In the Banana Country, only Mr Decent always tells the truth and only Mr Joke always tells lies. Everyone else has a probability of $p$ to tell a lie. One day, Mr Decent has decided to run for the President and told his decision to the first person who in turn told this to the second person. The second person then told this to the third person, and so on, till the $n^{th}$ person who told this news to Mr Joke. No one has been told this news twice in this process. Finally, Mr Joke announced Mr Decent's decision to everyone. What is the probability that Mr Joke's statement agrees with Mr Decent's intention?
Let $A_n$ be the probability that Mr Joke's statement agrees with Mr Decent's intention, and $D_n = 1-A_n$ be the probability that they disagree.
Mr Joke will give the correct statement if and only if the $n^{th}$ person tells him an incorrect statement. This probability depends on the statement of the $(n-1)^{th}$ person which all also influences the value of $A_{n-1}$ and $D_{n-1}$. The probability of the $(n-1)^{th}$ person to give the correct information equals $D_{n-1}$ and that of a wrong statement is $A_{n-1}$. Therefore, it should hold that $$A_n=pD_{n-1} + (1-p)A_{n-1}=p+(1-2p)A_{n-1}$$
This recursion can be rewritten as $$A_n-A_{n-1}=(1-2p)(A_{n-1}-A_{n-2})$$
Meanwhile, when $n=0$, i.e. Mr Decent shared his decision with Mr Joke directly, $A_0=0$. When $n=1$, $A_1=p$. Therefore $$A_n - A_{n-1} = p(1-2p)^{n-1}\implies A_{n-1}=A_n-p(1-2p)^{n-1}$$
Setting this back to the original recursion yields $$A_n=\boxed{\frac{1}{2}-\frac{1}{2}(1-2p)^n}$$