Originally Posted by Axel_Crak

Hehe Fiddy pretty easy Formula to verify sales!


More Building an Empire

The Poisson Distribution Revisited: The Mathematics of Signup Ratios

Updated Oct 16, 2001

"If your experiment needs statistics, then you ought to have done a better experiment." - Ernest Rutherford

I love being able to use mathematical or scientific methods to learn something about my adult sites. One of my favorite mathematical tools is the Poisson Distribution. The Poisson distribution was derived by the French mathematician Siméon Poisson in 1837. This particular distribution describes the probability that an event will occur under the conditions that the event occurring is very low and the number of opportunities for that occurrence is very high.

Doesn't that sound like the perfect tool for analyzing the distribution of pay site signups? A small probability of an event occuring (signup ratio) combined with many opportunities for that event occuring (many visitors).

The formula for calculating probabilities with the Poisson distribution is this:

Probability = (e^-m) * (m^x) / x! where m is the expected number of occurences and x are the various possible occurences (0,1,2,3, etc.)

Let me explain the notation:
e is a mathematical constant. If you have a scientific calculator, it will have a value for e. e is approximately 2.718

The caret (^) symbol means to raise it to a power. e^2 means "e squared" or "e*e". e^3 mean "e cubed" pr "e*e*e".

! means factorial. Multiply by all integers down to 1 .. so 3! = 3*2*1 and 4! = 4*3*2*1 , one exception: 0! = 1

So once again,

Probability = (e^-m) * (m^x) / x! where m is the expected number of occurences and x are the various possible occurences (0,1,2,3, etc.)

Note that to calculate the expected distribution of signups, you only need to know two numbers in order to use this formula: The average number of signups and the case in particular that you want to look at.

This can sound confusing in theory but is easier once you try a few examples. Let me take a concrete example and show you.

Example 1. Assume that you normally convert a program at 1 in 1000 and want to know what the odds are that you will get 0 signups out of your next 1000 hits. In our formula above m (the average value) is 1 (we get 1 out of that 1000 on average) and x (the particular occurence) is 0.

Plugging into the formula these values (m=1, x=0), we get:

Probability = (e^-m) * (m^x) / x!
Probability = (e^-1) * (1^0) / 0!
Probability = (.37) * (1) / 1!
Probability = (.37)
This means that if we expect 1 in 1000, that there is a 37% chance that we will go 0 for our next 1000.

Example 2. Let's try another example. If we normally convert 2 in 1500, what are the odds that we will get exactly 3 signups in our next 1500 hits. Now "m", the expected number of occurences is 2 and "x" is 3.

Plugging into the formula these values (m=3, x=2), we get:

Probability = (e^-m) * (m^x) / x!
Probability = (e^-2) * (2^3) / 3!
Probability = (.135) * (8) / 3*2
Probability = .18
There is an 18 % chance that we will get exactly 3 signups.

Example 3. Let's look at a full distribution. Say that we normally convert 1 in 1000 and want to know what the probability of getting exactly 0,1,2,3, and 4 signups. Here m = 1 and we will look at the cases of x being 0,1,2,3, and 4. Probability = (e^-m) * (m^x) / x!
Probability(x=0) = (e^-1) * (1^0) / 0! = .37
Probability(x=1) = (e^-1) * (1^1) / 1! = .37
Probability(x=2) = (e^-1) * (1^2) / 2! = .19
Probability(x=3) = (e^-1) * (1^3) / 3! = .06
Probability(x=4) = (e^-1) * (1^4) / 4! = .015

If you expect 1 in 1000, then the odds of getting exactly 0 are 37%. The odds of getting exactly 1 are 37%. The odds of getting exactly 2 ar 19%. The odds of getting exactly 3 are 6%. The odds of getting exactly 4 are 1.5%.

If you look carefully, you will notice that the 1000 from 1 in 1000 never appears in the formula. In other words, if you expect 1 in 10,000 the odds of getting exactly 0 are 37% also. The odds of getting 1 in 10,000 when you expect 10,000 is also 37%.

You will also notice when you are looking at the whole distribution for multiple cases (0,1,2,3,4,...) when you expect "1 in something" that you have a constant term

(e^-1) * (1^x) = .37
no matter what x is because one raised to any power is 1. This constant term is divided by a number that is x!. Since x! grows very fast, the probability of getting a large number of signups when only 1 is expected drops off very quickly as you try large values of x.

If you expect 1 signup in any amount of hits, the odds that you'll get exactly 1 signup in that amount of hits is 37% and the odds that you'll get exactly 0 is also 37%.

How to use it There are many ways that you might be able to use the Poisson Distribution in a useful way. Say you join a new webmaster program and you need to convert it at 1 in 1200 for it to be useful to you. You send 1200 visitors and want to know whether that is enough to have fairly tested it. From the example above, you know that you will get 0 out of any random 1200 37% of the time, a fairly large occurence, so you really need to stick it out a little longer.

How about if after 3600 visitors sent you still don't have a signup? You are still wondering whether you have sent enough traffic to have given it a fair chance. What are the odds that you will have gone 0 for 3600 by chance if the program is going to convert 1 in 1200 for you overall? Here you expected (needed) to have 3 signups (3600/1200) so m is 3. The case you are interested in is for x=0.

Probability = (e^-m) * (m^x) / x!
Probability(x=0) = (e^-3) * (3^0) / 0! = .05
In other words, in 5% of the cases where we look at any random set of 3600 visitors, we will have 0 signups. It's not impossible that you might end up doing 1 in 1200, but the odds have slipped pretty far by then.

No mathematical distribution can make a decision for you, but they can give you a very good idea of when enough is enough. The question then becomes, what probability level is enough for you? You might also invoke it for those cases when you suddenly have a really bad day and would like to know if it is within the realm of possibility that you are just having a bad day. A .05 probability means that it is going to happen once every 20 trials. So if you normally get 3 signups in 3600 hits sent per day, you are going to get 0 signups once every 20 days. This is not because there is anything wrong with your sponsor, it's just the way the distribution works. If you average 5 signups per day with a sponsor you will have a 0 signup day once every 5 months on average. They are rare, but they will happen.

Always remember that even a slight adjustment in your marketing approach might make 1 in 1200 become 1 in 900. A tool like the Poisson Distribution is useful only when all other factors stay constant. The poisson calculator

Sending enough hits to test a sponsor There's a concept in statistics called standard deviation. The idea is that 68% of the time a quantity will be within one standard deviation of the average. 95% of the time one expects to find the quantity within two standard deviations of the average. When a scientist measures some quantity they quote it with an error measurement which is one standard deviation. No measurement is certain. I will just refer to the standard deviation as the error, but you will know that technically I mean the standard deviation. I've also left a few technical details out such as how confident we are in our measurements because well, I'm only going to take this so far.

In the Poisson Distribution the error is simply the square root of the average. In other words, the error is the square root of the number of signups. If you send 4 signups to a sponsor the error is 2 signups because 2 is the square root of 4. You can say you measured 4 +/- 2 signups. With all other things being equal, 68% of the time you can expect to attain between 2-6 signups by sending the same number of visitors. You can take that further and say that 95% of the time you expect to attain between 0 and 8 signups for the same number of visitors sent.

If you send 10,000 visitors to a sponsor and get 9 signups, then 68% of the time when you send 10,000 visitors, you expect to attain between 6 and 12 signups (6 +/- 3) and 95% of the time you expect to attain between 3 and 15 signups. The interesting point of all this is that the error depends only on the number of signups and not the number of visitors. Most people think, ok, I'll send a certain number of visitors to a sponsor and see how many signups I get. What you should be doing instead is saying ok, I'll send a certain number of signups to a sponsor to test them.

Why don't we use the "bell curve" instead? You might guess that a graph of the signups for every 10,000 hits won't exactly look like a bell curve. This is because you can't get less than 0 signups but there is a small probability of getting a lot of signups. This is what skews the graph and makes it non-symmetrical. We must use the Poisson Distribution instead of the Normal Distribution, often refered to as the "bell curve". The error estimation method I used here is a little different than the true error estimate. They're very good though. The error estimates I used are symmetrical when we just said that they aren't. As one increases the number of signups measures towards infinity, the Poisson Graph eventually looks exactly like the Normal Distribution. Here the error estimates become the same. The Poisson Distribution is just a special case of the Normal Curve.

Originally Posted by MisterPeabody

OK I flunked math but thanks for this (I think).

FIDDY sales returning.

Originally Posted by wehateporn

As a very rough guide, you say you normally get 3 or 4 sales per hour, so your probability of getting no sales in an hour is quite low, but for the sake of argument let's say that we're looking at someone with less sales than you, and their probability of getting no sales in 1 hour is 0.5 (i.e. 1 in 2)

So, this is someone with far weaker sales than you, let's take a look at their rough probability of getting no sales in 12 hours.

No sales in 1 hour = 0.5 = 1 in 2

No sales in 2 consecutive hours = 1/2 * 1/2 = 1/4 = 1 in 4

No sales in 3 consecutive hours = 1/2 * 1/2 * 1/2 = 1 in 8

No Sales in 4 consecutive hours = etc = 1 in 16

No Sales in 5 = 1 in 32

No Sales in 6 = 1 in 64

No Sales in 7 = 1 in 128

No Sales in 8 = 1 in 256

No Sales in 9 = 1 in 512

No Sales in 10 = 1 in 1024

No Sales in 11 = 1 in 2048

No Sales in 12 = 1 in 4092


So with much weaker sales than MisterPeabodyCash one is likely to see this event of 12 consecutive hours with no sales 1 out of 4092 twelve hour periods, also known as 1 every 2048 days, that's just over once every 5 and a half years.

Whereas someone with better sales like yourself, let's say your probability of no sales in 1 hour is 0.3 (3/10)

It would look like this

No sales in 1 hour = 0.3 = 3 in 10

No sales in 2 consecutive hours = 3/10 * 3/10 = 9/100 = 9 in 100

No sales in 3 consecutive hours = 3/10 * 3/10 * 3/10 = 27 in 1000

No Sales in 4 consecutive hours = 81 in 10,000

No Sales in 5 = 243 in 100,000

No Sales in 6 = 729 in 1,000,000

..... blah blah blah

No Sales in 11 = 177,147 in 100,000,000,000

No Sales in 12 = 531,441 in 1,000,000,000,000

As this is all very approximate let's call it 500,000 in 1,000,000,000,000

= 1 in 2,000,000 (half days)

= 1 in 1,000,000 days (once every 2700 years)

i.e. lottery style odds (assuming all well with your servers, the Internet, banking system, third party processors etc)