Coin toss puzzle with 1 biased and 2 unbiased coins

Three coins are given: one two-head coin and two fair coins. You randomly choose a coin and the first three tosses give heads. What is the probability that 4-th toss is a head.

I have two solutions that give different results. Please help me to find the error.

1st solution: $$P(\text{first 3 tosses are heads})=\frac{2}{3}\times\frac{1}{8}+\frac{1}{3}\times 1=\frac{5}{12}$$

$$P(\text{first 4 tosses are heads})=\frac{2}{3}\times\frac{1}{16}+\frac{1}{3}\times 1=\frac{3}{8}$$

$$P(\text{4th is head} \mid \text{first 3 tosses are heads})=\frac{P(\text{first 4 tosses are heads})}{P(\text{first 3 tosses are heads})}=\frac{9}{10}.$$

2nd solution: $$P(\text{4th is head} \mid \text{first 3 tosses are heads})=$$ $$=\frac{2}{3}\times P(\text{4th is head} \mid \text{first 3 tosses are heads and coin is fair})+$$ $${}+\frac{1}{3}\times P(\text{4th is head}\mid\text{first 3 tosses are heads and coin is two-head})=\frac{2}{3}\times \frac{1}{2}+\frac{1}{3}\times 1=\frac{2}{3}.$$

• Partition the space according to whether or not you picked the unfair coin. Use Bayes' theorem to work out the probability you picked the unfair coin, given you got 3 heads. Then use the Law of total probability to compute the probability of getting a fourth head. – MattBecker82 Sep 12 '15 at 19:13
• In your 2nd solution, what general formula are you trying to apply? – MattBecker82 Sep 12 '15 at 19:41

Your second solution is erroneous. \begin{align} & \Pr(\text{4th is head} \mid \text{first 3 tosses are heads})\\[10pt] = {} & \Pr(\text{fair} \mid \text{1st 3 are heads}) \times \Pr(\text{4th is a head} \mid \text{fair & 1st 3 are heads}) \\[4pt] & {} + \Pr(\text{two-headed}\mid \text{1st 3 are heads} ) \\ & \phantom{{} + \Pr(\text{two-headed})} \times \Pr(\text{4th is a head} \mid \text{two-headed & 1st 3 are heads}) \\[10pt] = {} & \Pr(\text{fair} \mid \text{1st 3 are heads}) \times \frac 1 2 + \Pr(\text{two-headed}\mid \text{1st 3 are heads} ) \times 1. \tag 1 \end{align}

So we will find $\Pr(\text{fair}\mid\text{1st 3 are heads})$ and $\Pr(\text{two-sided} \mid\text{1st 3 are heads})$. \begin{align} & \frac{\Pr(\text{fair}\mid\text{1st 3 are heads})}{\Pr(\text{two-sided} \mid\text{1st 3 are heads})} \\[10pt] = {} & \frac{\Pr(\text{fair})}{\Pr(\text{two-headed})} \times \frac{\Pr(\text{1st 3 are heads} \mid \text{fair})}{\Pr(\text{1st 3 are heads} \mid \text{two-sided})} \\[10pt] = {} & \frac{2/3}{1/3} \times \frac{1/8}{1} = \frac 1 4. \end{align} If $p/(1-p) = 1/4$ then $p = 1/5$ and $1-p=4/5$.

Hence on line $(1)$ above we get $$\left( \frac 1 5 \times \frac 1 2 \right) + \left(\frac 4 5 \times 1\right) = \frac 9 {10}.$$

• I find the OP's explanation much simplier ! – krirkrirk Sep 12 '15 at 19:36

Your first solution uses a correct, direct approach.

The problem is in your second solution. You appear to be invoking a law of the form

$$P(A\mid B)=P(A\mid B\land C)P(C)+P(A\mid B\land \neg C)P(\neg C),$$

which is not a valid law. Try using the following instead:

$$P(A\mid B)=P(A\mid B\land C)P(C\mid B)+P(A\mid B\land \neg C)P(\neg C\mid B).$$