Box contains 2 different coins, one is chosen randomly and tossed n times, head came all n times. Probability that n+1 toss is a head too? A box contains two coins: a regular coin and one fake two-headed coin (P(H)=1). One coin is choose at random and tossed $n$ times. 
If the first n coin tosses result in heads, What is the probability that the $(n+1)^{th}$ coin toss will also result in heads?

My solution:
$$\text{Required Probability} = \frac{1}{2}\times \left(\frac{1}{2}\right)^n \times \frac{1}{2}+\frac{1}{2} \times 1^n \times 1$$

The probability of seeing $n$ heads in $n$ tosses of a fair coin is $2^{-n}$.  Thus the total probability of seeing $n$ consecutive heads is $$\frac 12\times 2^{-n}+\frac 12 \times 1$$ Therefore, given that you have in fact observed $n$ consecutive heads the new estimate for the probability that you have the fair coin is $$\frac {\frac 12 \times 2^{-n}}{\frac 12\times 2^{-n}+\frac 12 \times 1}=\frac {1}{1+2^n}$$  and the probability that you have the fake coin is therefore $$\frac {2^n}{1+2^n}$$  It follows that the probability that the next toss is also heads is $$\boxed {\frac 12\times \frac {1}{1+2^n}+\frac {2^n}{1+2^n}=\frac {1+2^{n+1}}{2+2^{n+1}}}$$
Note:  as $n$ goes to $\infty$ this tends to $1$, as it clearly should.
