Question

I have been thinking of this problem for some time now,I came across this in one of the lectures of Probabilistic graphical Model.https://class.coursera.org/pgm/lecture/preview

Question is pretty straight forward:

There are two coins one is normal other one is biased towards Heads 9 out of 10.

We pick one coin and toss it, the teacher’s claim is if Heads shows up the first time then probability of Head showing again would be high.

All we can tell that second coin is more likely to be selected by Bayes.
But in my opinion both should be independent

Can anyone help me with that?

Answer 1

To completely define the process, let us add that each coin has porbability 1/2 of being picked.

Let X1 and X2 be the first two tosses. “Heads shows up the first time” is written “X1=H”.

We have:

P(X2=H | X1=H) = P(X2=H & X1=H) / P(X1=H)

P(X1=H) = 1/2 × 1/2 + 1/2 × 9/10 = 14/20 = 0.7

P(X1=H & X2=H) = 1/2 × (1/2)² + 1/2 × (9/10)² = 53/100

So, P(X2=H | X1=H) = 53/70 ≃ 0.757 > 0.7

This means P(X2=H | X1=H), the probability of getting Heads at the second toss given that Heads showed up at the first, is higher than the probability of Heads showing up at the first toss.

To support intuition, you can consider another case:
we pick one coin and toss it, Tails shows up the first time. So you think that it’s “probably” not the biased coin that has been picked. Conversely, if Heads shows up, then having picked the biased coin is more probable. And if so, then getting Heads is more probable.

Or yet another process: you pick a coin, but not at random: you have a higher probability of picking the biased coin. Then you do a single toss. Heads more likely for this toss.