# Probability - Coin Toss [duplicate]

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

## marked as duplicate by José Carlos Santos, Leucippus, max_zorn, ArsenBerk, ShaileshOct 6 '18 at 10:58

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 $$E=.7\times 1+.3\times (2E+1)\implies \boxed {E=\frac 52}$$
• Well, sure. But there is no need to evoke any heavy machinery. If you are walking from $A$ to $C$ and must pass through $B$ then your expected time is the sum of your expected times from $A$ to $B$ and from $B$ to $C$. As simple as that. – lulu Oct 5 '18 at 16:01