## The Expose Narcs Equation

If you've ever wondered just how much cutting down the rate of narc'ing affects overall drug busts, then this page is for you. For those of you who don't know, when WANTF busts someone for drugs, that person is usually offered a deal: narc on several people or get arrested. The officers of WANTF know that most people are not enthused to narc, so they will oftentimes give a potential narc a lowball number of busts in which he must work, then jack up that number after the narc is already in too deep (so a "too late to turn back now" mentality kicks in).

If someone who is busted decides to take his arrest like an adult, then no other arrests are made as a result of that bust. If, however, someone decides to narc his way out of his charges, then the problem grows exponentially. In the following example, we have an original bust, in which the person who is busted has accepted a deal to narc on 3 people, and 2/3 (two-thirds) of those who are busted decide to narc. (Note that "yes" denotes someone who decided to narc, and "no" denotes someone who decided not to narc.)

If someone who is busted decides to take his arrest like an adult, then no other arrests are made as a result of that bust. If, however, someone decides to narc his way out of his charges, then the problem grows exponentially. In the following example, we have an original bust, in which the person who is busted has accepted a deal to narc on 3 people, and 2/3 (two-thirds) of those who are busted decide to narc. (Note that "yes" denotes someone who decided to narc, and "no" denotes someone who decided not to narc.)

At node 1 -- after the original bust -- there are 3 drug busts. Two of those who are busted decide to narc, while the other person has decided to be a standup guy. At node 2, there are 6 drug busts -- 3 from each narc. 4 of those who are busted decide to narc, while two decide not to narc. At node 3, there are 12 drug busts -- 3 from each of the 4 narcs. In total, after 3 iterations, there were 3 + 6 + 12 = 21 drug busts (7 arrests and 14 narcs), plus the original narc that got the whole thing started.

Now, suppose that we do everything the same, but this time, only 1/3 (one-third) of those who are busted decide to narc.

Now, suppose that we do everything the same, but this time, only 1/3 (one-third) of those who are busted decide to narc.

At node 1 -- after the original bust -- there are 3 drug busts. One of those who are busted decides to narc, while the other two people have decided not to narc. At node 2, there are 3 drug busts -- one decides to narc, and two decide not to narc. At node 3, there are 3 drug busts. In total, after 3 iterations, there were 3 + 3 + 3 = 9 drug busts (6 arrests and 3 narcs), plus the original narc that got the whole thing started.

By decreasing the rate of narc'ing from 2/3 to 1/3, drug busts decreased by (21 - 9)/21 = 57%. In reality, this process goes on for more than 3 iterations, but the illustration above was done for the sake of simplicity. The longer this process goes on, the more pronounced is the impact of lowering the narc rate on drug busts (see below for an explanation on why that 57% figure climbs to 73% after 4 iterations).

Below is The Expose Narcs Equation, which will give you an idea of just how quickly the problem of narc'ing can get out of control.

By decreasing the rate of narc'ing from 2/3 to 1/3, drug busts decreased by (21 - 9)/21 = 57%. In reality, this process goes on for more than 3 iterations, but the illustration above was done for the sake of simplicity. The longer this process goes on, the more pronounced is the impact of lowering the narc rate on drug busts (see below for an explanation on why that 57% figure climbs to 73% after 4 iterations).

Below is The Expose Narcs Equation, which will give you an idea of just how quickly the problem of narc'ing can get out of control.

So, in the 4th node, if 2/3 of those busted decide to narc, there will be 3(3 x 2/3)^(4-1) = 3(2)^3 = 3 x 8 = 24 busts. In total, there would be 3 + 6 + 12 (from above) + 24 = 45 drug busts by the 4th node.

In the 4th node, if 1/3 of those busted decide to narc, there will be 3(3 x 1/3)^(4-1) = 3(1)^3 = 3 x 1 = 3 busts. In total, there would be 3 + 3 + 3 (from above) + 3 = 12 drug busts by the 4th node, which is a decrease in drug busts of 73%.

Because adding up the number of busts node by node can be a pain in the ass, below we have the formula for adding the sum of a geometric series (remember learning that in Calculus 2?).

In the 4th node, if 1/3 of those busted decide to narc, there will be 3(3 x 1/3)^(4-1) = 3(1)^3 = 3 x 1 = 3 busts. In total, there would be 3 + 3 + 3 (from above) + 3 = 12 drug busts by the 4th node, which is a decrease in drug busts of 73%.

Because adding up the number of busts node by node can be a pain in the ass, below we have the formula for adding the sum of a geometric series (remember learning that in Calculus 2?).

Using the example with a narc rate of 2/3, b = 3, and r (the common ratio) = 2 (since we started with 3 busts, then 6, then 12, then 24, ... [the number of busts keeps doubling, so the common ratio is 2]). That means, after the 5th iteration the cumulative number of busts will be:

3[(1 - 2^5)/(1 - 2)] = 3[(1 - 32)/(1 - 2)] = 3(-31/-1) = 3(31) = 93

If we cut the narc rate down to 1/3, then r = 1 (since we started with 3 busts, then 3 busts, ... ). That's okay, because those of you who took calculus should remember L'Hopital's Rule. Since, when r = 1, we have an indeterminate form, so we simply take the derivative of the numerator and denominator with respect to r. When we do that, we get b[n(r)^(n-1)], which is simply bn (because r = 1, and one raised to any exponent is one), thus after the 5th iteration, we get 3 x 5 = 15 drug busts in total -- a reduction of 84%!

The figures above were used for the sake of simplicity, but I believe that we can assume that the narc'ing rate is 60%, and we can cut that to 40%, so below is a more real-life example (don't get tripped up by the fractional number of people, since that effect goes away when dealing with large numbers):

The number of drug busts, assuming 3 busts per narc and a 60% narc rate, after the 3rd iteration is:

3(3 x .60)^2 = 3(1.80)^2 = 9.72 (and the common ratio is 3 x .60 = 1.80)

If the narc rate drops to 40%, then after the 3rd iteration, the number of busts is:

3(3 x .40)^2 = 3(1.20)^2 = 4.32 (and the common ratio is 3 x .40 = 1.20)

The cumulative number of busts after the 3rd iteration is:

3[(1 - 1.80^3)/(1 - 1.80)] = 3[(1 - 5.832)/(1 - 1.80)] = 3(-4.832/-.80) = 18.12 for a 60% narc rate, and

3[(1 - 1.20^3)/(1 - 1.20)] = 3[(1 - 1.728)/(1 - 1.20)] = 3(-.728/-.20) = 10.92 for a 40% narc rate -- a reduction of 40% after just 3 iterations! (55% after 4 iterations and 67% after 5 iterations).

You might notice that it doesn't matter what the number of busts per narc is -- ultimately, the reductions stated above will remain the same, regardless if there are 3 busts per narc or 5 busts per narc (or however many busts per narc). The important factors are the old narc rate, the new narc rate, and the number of nodes (iterations).

We've got more to say about the implications of The Expose Narcs Equation; however, we'll save that for another day. For now, it's enough that you understand that cutting the rate of narc'ing, even a little bit, can have a profound impact on the number of drug arrests.

3[(1 - 2^5)/(1 - 2)] = 3[(1 - 32)/(1 - 2)] = 3(-31/-1) = 3(31) = 93

If we cut the narc rate down to 1/3, then r = 1 (since we started with 3 busts, then 3 busts, ... ). That's okay, because those of you who took calculus should remember L'Hopital's Rule. Since, when r = 1, we have an indeterminate form, so we simply take the derivative of the numerator and denominator with respect to r. When we do that, we get b[n(r)^(n-1)], which is simply bn (because r = 1, and one raised to any exponent is one), thus after the 5th iteration, we get 3 x 5 = 15 drug busts in total -- a reduction of 84%!

The figures above were used for the sake of simplicity, but I believe that we can assume that the narc'ing rate is 60%, and we can cut that to 40%, so below is a more real-life example (don't get tripped up by the fractional number of people, since that effect goes away when dealing with large numbers):

The number of drug busts, assuming 3 busts per narc and a 60% narc rate, after the 3rd iteration is:

3(3 x .60)^2 = 3(1.80)^2 = 9.72 (and the common ratio is 3 x .60 = 1.80)

If the narc rate drops to 40%, then after the 3rd iteration, the number of busts is:

3(3 x .40)^2 = 3(1.20)^2 = 4.32 (and the common ratio is 3 x .40 = 1.20)

The cumulative number of busts after the 3rd iteration is:

3[(1 - 1.80^3)/(1 - 1.80)] = 3[(1 - 5.832)/(1 - 1.80)] = 3(-4.832/-.80) = 18.12 for a 60% narc rate, and

3[(1 - 1.20^3)/(1 - 1.20)] = 3[(1 - 1.728)/(1 - 1.20)] = 3(-.728/-.20) = 10.92 for a 40% narc rate -- a reduction of 40% after just 3 iterations! (55% after 4 iterations and 67% after 5 iterations).

You might notice that it doesn't matter what the number of busts per narc is -- ultimately, the reductions stated above will remain the same, regardless if there are 3 busts per narc or 5 busts per narc (or however many busts per narc). The important factors are the old narc rate, the new narc rate, and the number of nodes (iterations).

We've got more to say about the implications of The Expose Narcs Equation; however, we'll save that for another day. For now, it's enough that you understand that cutting the rate of narc'ing, even a little bit, can have a profound impact on the number of drug arrests.