Finance expert Janet Tavakoli thinks so, on the basis of the fact that two off-duty police officers were victimized (one murdered, one subjected to a robbery attempt) on the same day:

If crime statistics in Chicago are as low as Mayor Rahm Emanuel claims, [two victimizations of off-duty cops on one day] should be extremely unlikely. It's so unlikely that everyone should challenge the narrative about Chicago's crime being low or even being managed in any meaningful way.

... 6/1,000 [Chicagoans] will be victims of robbery and/or murder [in a given year]. ...

Full time law enforcement staff was 12,799 of which 12,092 were officers, or 4.47 officers / 1,000 residents. I do not know what percent are off duty at any time, but obviously the number of off-duty officers is lower [than the total number]. Let's put the stats in Rahm's favor and make it more likely for an off-duty officer to be a victim, say 3 officers/1,000 residents are off-duty.

How likely is it that a person is both a victim AND an off-duty officer (assuming independence)?

Using the above statistics, it's not likely at all: (6/1000) * (3/1000) = 0.000018 or 0.0018 percent. In other words, there's only a 0.0018 percent, less than two thousandths of a percent probability of this occurring.

Now what is the probability of two off-duty officers being involved in a random robbery and/or murder crime incident? Much lower.

Now what is the probability of this happening to two off-duty officers on the same day? Even lower.

This is not the right way to do this math. She's calculating the odds that an individual person will be both an officer and a victim over the course of a year, and then taking the result to mean something about the odds of two officers being victimized on the same day out of a huge population.

Because we're looking at off-duty officers only, let's say that instead of a 6/1,000 annual rate of victimization, officers have a 4/1,000 rate of off-duty victimization. (This mimics the author's reduction from 4.5 to 3.) This means that a population of 12,000 officers will have 48 victimizations a year -- almost one a week.

What are the odds of two of these victimizations falling on the same day? It's easiest to calculate the odds of them all falling on different days, and then subtract the result from 1. The first victimization is guaranteed to fall on its own day, the second has a 364/365 chance of not falling on the same day as the first, the third has a 363/365 chance of having its own day assuming the first two days are different, and so on, down to 318/365. To get the odds that our series of victimizations will clear all of these hurdles -- the odds that every single incident will happen on its own day -- we multiply them together.

Subtract the result from 1, and we get ... a 96 percent chance that there will be at least one match. It's practically guaranteed, based on the author's numbers, that at some point during the year two off-duty cops will be victimized on the same day.

It's very similar to the "birthday problem." Once you have more than 23 people in a room, the odds are above 50-50 that two of them share the same birthday. Similarly, once you have more than 23 victimizations of off-duty cops in the same city and the same year, the odds are better than not that two will occur on the same day. From there the odds go up rapidly:

It's quite possible that stat-focused policing makes departments undercount crimes. This particular anecdote just isn't evidence that it's happening in Chicago.

Robert VerBruggen is editor of RealClearPolicy. Twitter: @RAVerBruggen