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I solved this problem but my solution is apparently incorrect.

Let $A$ and $B$ denote the event that we chose the fair and biased (both sides are heads) coin, respectively. $P(A) = \frac{2}{3}$, $P(B) = \frac{1} {3}$. Let $HH$ denote event of getting $2$ heads.

Then

\begin{align} P(A|HH) = \frac{P(HH|A)P(A)}{P(HH)} \\ = \frac{\frac{1}{4}\cdot\frac{2}{3}}{\frac{1}{4}\cdot\frac{2}{3} + 1\cdot\frac{1}{3}} \\ = \frac{1}{3} \\ P(B|HH) = \frac{P(HH|B)P(B)}{P(HH)} \\ = \frac{\frac{1}{3}}{\frac{1}{4} \cdot \frac{2}{3} + 1\cdot\frac{1}{3}} \\ = \frac{2}{3} \\ \end{align}

Let $3H$ denote the event of getting a $3^{\text{rd}}$ head. We can essentially treat $A$ as $1$ single fair coin chosen with probability $\frac{1}{3}$ and $B$ as $1$ single biased coin chosen with probability $\frac{2}{3}$.

\begin{align} P(3H) = P(3H|A)P(A) + P(3H|B)P(B) \\ = \frac{1}{2} \cdot \frac{1}{3} + \frac{2}{3} = \frac{5}{6} \end{align}

But apparently the solution is $\frac{3}{4}$. Where did I go wrong in my thought process?

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  • $\begingroup$ What is the probability to get H for the biased coin? From your computation, it seems like you have $P(HH|B) = 1/3$. $\endgroup$
    – Raoul
    Aug 16, 2020 at 22:50
  • $\begingroup$ @Raoul No. It's 1. Are you referring to the line starting with $P(B|HH)$? The numerator on the RHS is $P(HH|B) * P(B) = 1 * 1/3$ $\endgroup$ Aug 16, 2020 at 22:58
  • $\begingroup$ Indeed, I did not think this through. Then it looks right. Do you have the solution where this $3/4$ is found? $\endgroup$
    – Raoul
    Aug 16, 2020 at 23:14
  • $\begingroup$ @user5965026: A simulation supports your answer of ${\large{\frac{5}{6}}}$. $\endgroup$
    – quasi
    Aug 16, 2020 at 23:16
  • $\begingroup$ You're sure that the biased coin always comes up heads? $\endgroup$
    – mjqxxxx
    Aug 16, 2020 at 23:19

1 Answer 1

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You need to be precise describing the experiment. It looks like you are saying that (1) a single coin is chosen at random from three coins and that coin is flipped three times, and (2) two of the three coins are fair, and the third always comes up heads. And you want to know, given that the first two flips were heads, what is the probability that the third flip is also a head?

If that description is right, then your Bayesian analysis is also correct. The probability that you selected the biased coin is $2/3$, and given that, the probability of a third head is $(1/2)(1/3) + (1)(2/3)=(5/6)$. So, if the "correct" answer is $3/4$, your description of the experiment must be wrong. An alternate experiment might be to choose a coin at random, flip it, and replace it, three times. In that case, the probability of a third head would be just $(1/2)(2/3) + (1)(1/3) = (2/3)$. So that's not right either.

Yet another guess would be that the experiment is to flip each coin once, in a random order. In that case, you have either flipped the two fair coins, or a fair coin and then the biased coin, or the biased coin and then a fair coin. $$ P(2H\vert AB)=P(2H\vert BA)=1/2; \qquad P(2H\vert AA)=1/4. $$ $$ P(AB\vert 2H)=P(BA\vert 2H)=\frac{P(2H\vert AB)P(AB)}{P(2H\vert AB)P(AB) + P(2H\vert BA)P(BA) + P(2H\vert AA)P(AA)}=\frac{(1/2)(1/3)}{(1/2)(1/3)+(1/2)(1/3)+(1/4)(1/3)}=\frac{2}{5}. $$ $$ P(AA\vert 2H)=\frac{1}{5}.$$ In which case, $P(3H\vert 2H)=(4/5)(1/2)+(1/5)=(3/5)$. Which is still not right. So I'm out of guesses :).

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