × 0.3 = $6,000.

Your loss in option 1 is -$5,000 × 0.7 = -$3,500.

Remember, if the investment doesn't do well, you lose the whole $5,000.

Adding these together, you get +$2,500.

In the second option, you could make $8,000 ($10,000 - $2,000), but there is only a 2/5 probability of making that amount. That is $8,000 × 2/5 = $3,200.

Your loss could be -$2,000 × 3/5 = -$1,200.

Adding together these amounts, you get $2,000 for an expected payoff (value).

Clearly, $2,500 > $2,000, so the first option is better.

Does the logic make sense now? If not, where am I losing you?

Answer 2

Let the random variable X represent your return on the first investment. If I understood the problem description correctly, then there are two possibilities -- either X takes the value -$5000 (you invest $5000 but the investment doesn't "pay off") or X takes the value $20000 (you earn $25000 less your initial investment).

If X is a random variable taking one of n values x₁,..., x_n then the expected value is:

E[X] = x₁ P(X = x₁) + ... + x_n P(X = x_n)

so for this problem

E[X] = -$5000 · P(X = -$5000) + $20000 P(X = $20000)

= -$5000 · (7/10) + $20000 * (3/10)

= $2500

Similarly, if Y represents the return on the second investment, then Y = -$2000 or Y = $8000 and

E[Y] = -$2000 · P(Y = -$2000) + $8000 P(Y = $8000)

= -$2000 · (3/5) + $8000 · (2/5)

= $2000

Since E[X] > E[Y], you should choose the first investment opportunity.

All this is based on the assumption you can't get your initial investment back -- if you could then it would change your answer.

Answer 3

Hi Teresa. For two-case probability and expected value type questions like this, we are calculating how much we can expect to make in the future, based on two variables - magnitude of the opportunity and the likelihood that it will occur.  For this particular question, we look at the formula for expected value, E[x], given the probability, p(x), of earning x dollars:

E[x] = p(x)*x

For example, if I flipped a fair coin and promised you $20 if a heads came up, you can expect to earn:

E[x]= (.5) * ($20) = $10    (Notice .5 is the probability of a heads facing up)

Now for your example, there is a slight catch; for each opportunity, you are required to invest an initial amount (there is a 100% chance that you pay this initial amount) or E[x]=(1)* (- initial_investment) = - initial_investment (notice the negative sign because you are paying, not earning, this amount).

 

After calculations, we see E₂[x] > E₁[x], and we would rather choose Opportunity 2 because it is expected to earn a larger sum of money.