hay guys

this is a continuation from
http://www.blackhatworld.com/blackha...doing-ppc.html

we're still going to be answering the same question of "how do i know when to stop spending money on an unprofitable keyword", only this time we'll be able to calculate for more than 0 conversions.

if you look closely, you'll notice that the previous equation i posted only works for 0 conversions, because i had no idea how to calculate for more than that at the time lol. but now i do, so i thought i'd share it with you guys and save you even more money on your ppc campaigns.

the following is gonna assume you've read the previous thread, so if you haven't, none of it's probably going to make any sense =P

anyway, turns out the actual equation to calculate P (probability of a keyword being profitable) is a poisson distribution.
http://en.wikipedia.org/wiki/Poisson_distribution


in this case, k is the number of conversions you currently have on your keyword, and λ is your required number of conversions to break even with a 0% roi.
using the variables from the previous thread, λ = cRate * clicks

in the previous thread, k was just equal to 0, and you'll see the poisson equation above simplifies to the equation in the previous thread when k=0
P = ( λ^0 * e^(-λ) ) / 0! = e^(-cRate * clicks)
it's not the same equation, since the other one was just an approximation, but the results are nearly identical.

so anyway, we just gather the same variables again and plug them in.
cRate = bid / payout
k = conversions made so far
clicks = clicks made so far
λ = cRate * clicks

then calculate p by plugging in
p = ( λ^k * e^(-λ) ) / k!
(same as the equation in the white box above)

as an example, lets say you have an offer that pays out $30 per conversion. you've made 2 conversions so far, your average cpc is $0.80 per click, and you've received 200 clicks so far.
then your requred conversion rate is cRate = 0.80 / 30 = 0.027
you had 2 conversions so far, so k = 2
clicks = 200
plug in the numbers to get λ = 0.027 * 200 = 5.4
and finally, p = ( 5.4^2 * e^(-5.4) ) / 2! = 0.066

now this isn't the same p as the previous thread. to get P (probability of keyword being profitable), you need to repeat this calculation for every k down to 0, and then sum up the values.
so for k=2, p(2,5.4) = 0.066
for k=1, p(1,5.4) = ( 5.4^1 * e^(-5.4) ) / 1! = 0.024
and for k=0, p(0,5.4) = ( 5.4^0 * e^(-5.4) ) / 0! = 0.005

so finally, P = 0.066 + 0.024 + 0.005 = 0.095

this means that given the data currently available to you, you can estimate that the keyword has around a 9.5% chance of ever being profitable. imo this is kinda right on the line, so i might continue running it for a while until P drops even lower.

think of it this way, nothing in ppc is ever 100% certain. if you decide to axe every keyword that has below a 10% probability of ever being profitable, then what you're doing is throwing away 1 in every 10 profitable keywords. unfortunately this is unpreventable as you can never get P = 0, it just drops lower with the more money you throw at it, so you have to use your own judgement there. i usually cut off keywords that drop below 5%, meaning that i'm willingly throwing away 1 in 20 profitable keywords, because the money i save from testing is worth it in the end.

get the math down right and you don't need to throw away hundreds of dollars when testing out new keywords. if you know how to not lose money, you can make the initial unprofitable testing phase extremely short and start seeing positive profits very quickly.

good luck and have fun
o/

wish I learnt this kind of math in high school

wow, i love this really .i was always excellent at maths

man i used to curse my probability teacher when she used to teach me this.
but now i am actually thankful to her.
thanks for the detailed analysis.

Originally Posted by secretboy08

man i used to curse my probability teacher when she used to teach me this.
but now i am actually thankful to her.
thanks for the detailed analysis.

haha you are soo right now we know for what the math studies are useful heheh

thanks at the OP, learnt a great new way to save money

Here is another discussion of the poisson distribution vis a vis sign up ratios, by one of the first $100million internet pioneers.

http://www.buildinganempire.com/poisson2.html

thanks a lot mate!