# Probability of having chosen a fake coin with two heads knowing the result of multiple coin flipping

You have $$65$$ coins: $$64$$ are normal coins with heads and tails, both with a $${1\over2}$$ chance, whereas one coin is fake and has two heads.

You randomly choose a coin and flip it $$6$$ times: all of them you get heads. What's the probability the coin you picked is the fake one?

Here's what I've tried. There is one specific step I'm not too confident about.

Let $$A$$ be the event "picking the fake coin," $$B$$ be the event "getting $$6$$ heads out of $$6$$ attempts."

$$P(A) = {1\over65}$$ $$P(B) = p\{HHHHHH\} = {1\over{2^6}} = {1\over64}$$ $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

At this point, I'm having difficulties calculating $$P(A \cap B)$$: intuitively, that would be the probability of both picking the fake coin and getting six heads in a row. My guess would be that $$P(A \cap B) = \frac{1}{65\cdot64}$$.

Then, $$P(A | B) = {1\over65}$$, which would mean that the two events are stochastically independent.

Is this the correct answer? What am I missing here?

The probability that you chose a real coin and got $$6$$ heads ($$\neg A\land B$$) is $$\frac{64}{65}×\frac1{64}=\frac1{65}$$. The probability that you chose the two-headed coin and got $$6$$ heads ($$A\land B$$) is also $$\frac1{65}$$ (more formally $$\frac1{65}×1$$). Thus, by total probability the chance that you picked the fake coin is $$\frac12$$.