Expected value in rock paper scissor Two players play rock, paper, scissors for 3 wins ( one player wins 3 times). What is the expected value of the number of rounds?
I tried: for the n-th games probability : $$ p(x_n)=2\frac{\binom{n}{3}}{3^n} $$
Therefore my $E(x)\approx 1.8733$ which is too low...

As suggested in comments, let's consider the states the game can be in after each round. $S(x,y)$ is the state where player A has won $x$ rounds so far and player B has won $y$ rounds so far. A tie round keeps the game in the same state. And let $T(x,y)$ be the distribution of the number of turns remaining in the game at state $S(x,y)$.
Since the rules and stop condition are symmetric for both players, $T(y,x) = T(x,y)$. When either win count reaches $3$, our overall count is done: $T(x,3)=T(3,y)=0$. When $x<3$ and $y<3$, we get the recursion
$$ E\{T(x,y)\} = 1 + \frac{1}{3} E\{T(x,y)\} + \frac{1}{3} E\{T(x+1,y)\} + \frac{1}{3} E\{T(x,y+1)\} \\
E\{T(x,y)\} = \frac{3}{2} + \frac{1}{2} E\{T(x+1,y)\} + \frac{1}{2} E\{T(x,y+1)\} $$
From here the expected values are easy to compute, going from largest $x$ and $y$ to smallest:
$$E\{T(2,2)\} = \frac 32 \\
E\{T(2,1)\} = \frac 32 + \frac 34 = \frac 94 \\
E\{T(2,0)\} = \frac 32 + \frac 98 = \frac{21}{8} \\
E\{T(1,1)\} = \frac 32 + \frac 98 + \frac 98 = \frac{15}{4} \\
E\{T(1,0)\} = \frac 32 + \frac{21}{16} + \frac{15}{8} = \frac{75}{16} \\
E\{T(0,0)\} = \frac 32 + \frac{75}{32} + \frac{75}{32} = \frac{99}{16}$$
The last line is the expected total number of rounds for the whole game, $99/16$.
