Flip 98 fair coins and 1 HH coin and 1 TT coin 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 $HH => 1/99$
 A: 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
$$P(HH \mid H) = \dfrac{P(HH \cap H)}{P(H)} = \dfrac{\text{# of ways to pick H and get the HH coin}}{\text{# of ways to pick H}} = \dfrac{2}{100}$$
A: 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 $$\frac{1*\frac{1}{100}}{\frac{1}{100}+\frac{1}{2}\frac{98}{100}}$$
$$ = \frac{1}{50}$$
A: Using Bayes' theorem is obviously the correct way to answer the question, but I think a good / valid layman's term answer is:


*

*There are $200$ faces, and $100$ of those are heads ($98$ from the fair coins, $2$ from the $HH$ coin);

*If I'm seeing a heads, it means I saw one of the $100$ faces that could be heads;

*Since $2$ of those $100$ heads faces are provided by the $HH$ coin, if I'm seeing heads, $2/100$ or $1/50$ is the probability that I'm seeing a face of the $HH$ coin.

