Is This the Biased Coin? Part 1

Let $HH$ be the event of getting two heads in a row, and let $HHH$ be the event of getting three heads in a row. Let $B$ be the event of choosing the biased coin, so that $B'$ is the event of choosing the fair coin. Thus $\begin{aligned} P[HH] & = \; P[HH|B]P[B] + P[HH|B']P[B'] \; = \; 1 \times \tfrac12 + \tfrac14 \times \tfrac12 \; = \; \tfrac58 \\ P[HHH \cap HH] \; = \; P[HHH] & = \; P[HHH|B]P[B] + P[HHH|B']P[B'] \; = \; 1 \times \tfrac12 + \tfrac18 \times \tfrac12 \; = \; \tfrac{9}{16} \end{aligned}$ so that $P[HHH|HH] \; = \; \frac{P[HHH \cap HH]}{P[HH]} \; = \; \frac{\frac58}{\frac{9}{16}} \; = \; \tfrac{9}{10}$ making the answer $\boxed{90}$ %.

In the beginning, there was a $\frac{1}{2}$ chance that the coin you picked had a $\frac{1}{2}$ chance of landing on heads, and a $\frac{1}{2}$ chance that the coin would land on heads with probability 1. Therefore, the probability of getting heads in the beginning is $\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times 1$ , which equals $\frac{3}{4}$ .

So, there was a $\frac{1}{4}$ chance that the first flip would've resulted in tails. If the first flip resulted in tails, then you certainly chose the fair coin. We can call the probability that you chose the fair coin if the first flip resulted in heads $p$ , and solve for it. Since the overall probability that you chose the fair coin in the beginning was $\frac{1}{2}$ , we could form the equation $\frac{1}{4} \times 1 + \frac{3}{4} \times p = \frac{1}{2}$ . Solving for $p$ gives $\frac{1}{3}$ , meaning that after the first flip resulted in heads, there was a $\frac{1}{3}$ chance you chose the fair coin. Therefore, the probability of getting heads on the second flip was $\frac{1}{3} \times \frac{1}{2} + \frac{2}{3} \times 1$ , which equals $\frac{5}{6}$ .

We can apply the same process to find the answer to the problem. There was a $\frac{1}{6}$ chance that the second flip would've resulted in tails, and if it did, you certainly picked the fair coin. Let's call the probability of getting heads on the third flip provided the second flip resulted in heads $q$ . Like I said, there was a $\frac{1}{3}$ chance that you picked the fair coin after the first flip resulted in heads. So, $\frac{1}{6} \times 1 + \frac{5}{6} \times q = \frac{1}{3}$ . Solving gives $q=\frac{1}{5}$ . So, after the first two flips resulted in heads, there was a $\frac{1}{5}$ chance you picked the fair coin, and a $\frac{4}{5}$ chance you picked the biased coin. Therefore, the answer is $\frac{1}{5} \times \frac{1}{2} + \frac{4}{5} \times 1 = 0.9$ , which is $\boxed{90}$ as a percentage.