Confused about relevance of Bayes' theorem to my problem

I've spent the morning teaching myself about Bayes's theorem, because I assumed that it was required to help me solve my problem. However, the answer I've ended up with is the same as I would have got if I had used my naive approach prior to understanding Bayes's rule.

I'll explain my problem:

Suppose there are four possible methods of purchasing some item, $A_1, A_2, A_3, A_4$. These methods have different levels of attractiveness to customers. Over the previous month, for example, 14,456 people have used one of the methods. Thus, knowing how many people used each of the methods, one can clearly see the probabilities that customer has used one of the methods:

• $A_1 = 1500 / 14456 = 0.103$
• $A_2 = 11000 / 14456 = 0.76$
• $A_3 = 56 / 14456 = 0.0038$
• $A_4 = 1900 / 14456 = 0.131$

Customers are pretty fickle though, and most of the time they don't go through with the purchase. In fact, the number of customers who actually convert are pretty low. It's dependent also on the different method that they are using to make the purchase (imagine some methods are easier than others). In Bayesian language, I'm talking about the probability of a conversion given the method as $P(B | A_k)$. Using our example past month, we also know these probabilities, because we can see the number of conversions for each of the methods. There were varying conversion numbers for customers using each method.

• $P(B|A_1) = (76/1500) * A_1 = 0.0051$
• $P(B|A_2) = (24/11000) * A_2 = 0.0015$
• $P(B|A_3) = (10/56) * A_3 = 0.00065$
• $P(B|A_4) = (34/1900) * A_4 = 0.0022$

Now, let's suppose that sometimes the method we use to track the methods of purchase, and the number of conversions therein is a bit clunky and doesn't always work. Suppose a 145th conversion is in the system, but we don't know which method was used to make the purchase; we want to estimate this using the available data.

Bayes' theorem tells me I need to calculate:

• $P(A_1 | B) = 0.5277$
• $P(A_2 | B) = 0.1666$
• $P(A_3 | B) = 0.0694$
• $P(A_4 | B) = 0.2361$

Great! So I now know that there was just over a 52% chance that the 145th customer purchased using method 1.

However, this involved quite a bit of maths in making the $P(A_k | B)$ calculations. When I looked at the numbers that I got out I realised that if I had just divided the number of conversions for each channel by the total number of conversions, I got exactly the same answers.

• 76 / 144 = 0.527
• 24 / 144 = 0.166
• 10 / 144 = 0.069
• 34 / 144 = 0.236

I guess I'm a bit surprised that I didn't need to go to all the effort with the Bayes's Rule stuff, so I don't quite understand situations where I'll need to use the rule, and not just do other summary calculations.

• Note that use of Bayes theorom does not automatically warrant calling it bayesian statistics. One problem is that at this moment you just have estimates of the underlying probability and no feeling for the uncertainty of your estimation. +1 for the detailled writeup. – Erik Aug 31 '12 at 8:46

The mistake lies in your calculation for $P(B |A_i)$. It should actually be something like $P(B|A_1)=76/1500$, and then $P(B,A_1)=P(B|A_1)P(A_1)=76/1500 * 1500 / 14456=76/14456$.

The marginal of $B$ is then $P(B)=\sum_{i=1}^4 P(B,A_i)=144/14456$. Finally, $P(A_i|B)=\frac{P(B,A_i)}{P(B)}$. And yes, $P(A_1|B)=76/144$. This final formula is a form of Bayes theorem. The reason you did not need the calculations earlier is because you already had the joint probabilities $P(A_i,B)$, and you never needed conditionals like $P(B|A_1)$.

• This is essentially the correct answer, but the fact is that tomtom does not have any probabilities at all but just observed frequencies, which are just an estimate of the underlying probability. – Erik Aug 31 '12 at 8:45
• Who calls the final form "a form of Bayes theorem"? All I've ever heard is that it is the definition of conditional probability. – JiK Aug 4 '17 at 14:11
• @JiK Bayes theorem follows from the definition of conditional probability. Is there a question there, or are you just nitpicking? He's going from P(B|A) to P(A|B), which is that Bayes theorem does. – highBandWidth Aug 10 '17 at 1:36

You can read the conditional probability from the cross-tabulation directly, so Bayes'Theorem is probably an overkill...

A classical use for Bayesian statistics is in diagnosing disease.

Suppose you have a test that gives a positive result 90% of the time if someone has the disease. The test also gives a false positive 5% of the time if someone does not have the disease.

Further you know that in the general population 1 in 100,000 people have the disease.

Someone comes into your office and you want to know if they have the disease. Perhaps the only way to know for certain if they have it is during an autopsy after they die, so the best you can do right now is run the test.

The key to understanding the Bayesian nature of this situation is that you have no way of observing directly what you want to know (does he have the disease?) The best you can do is observe the test results.

Thus the problem becomes, "What is the probability the person has the disease given a positive test result?

• Your answer has no relation or explicit link to the specific question. – Erik Aug 31 '12 at 8:42