Coin - Real or Biased

Question

There are two coins: Fair Coin: Heads and Tails with 50% Probability, and Biased Coin: Always lands on Heads.  Dealer uses either coin with equal probability.

We see that the dealer tosses the coin, and it comes in a sequence as H, H, T.  We can immediately infer that the dealer is using the fair coin (because biased coin never lands on Tails).
 
In next run, the dealer tosses the coin, and it comes in a sequence as H, H, H.  What is the probability that the dealer is using the biased coin.

The dealer does not switch coins during a single run.

Answer 1

Let's define the following events:

A: Dealer is using the fair coin.

B: Dealer is using the biased coin.

O: We observe the sequence H, H, H.

We want to calculate the probability of event B given O, i.e., P(B|O).

By Bayes' theorem, we have: P(B|O) = P(O|B) * P(B) / P(O)

P(O|B) = 1

P(B) = 0.5

P(O) = P(O|A)*P(A) + P(O|B)*P(B)

We need to calculate P(O|A) and P(O|B):

P(O|A) = P(H, H, H|A) = P(H|A)^3 = 0.5^3 = 0.125

P(O|B) = P(H, H, H|B) = 1

P(O) = P(O|A)*P(A) + P(O|B)*P(B) = 0.125 * 0.5 + 1 * 0.5 = 0.5625

Now we can use these values for the Bayes theorem:

P(B|O) = P(O|B) * P(B) / P(O) = 1 * 0.5 / 0.5625 = 0.8889

Answer 2

Here is my solution.

Set A represent the event that use biased coin, and B represent the event that observed H,H,H.

We need to calculate P(A|B) = P(B|A) * P(A) / P(B)

For P(B|A), it should be 1.

For P(A), I think it should be 1/2 in intuition.

For P(B), we can divide it into two parts. If it is fair coin, it should be (1/2)^3; if it is biased coin, it should be 1.

So, P(B) = 1/2 * (1/2)^3 + 1/2 * 1

P(AA|B) = P(B|A) * P(A) /P(B)

= 1 * 1/2 / (1/16 + 1/2)

= 8/9

And 8/9 ≈0.8889