There are three offers

Offer A - $ 5 probability of redemption A - P (A) = 0.5

Offer B - $ 4 Probability of redemption B – P(B) = 0.6

Offer C - $ 3 probability of redemption C – P(C) = 0.7

If I send only offer A, Expected Value is 0.5 * 5 = $2.5

If I send only offer B, Expected Value is 0.6 * 4 = $2.4

If I send only offer C, Expected Value is 0.7*3 = $2.1

Expected value of sending All three ?? is it E(A)+E(B)+E(C) = 7 ? Am I doing something wrong here.

What will be the overall expected value if I send all three offers. The customers can redeem any combinations of offers ie only A, only B ,only C, Both AB, both BC, both CA, ABC and none


2 Answers 2


There are some things unclear about the question, but you should review basic properties of expectation, which I would guess you're expected to use here.

For example, see wikipeda to review properties of the expected value.

You might also need to consider other properties, such as the law of total expectation.


A good place to start is to realise that the sum of all probabilities should be 1, this is a basic property of probabilities. In your case P(A) + P(B) + P(C) + P(AB) + P(BC) + P(CA) + P(ABC) = 1. This must be true, since the customer does not have any other choice! You might want to consider adding P(D), the chance that no offer is taken.

The expectation is then the sum over all events of (probability * the dollar value).

  • $\begingroup$ This is a difficult (actually, impossible) way to solve an easy problem: expectations add, as stated in the question. It is impossible because you require the calculation of unknown probabilities. $\endgroup$
    – whuber
    May 7, 2015 at 22:08

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