There are 4 fair coins and 1 unfair coin that has only heads. We choose a coin and flip it three times. The result is HHH. What is the probability that the fourth flip is H?


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We would expect the probability to be greater than $\frac{1}{2}$. We must solve for $P(\text{Heads})$. We have

$$P(\text{Heads})=P(\text{Heads}|\text{fair})\cdot P(\text{fair})+P(\text{Heads}|\text{unfair})\cdot P(\text{unfair})$$

However, we cannot just say $P(\text{fair})=\frac{4}{5}$ because we are given that the first $3$ tosses are heads.

$$P(\text{fair|}HHH)=\frac{P(\text{fair} \cap HHH)}{P(HHH)}=\frac{\frac{4}{5}\cdot\frac{1}{2}^3}{\frac{4}{5}\cdot\frac{1}{2}^3+\frac{1}{5}\cdot\left(1\right)^3}=\frac{1}{3}$$

So we have that the probability that the coin was fair when we rolled those $3$ heads was $\frac{1}{3}$ and the probability that it was unfair is $\frac{2}{3}$.

Thus we have

$$\begin{align*} P(\text{Heads}) &=P(\text{Heads|fair})\cdot P(\text{fair})+P(\text{Heads|unfair})\cdot P(\text{unfair})\\\\ &= \frac{1}{2}\cdot\frac{1}{3}+1\cdot\frac{2}{3}\\\\ &=\frac{5}{6} \end{align*}$$

Which agrees with Nicola's answer.


First, use Bayes' rule to determine the probability of having selected the unfair coin. Call $UC$ the event "we selected the unfair coin" and $FC $ "we selected a fair coin".

$$P(HHH) = P(HHH|UC)P_0(UC) + P(HHH|FC)P_0(FC) = 1\times\frac{1}{5}+\frac{1}{8}\times\frac{4}{5} = \frac{3}{10}$$ $$P(UC|HHH) = \frac{P(HHH|UC) P_0(UC)}{P(HHH)} = \frac{1}{5}\times\frac{10}{3} = \frac{2}{3}$$ $$P(FC|HHH) = 1-P(UC|HHH)=\frac{1}{3}$$

So we have:

$$P(H|HHH) = P(H|HHH,UC)P(UC) + P(H|HHH,FC)P(FC) = \frac{2}{3}\times1 +\frac{1}{3}\times\frac{1}{2} = \frac{5}{6} $$


$ p(H|HHH) = p(H|unfair) + 4 \cdot p(H|fair)= p(unfair) + p(fair) \cdot p(H) = \dfrac{1}{5} + \dfrac{4}{5} \dfrac{1}{2} = \dfrac{6}{10}$

  • 1
    $\begingroup$ This is not correct. We are given that the first $3$ flips are heads. $\endgroup$ – Remy Dec 4 '17 at 22:14
  • $\begingroup$ thanks, now it is possible that it is correct, I had not understood the problem $\endgroup$ – Mario Dec 4 '17 at 22:23
  • $\begingroup$ Not quite correct. See the other answer above. $\endgroup$ – Remy Dec 4 '17 at 22:25

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