# $2$ fair coins, $1$ biased coin $-$ probability of getting a $3^{\text{rd}}$ head after $2$ consecutive heads?

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?

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

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 :).