Problem 1: random unfair coin

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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?


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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