One about flipping coins. Coin A has 0.6 probability of being heads, Coin B has a 0.3 probability of being heads and Coin C has a 0.1 probability of heads. Given that a coin is flipped and lands on heads, what is the probability that when it is flipped again it will also be heads.
I'm not sure how I can apply Bayes' rule here.
I did a Monte Carole simulation the answer come out to be around 0.46, I am not sure how to give a analytical answer.
Best Answer
As a way to get intuition for such problems: suppose we toss each coin $10$ times. Then we expect to get $10$ Heads... $6$ Heads from $A$, $3$ from $B$, and $1$ from $C$. Accordingly if our prior was that our mystery coin could be any of the three equally, then seeing Heads now makes us rethink. Now Bayes' tells us that there is a $.6$ probability the coin was $A$, $.3$ that it was $B$, and $.1$ that it was $C$.
Note: the fact that these numbers coincide with the weightings is a consequence of the fact that the weightings add to $1$, it is not generally true.
Thus the probability that tossing the coin again yields another $H$ is: $$.6^2+.3^2+.1^2=.46$$
Which, unsurprisingly, lines up with the result of your Monte Carlo simulation.