Player A tells the truth $4$ out of $9$ times. A card is drawn from pack of $52$ cards, and A reports that there is diamond. What is probability that actually there was a diamond?
Probability of truth is $4/9$ probability of lie is $5/9$
Probability that he reported there is diamond is $$\frac{4}{9}\cdot\frac{13}{52}+\frac{5}{9}\cdot\frac{39}{52} = \frac{19}{36}$$
Applying Bayes' theorem, the probability of actual diamond is:
$$\frac{\frac{4}{9}\cdot\frac{13}{52}}{\frac{19}{36}}$$
I don't know if I'm right or wrong, can anyone verify it? Thanks in advance
Best Answer
If a Diamond is reported, one of two things happened. Diamond+truth or Other+lie.
Those probabilities are $\frac{1}{4}\times\frac{4}{9}$ and $\frac{3}{4}\times\frac{5}{9}$ or $\frac{1}{9}$ and $\frac{5}{12}$
Thus the probability of it actually being a diamond would be $$\frac{\frac{1}{9}}{\frac{1}{9}+\frac{5}{12}}=\frac{4}{19}$$
I think your (4/9)(13/52)+(5/9)(39/52) calculation was incorrect.