# Calculating probability. Question on fair coins.

I have two fair coins with me, one having both side head and other having one head and one tail. I toss any one of these coins for the first time(with my eyes closed, and then i open my eyes to see whether it's head or tail), if I get tail it is confirm that it is H-T coin. If i get head i'll toss it again to confirm whether it is H-H coin or H-T coin. For the second time if i get tail it is confirm that it's H-T coin, if head tossing again. Like this i tossed for six times. And i got head in all the six times. Then I confirm that it is H-H coin. What is the probability that my answer is correct?

The simplest way to compute the conditional probability that your inference is correct given that you observed six heads in a row is to use Bayes' theorem. Let $H$ represent the event that the coin you flipped is the two-headed coin; and $S$ represent the event that you obtained six heads in six trials. Then by the law of total probability, $$\Pr[S] = \Pr[S \mid H]\Pr[H] + \Pr[S \mid H^c]\Pr[H^c] = 1 \cdot \frac{1}{2} + \frac{1}{2^6} \cdot \frac{1}{2} = \frac{65}{128},$$ where $H^c$ is the complementary event (that the coin you flipped is the head/tail coin). This is because, given the coin is two-headed, you are certain to observe all heads, so $\Pr[S \mid H] = 1$. If the coin is head/tail, then the probability of observing all heads in 6 trials is $(1/2)^6$. And since you are equally likely to have chosen either type of coin, $\Pr[H] = \Pr[H^c] = \frac{1}{2}$.

Then by Bayes' theorem, $$\Pr[H \mid S] = \frac{\Pr[S \mid H]\Pr[H]}{\Pr[S]} = \frac{1/2}{65/128} = \frac{64}{65}.$$

We will use the formal machinery of conditional probabilities. We assume that when we pick a coin to toss, we choose the one-headed coin and the two-headed coin with equal probabilities.

Let $T$ be the event that we picked the two-headed coin, and let $S$ be the event we get $6$ heads in a row. We want the conditional probability $P(T|S)$. By the usual definition of conditional probability, we have $$\Pr(T|S)=\frac{\Pr(T\cap S)}{\Pr(S)}.$$ We compute the probabilities on the right.

There are two ways that $S$ can happen: (i) We picked the two-headed coin, and got $6$ heads, or (ii) we picked the ordinary coin, and got $6$ heads.

The probability of (i) is simply $\frac{1}{2}$. The probability of (ii) is $\frac{1}{2}\cdot \frac{1}{2^6}$. It follows that $\Pr(S)=\frac{1}{2}+\frac{1}{2}\cdot \frac{1}{2^6}$.

The probability $\Pr(T\cap S)$ has already been dealt with: it is just $\frac{1}{2}$.

Divide, and simplify.