Coinflipping game and number of expected throws The game is as follows: a coin is flipped and if the result is head, player $A$ gets a point. If the result is tails, player $B$ gets a point. This games goes on untill one of two players has two points ahead of the other player. 
The questions are:


*

*The probability that a game takes exactly 20 turns

*The number of expected throws


For the first question, I tried to look at lower number cases: 


*

*the chance of the game ending in two throws is $0.5$: there are four possible sequences of heads and tails and only 2 result in a two-points lead for one player. 

*the chance of the game ending in three throws is zero: all possible sequences either result in the game ending after two throws or one of the players having one point ahead.


I tried doing the case of four throws, here the first two throws should be alternate (or the game is finished in two throws), the last two throws should have the same result. This gives four possibilities over a total of 16 possible sequences of throws. So the result is $0.25$.
The game can't end in 5 throws: after three throws the possible scores are $3-0$ (game ended) or $2-1$. The only possible outcomes then are $3-1$ (game ended) or $2-2$ and from there $3-2$ (or alternate, either case does not result in a $2$-point lead).  
I am not sure how to generalise this pattern to the case with $20$ throws. I am also not sure if what I have so far is correct. Could anyone give me a hint?
For the expected number of throws, I could use a hint as well, I don't really have an idea how to start that problem...
 A: These are well handled through states.  Since we don't care who wins, there are only three states:  $0,1,2$ according to whether they are tied, one player is up a point, the game is over.  We note that state $0$ can only move to state $1$ and that state $1$ moves to state $0$ or $2$ with equal probability.  Thus the only path to a in round $20$ must be $$0\mapsto 1\mapsto 0 \cdots \mapsto 0 \mapsto 1 \mapsto 2$$
In that sequence you end in state $1$ in every odd numbered round (so ten times) and each time you transition out in a specified way.  Hence the probability of that path is $$\left(\frac 12\right)^{10}$$
For the second we work recursively.  Let $E_i$ denote the expected number of rounds it will take if you are starting in state $i$.  The answer we want is $E_0$. Of course $E_2=0$. Since state $0$ can only go to state $1$ we see that $$E_0=E_1+1$$
If you are in state $1$ you go to state $0$ or state $2$ with equal probability.  Thus $$E_1=\frac 12\times 1+\frac 12\times (E_0+1)\implies 2E_1=2+E_0$$
We solve those equations simultaneously to get $$E_0=4$$
