Flip 98 fair coins and 1 HH coin and 1 TT coin

Question

Flip $98$ fair coins and $1 \ HH$ coin and $1 \ TT$ coin. Given that you see an $H$, what is the probability that it was the $HH$ coin.

Applying Bayes Theorem, :

$$P(HH|H) = \frac{P(H|HH) * P(HH)}{P(H)}$$

$P(H|HH) = 1$

$P(HH) = \frac{1}{100}$

$P(H) = \frac{100}{200} = \frac{1}{2}$

So I get $P(HH|H) = \frac{1}{50}$

1) Is this the correct answer?

2) What's wrong with the 'intuitive' answer? I.e. You see $1\ H$, so we only have $99$ possibilities remaining. Of these $99$, only $1$ of them is $HH => 1/99$

Answer 1

Answer:

Probability that you see a head given it is a faircoin = P(H)*P(Faircoin) $$= \frac{1}{2}\frac{98}{100}$$

Probability that you see a head given it is a HH coin = P(H)*P(HH coin) $$= 1*\frac{1}{100}$$

Using Bayes' theroem,

Probability that it is HH coin given it is a Head $$\frac{1*\frac{1}{100}}{\frac{1}{100}+\frac{1}{2}\frac{98}{100}}$$ $$ = \frac{1}{50}$$

Answer 2

For an "intuitive" answer, there are 100 ways to pick a coin and see heads. Only two of these will result in choosing the HH coin. So the answer is

$$P(HH \mid H) = \dfrac{P(HH \cap H)}{P(H)} = \dfrac{\text{# of ways to pick H and get the HH coin}}{\text{# of ways to pick H}} = \dfrac{2}{100}$$