I encountered this problem:

We have three coins, two fair coins and one coin with heads on each face. One coin is picked randomly among those three coins and is flipped two times. We see the sequence: H, H. What is the probability of obtaining heads if we flip this coin one more time?

To tackle it, here's my reasoning:

  1. Find the probabilities of the coin being fair and unfair conditional on the sequence: H, H.
  2. Find the probability of getting a heads in the additional flip conditional on the sequence: H, H.

1) Using Bayes formula we have: \begin{align} P(fair|HH) &= \frac{P(HH|fair)P(fair)}{P(HH|fair)P(fair)+P(HH|unfair)P(unfair)} \\ &= \frac{\frac{1}{4}\frac{2}{3}}{\frac{1}{4}\frac{2}{3}+1\cdot\frac{1}{3}} \\ &= \frac{1/6}{1/6+1/3} \\ &= \frac{1}{3} \end{align} We get at the same time: \begin{align} P(unfair|HH) = 1 - 1/3 = 2/3 \end{align}

2) Let's denote $A$ the event of obtaining of heads at the additional flip. We have: \begin{align} P(A) = P(A|fair)P(fair) + P(A|unfair)P(unfair) \end{align} and conditioning on the sequence $H,H$ we get: \begin{align} P(A|HH) &= P(A|fair,HH)P(fair|HH) + P(A|unfair,HH)P(unfair|HH) \\&= P(A|fair)P(fair|HH) + P(A|unfair) P(unfair|HH) \end{align} We then get: \begin{align} P(A|HH) &= \frac{1}{2}\frac{1}{3} + 1\cdot\frac{2}{3} \\ P(A|HH) &= \frac{5}{6} \end{align}

Is my reasoning correct? Is there maybe a quicker way to get the result?


Your answer is correct. A shorter way of writing it: Let $H_n$ denote the event that we get $H$ at the $n$-th coin toss. Hence, we want the probability $$P(H_3 \vert (H_1 \cap H_2))=\frac{P(H_1 \cap H_2 \cap H_3)}{P(H_1 \cap H_2)}=\frac{P(H_1 \cap H_2\cap H_3 \vert fair)\cdot P(fair)+ P(H_1 \cap H_2\cap H_3 \vert unfair)\cdot P(unfair)}{P(H_1 \cap H_2 \vert fair)\cdot P(fair)+ P(H_1 \cap H_2 \vert unfair)\cdot P(unfair)}=\frac{(\frac{1}{2})^3 \cdot \frac{2}{3}+1^3\cdot \frac{1}{3}}{(\frac{1}{2})^2 \cdot \frac{2}{3}+1^2\cdot \frac{1}{3}}=\frac{5}{6}.$$


Similar concept, but perhaps simpler:

Say the coins are $A,B,C$ with $C$ the unfair one.

There are three ways to get the outcome $HH$.

You could pick $C$ (probability $\frac 13$) and then get $HH$ (probability $1$).

You could pick $A$ (probability $\frac 13$) and then get $HH$ (probability $\frac 14$).

You could pick $B$ (probability $\frac 13$) and then get $HH$ (probability $\frac 14$).

Thus the total probability of getting $HH$ is $$\frac 13+2\times \frac 1{12}=\frac 12$$

Of this, $\frac 13$ is explained by having chosen the unfair coin so the probability that you have the unfair coin is $\frac 23$ and the probability that you have one of the fair coins is $\frac 13$. Thus the probability that the next toss is $H$ is $$\frac 23\times 1+\frac 13\times \frac 12=\frac 56$$

  • $\begingroup$ $\frac 23\times 1+\frac 13\times \frac 12=\frac 56$ Is this part not clear at all? ahh $\endgroup$
    – heiverjust
    Jun 8 at 19:38
  • $\begingroup$ @heiverjust Is it clear? The computation has two parts. Letting $p_u$ be the probability you have the unfair coin, and $p_f=1-p_u$ the probability you have the fair coin, we first use Bayes' Theorem to find $p_u, p_f$. Then we note that the probability that the next one is $H$ is $p_u\times 1+p_f\times \frac 12$. $\endgroup$
    – lulu
    Jun 8 at 20:29

Your reasoning is correct and that's the answer I got as well. It's pretty much the simplest way I know how to get it. You can use Bayes Theorem to get the unfair probability directly, but that doesn't cut orders of time off of the calculation or anything.


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