Probabilty of next two tossing coin if we know the result from small experiment 
We toss a coin 20 times which have heads $12$ times and tails $8$ times, what is the probability for the next two tossing that it will be two heads?

I don't know when I should start? And how to find this probability?
 A: If you have a fair coin, tossing heads or tails after some specific configuration has still a $50-50$ chance. Using the right terminology, each toss is independent of the others. That means that the coin has "no memory" i.e. it does not care what its previous flips were. Tossing a fair coin always yields with $50\%$ chance heads and yields tails with the remaining $50\%$. This thinking that after $n $ consecutive heads we should have tails (and vice-versa) is called the gambler's paradox I believe.
A: If the coin is guaranteed to be fair, then future results are independent of previous results.   In that case, we may discard the test as irrelevant to the problem, and focus on the question of: What is the probability of throwing two consecutive heads with two tosses of a fair coin?

If the coin is suspected to be biased, then the twenty toss test was performed to provide an estimate of that bias.   The problem is: what is the conditional probability of obtaining two heads with two further tosses given that we obtained those results.
Then your measure of probability for obtaining the event given the results, will depend on what you assume about the distribution of the bias.
$$\mathsf P(E\mid R) = \int_0^1 \mathsf P(E\mid B=p)~f_{B\mid R}(p)\operatorname d p = \dfrac{\int_0^1 \binom{20}{12}p^{14}(1-p)^{8}~f_B(p)\operatorname d p}{\int_0^1 \binom{20}{12}p^{12}(1-p)^8~f_B(p)\operatorname d p}$$
So, what might you assume about the distribution of the bias?
A: Coin tosses are not dependent on each other, so the probability that the next two tossings both being heads will just be the probability for getting a head squared. 
In case of a fair coin, the result would be $(\frac 12) ^2=\frac 14$
