Navigation Panel:             (These buttons explained below) Question Corner and Discussion Area

# Interpreting the "Expected Number"

Asked by Steve Yuen on Monday May 27, 1996:
We are having a problem with the solution to the following problem.

QUESTION:

The DJ of the local radio show conducts a contest in which the listeners phone in the answer to the contest question. If the DJ estimates that the probability of a caller giving the correct answer is 0.28,

a) What is the probability that the third caller is the winner?
b) What is the expected number of calls BEFORE a correct answer is received?

Part A

Calculation: 0.72 X 0.72 X 0.28 =0.145152

Therefore, the probability that the third caller is the winner is 0.145152

Part B

The formula of the expected waiting time is E(X)=q/p (Geometric Distribution). p is the probability of success on each trail. q =1-p is the probability of failure on each trail. X=0,1,2,. . .

Calculation: By the formula E(X)=q/p

The expected waiting time = 0.72/0.28 is about 2.571428571

Here is where the interpretation problem comes

Since the expected waiting time is greater than 2.5 then we expect the third caller to get the correct answer. Therefore, the expected number of the calls before a correct answer = 2

Solution # 2

Since the expected waiting time (average number of failures before success) is closer to 3 therefore more often than not the fourth caller will be the first successful one. In this case we have waited through 3 calls.

Which (if any) do you think is the correct answer? (2 or 3)

I think the issue here is the meaning of "expected number of calls". The correct answer to the question that was asked is the number 2.57... that you calculated; that's the expected number of calls.

It's important to realize that this "expected number" is not an integer (so it's wrong to round it to either 2 or 3); it is not the actual outcome of any contest. Rather, it means that if you run the contest a large number of times (say around 100 times), you'd expect the total number of calls (failures before success) for all these contests to be around 257; if you ran it 1000 times, you'd expect the number to be around 2571; and so on.

It's like saying that the average number of children per family is 2.5; that doesn't mean that "the average family has 2.5 children". It doesn't mean that a typical family you meet on the street will have 2.5 children; that can't happen, since families have whole numbers of children. Instead, it refers to the average over all families.

Both Solution #1 and Solution #2 are attempting to interpret "expected value" as "what do I expect will happen in one particular contest?", and the expected value doesn't say what you might think it says about that. Here are three examples to illustrate this:

1. Suppose that half of the time there are 2 calls and the other half of the time there are 3 calls. The expected number of calls is 2.5.

2. Suppose that three quarters of the time there are 2 calls and one quarter of the time there are 4 calls. The expected number of calls is (3/4)(2) + (1/4)(4) = 2.5.

3. Suppose that one quarter of the time there is 1 call and three quarters of the time there are 3 calls. The expected number of calls is (1/4)(1) + (3/4)(3) = 2.5.

In each case there is the same expected number of calls. But what's most likely to happen is very different for each situation: in situation 2, it's most likely that there'll be only 2 calls (this happens 75% of the time), while in situation 3, it's most likely there'll be 3 calls (this happens 75% of the time). So the expected value does not tell you what's most likely to happen.

To put it another way: suppose you got 1 call 99% of the time, and 99901 calls 1% of the time. The expected number of calls is 1000, even though the most likely outcome is only 1 call, with 99% probability! A large number of calls occurring with low probability can contribute just as much to the expected value as a few number of calls occurring with high probability.

If you really want to think about what's most likely to happen in the DJ example, you need to think about the individual probabilities and not the expected number:

Probability of 0 calls before first successfull call: 0.28
Probability of 1 call before first successfull call: 0.2016
Probability of 2 calls before first successfull call: 0.1451. . .
Probability of 3 calls before first successfull call: 0.1045. . .
Probability of 4 calls before first successfull call: 0.0752. . .
and so on.
So the most likely outcome is that the first call is successful, the next most likely outcome is that there's 1 unsuccessful call, and so on. The expected number of unsuccessful calls (the average number if you run the contest many times) is 2.57. . . .

[ Submit Your Own Question ] [ Create a Discussion Topic ]

This part of the site maintained by (No Current Maintainers)
Last updated: April 19, 1999
Original Web Site Creator / Mathematical Content Developer: Philip Spencer
Current Network Coordinator and Contact Person: Joel Chan - mathnet@math.toronto.edu

Navigation Panel:             