## Do I have an evil eye? (A bayesian proof that I don’t)

My mom says that I have an evil eye. Her claim is based on some evidences. One of the evidences are as follows:

Once I was in car, in a highway with my family. A nice car took us over speeding. I said something about the car, and a few minutes later we saw the car had an accident at the side of the highway.

So does that mean I have an evil eye? Isn’t the probability of someone talking about a car and the car crash a few minute later rare enough to be considered weird?

I would argue NOT! And prove it mathematically. Read on.

Having no additional knowledge, the P of me talking about a car and the car crashes in a few minutes is close to zero. However, there are a few more points to consider and the situation is not that simple. I would re-phrase the problem as:
“Having a speeding car that passed me and has had an accident, what is the probability that I have talked about it?”

Let’s gather some information. Please note that we limit the world to the cars that have passed me in the highway during my life:

First, the car that has had an accident has been speeding. Let’s say that P of a car that has accident in highway, have also been speeding is about 0.4. It is reasonable to assume most of the cars that crash in highways have been speeding. We call this P(S|X) = 0.4 (X for accident and S for speeding).

Second, I have talked about a car that took us over in highway, speeding. Actually I know this part! The P of me saying something about a car that takes us in a highway speeding is about 0.1. Let’s call this P(T|S) = 0.1. (T for talk and S for speeding).

Last thing I need is the P of a car being speeding having I talked about it in highway. P(S|T) which by experience I can say it is about 0.15.

So, the final question is formulated as what is the probability of me, speaking about a car having it has been speeding, and have had an accident?

Using the Bayes rule, we come up with the following formula:

P(T|S,X) = [P(T|S).P(S|X,T)] / [P(T|S).P(S|X,T) + P(~T|S).P(S|~T,X)] = [0.1*0.4*0.15]/[0.1*0.4*0.15+0.9*0.4*0.85] = 0.019

So, the probability of the accident which my mom considered extremely rare is actually more than 1.9%! Not so rare all of a sudden!

## The answer for the previous challenge

This is the question in my last post:
If we have two coins which one is double headed, and we toss a coin and is head, what is the probability for the other side to be a tail?

If your answer is 1/2, then you are wrong (as I was). I thought this is the same problem as: what is the probability of a selected coin being a normal coin which is 1/2.

I was told that the answer is 1/3 and for some time I was convinced that 1/3 is the right answer but after thinking more about the problem. I thought again in calm and again realized how naive I was the first time.

To understand the problem better let’s rephrase it as below:
We have two boxes one contains a pair of black beans and the other contains a black bean and a white bean. We pick a bean from a box and it is black. What is the probability of “picking another bean from the same box and it being white”?
Look at the figure below: After taking a black bean out, the bean might be either from the box with  white bean in it or from the box with black bean in it or again from the box with another black bean in it.

Sometimes formulas come to rescue.

P(H) is the probability of getting the Head. And P(H) = 1 – P(T). There is no conditional probability here. If we translate the problem as: Having a head out, what is the probability of NOT having another head.

Having a head out P(H) = 2/3 and P(T)=1-P(H)=1/3.