She smiles for a few minutes, then drifts out to the parking lot for a smoke, and slides unobtrusively into her car. Pausing only to pick up her Goya, she heads up the highway at legal speeds toward Vancouver. Where, we need not doubt, she has a prepared alternate identity waiting for her.Explanation
Nothing we know about Janice suggests that she panics easily. On the contrary, we get the impression of a cool customer. She will therefore probably weigh the odds of discovery, and if they are on her side, she will tough it out. If not, she will move on to the next thing. We are thus asking, given some risk of discovery, is that risk less than 50%? This is the kind of question we can answer.
Ignoring some fine points, to which we will return in a moment, the problem looks like this. The proportion of kited invoices in the whole file of invoices is 28/4000 = 0.0070. Janice has taken care to be inconspicuous. The auditor's unit of observation is a sample of 150 invoices; not many. Every random sample of that size might be expected to turn up a different mix of things, including different amounts of kited invoices. But the average of such samples ought to be the sample size times the rate of occurrence, or (150)(0.007). This gives us r = 1.05. All Janice needs to do is to look up the Poisson table for r = 1.05, and ask: what is the chance that one such sample will contain zero kited invoices? In fact, she does not even need to do that, since the fact that the rate is a little over 1 tells her than 1 is the likeliest outcome. 1 means that discovery is her likeliest outcome. Janice's decision is accordingly made for her. Done.
All this took only about two seconds, and Janice did it in her head. But those who are less practiced at this stuff may like to see the first few rows of the table for r = 1.05. They look like this:
r = 1.05 p(0) 0.3499 p(1) 0.3674 p(2) 0.1929 p(3) 0.0675
The main point is that the chance of zero kited invoices (which is Janice's chance of getting through the audit undiscovered) is a paltry 0.3499, or 35%. Another way to read the strategy off the table is to note that the likeliest number of kited invoices that will be discovered is 1. It is of no consequence to Janice that the chance of more invoices being discovered dwindles very rapidly. One is enough to do her in. What is make or break for her is the probability of zero invoices: p(0).
This is not a Poisson problem, and it is not really a Binomial problem solved with Poisson methods; it is a Hypergeometric problem solved with a Poisson simulation of the Binomial shortcut to the Hypergeometric solution. Hypergeometric problems are not amenable to quick and easy calculation. Those problems arise when we take items from a closed set (like fish from a lake, or cards from a deck) without putting them back before taking the next item, so that the probability of getting one particular kind of thing changes constantly over the course of the problem. The Binomial shortcut works in cases where, even though the set of things drawn from is finite, it is large, and so the odds do not change very much during the problem. And the Poisson simulation of the Binomial shortcut works for the usual reasons that apply to these things: basic probability small, and number of samples large.
Auditors, we understand, like the Poisson distribution because it can acceptably mimic the efficiency of sampling without replacement in finite but large collections of financial records. That situation constantly presents itself in the course of an auditor's workday.
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