Probability GIVEN I already know some event has occurred

[jnscollier]

I ask 14 people to buy something, 2 say yes, 12 say no.

Given this happened, what is the probability that the next person I ask will buy something? So that would make it 3 yes, 12 no, 15 total.

Alternatively, what if I ask 14 people, 0 people buy.

What is the probability that the next person I ask will buy given 0/14 have bought currently?

This isn't a homework problem, I work at a marketing agency and am trying to predict online conversions given the fact I already know some data (clicks, conversions) ex: 14 clicked, 2 people converted, 12 didn't.

I'm not sure how to approach this problem because I'm trying to find some calculation given I already know something.

Thanks

 

[staassis]

"I ask 14 people to buy something, 2 say yes, 12 say no.

Given this happened, what is the probability that the next person I ask will buy something? So that would make it 3 yes, 12 no, 15 total."

The estimate of the probability is 2 / 14. The estimate will have high standard error, since 14 is a small sample size.


"Alternatively, what if I ask 14 people, 0 people buy.

What is the probability that the next person I ask will buy given 0/14 have bought currently?"

If 0 people buy out of 14, then we are dealing with estimating the probability of an extremely rate event. In this case, a sample size of the order of 500-1000 would be necessary to say anything meaningful about the probability.

 

[jnscollier]

"I ask 14 people to buy something, 2 say yes, 12 say no.

Given this happened, what is the probability that the next person I ask will buy something? So that would make it 3 yes, 12 no, 15 total."

The estimate of the probability is 2 / 14. The estimate will have high standard error, since 14 is a small sample size.


"Alternatively, what if I ask 14 people, 0 people buy.

What is the probability that the next person I ask will buy given 0/14 have bought currently?"

If 0 people buy out of 14, then we are dealing with estimating the probability of an extremely rate event. In this case, a sample size of the order of 500-1000 would be necessary to say anything meaningful about the probability.

No, you are getting too technical in your thinking. I'm using these numbers as a 'for example'. The underlying fundamental question I have is how to calculate the conditional probability using these numbers. The ACTUAL numbers are much much larger. This is a simplification of the real-life problem.

So, to modify the question, forget about sample size for a moment. I'm more interested in the calculations.

Thanks anyways.

 

[staassis]

I am not getting too technical because this is simple, 7-th grade, arithmetics. There is no simpler mathematics than my calculations... Again, you cannot say anything accurately if your sample size is small. Does not matter whether you phrase it as an example, advice or something else. What you want to use this for is irrelevant. 2 + 2 does not equal 5, no matter how easy going you are about the whole thing.

 

[jnscollier]

I am not getting too technical because this is simple, 7-th grade, arithmetics. There is no simpler mathematics than my calculations... Again, you cannot say anything accurately if your sample size is small. Does not matter whether you phrase it as an example, advice or something else. What you want to use this for is irrelevant. 2 + 2 does not equal 5, no matter how easy going you are about the whole thing.

I still don't think you comprehend staassis. You're getting hung up on the low sample size.

Anyone else know how to use conditional probability mathematics to compute this in an example using the SAMPLE DEMO NUMBERS that staassis seems to be hung up on?

 

[staassis]

Check out my home page to see who are talking to. Good luck.

 

[Dason]

Without prior information I would probably use a bayesian approach to updating this probability. You can find some details here: http://en.wikipedia.org/wiki/Rule_of_succession

It basically amounts to using: (s+1)/(n+2) as your estimate where s is the total number of successes so far and n is the total number of observations so far. Now you can use different 'prior' distributions to arrive at slightly different estimates but if you have a large sample size then it won't make much of a difference anyways.

Check out my home page to see who are talking to. Good luck.

An unnamed PhD. Oh well. Throwing out your credentials isn't much of a help when somebody is just asking for help. You didn't actually explain anything.

 

[staassis]

Without prior information I would probably use a bayesian approach to updating this probability. You can find some details here: http://en.wikipedia.org/wiki/Rule_of_succession

It basically amounts to using: (s+1)/(n+2) as your estimate where s is the total number of successes so far and n is the total number of observations so far. Now you can use different 'prior' distributions to arrive at slightly different estimates but if you have a large sample size then it won't make much of a difference anyways.

An unnamed PhD. Oh well. Throwing out your credentials isn't much of a help when somebody is just asking for help. You didn't actually explain anything.

Dason, your estimate (s+1) / (n+2) is extremely biased in the case of rare events. It means that you are implicitely using a very rough prior, for no apparent reason. You should have mentioned that to the user (in plain English)... Also, you do realize that he does not even know college math that well. So you just confused him with the word "prior". You did not actually explain anything. Also, formula (s+1) / (n+2) is BIASED and is especially bad in HIS case of 0 out of 14...... It is your duty to draw his attention to the fact that running any well-justified formula on a handful of points will only lead to inaccurate estimates, false confidence and wrong decision making. Seriously, would YOU bet $100 on applying your fomula to his 14 data points?

I did provide the user with most statistically efficient (accurate) formula, which is s/n. But every statistical method has its limitations, which must be clearly stated, especially for not quantitatively fit users...

 

[jnscollier]

Thanks Dason and Staassis. Either way, it sounds like there is no advanced statistical formula to predict try and predict the probability of the next outcome (they purchase or don't). It sounds like both of you are saying essentially something similar which is it's simply s / n or (s + 1)/ (n + 2) (which doesn't really get impacted when a large sample is present).

I was hoping that there was some more advanced formula that I didn't know about for predicting the next outcome given pre-existing data.

Anyways, thanks to both of you for your help.