# Problem 1: random unfair coin

** Published:**

There is a fair coin (one side heads, one side tails) and an unfair coin (both sides tails). You pick one at random, flip it 5 times, and observe that it comes up as tails all five times. What is the chance that you are flipping the unfair coin?

### Solution:

We will use Bayes’ theorem and the law of total probability. Define the following events:

- $A$ = obtaining 5 tails,
- $F$ = the coin is fair,
- $U$ = the coin is unfair.

Note that $F$ and $U$ are exclusive events. By the law of total probability, we have that

\[\mathbb{P}(A) = \mathbb{P}(A|U)\mathbb{P}(U)+ \mathbb{P}(A|F)\mathbb{P}(F).\]Note that if the coin is unfair, then all we would observe are tails, hence the first term is $1/2$. On the other hand, if the coin is fair, then the probability of obtaining five tails in a row is $(1/2)^{-5}$, and then the second term is $(1/2)^{-6}$. Now we compute $\mathbb{P}(U|A)$ using Bayes’ theorem:

\[\mathbb{P}(U|A) = \dfrac{\mathbb{P}(A|U)\mathbb{P}(U)}{\mathbb{P}(A)}.\]Putting everything together we obtain

\[\mathbb{P}(U|A) = \dfrac{\mathbb{P}(A|U)\mathbb{P}(U)}{\mathbb{P}(A|U)\mathbb{P}(U)+ \mathbb{P}(A|F)\mathbb{P}(F)} = \dfrac{1\cdot\frac{1}{2}}{1\cdot \frac{1}{2}+(\frac{1}{2})^5\cdot\frac{1}{2}} = \frac{32}{33}\]
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