I've an unfair coin which has the probability of the head =70%, I toss the coin the first time and I get a tail. Now how many time do I need to flip the coin so that the number of heads is equal to the number of tails.
My solution: As I already have one tail, I need n-1 tail and n heads. If I have 3 coin tosses, I expect 2.1 expectations of heads and 0.9 of tails expectation. So I answered 3. I'm not sure if this is the right way to solve, feel free to clarify anything that seems unclear
Suppose the answer you seek is $E$. Consider what happens on the first toss. Either you get $H$ and can stop or you get $T$. In the latter case, you now have twice as long to wait as you must first get back to a deficit of just $1$ and then pass to even. Each of those steps is expected to take $E$ turns
Considering the probabilities of each scenario we get $$E=.7\times 1+.3\times (2E+1)\implies \boxed {E=\frac 52}$$