I recently stumbled upon a question and I can't seem to find the answer. Here it is:
You are given a coin. You estimate that there is a probability of 3/4 that the coin is a fair coin (1/2 chance of obtaining tail and 1/2 chance of obtaining head) and a probability of 1/4 that the coin is rigged (3/4 chance of obtaining tail and 1/4 chance of obtaining head). Knowing that you estimated the probability that the coin is rigged with probability 1/4, what is the probability that after 2 flips you estimate the probability that the coin is rigged with probability more than 1/4.
Thank you in advance for your replies!
Let $F$ be the event that the coin is fair. Let $R = F^C$ be the event that the coin is rigged. Let $HH,HT,TH,TT$ be the events that the coin flips the sequence shown.
$$P(HH|F) = P(HT|F) = P(TH|F) = P(TT|F) = \dfrac{1}{4}$$
$$P(HH|R) = \dfrac{1}{16}, P(HT|R) = P(TH|R) = \dfrac{3}{16}, P(TT|R) = \dfrac{9}{16}$$
$$P(HH) = P(HH|F)P(F)+P(HH|R)P(R) = \dfrac{3}{16}+\dfrac{1}{64} = \dfrac{13}{64}$$
$$P(HT) = P(TH) = P(HT|F)P(F)+P(HT|R)P(R) = \dfrac{3}{16}+\dfrac{3}{64} = \dfrac{15}{64}$$
$$P(TT) = P(TT|F)P(F)+P(TT|R)P(R) = \dfrac{3}{16}+\dfrac{9}{64} = \dfrac{21}{64}$$
$$P(R|HH) = \dfrac{P(HH|R)P(R)}{P(HH)} = \dfrac{\dfrac{1}{64}}{\dfrac{13}{64}} = \dfrac{1}{13}$$
$$P(R|HT) = P(R|TH) = \dfrac{P(HT|R)P(R)}{P(HT)} = \dfrac{\dfrac{3}{64}}{\dfrac{15}{64}} = \dfrac{1}{5}$$
$$P(R|TT) = \dfrac{P(TT|R)P(R)}{P(TT)} = \dfrac{\dfrac{9}{64}}{\dfrac{21}{64}} = \dfrac{3}{7}$$
So, the only case where you would estimate the coin is rigged to have a greater than $\dfrac{1}{4}$ chance is when it flips $TT$. This occurs $\dfrac{21}{64}$ of the time.