Two players A and B are playing the final of chess championship which contains a series of matches. Draw is not considered
The probability that A wins is 2/3.
The probability that B wins is 1/3.
The winner will be A if wins at least 12 games and the winner will be B if wins at least 6 games.
we're given that A has won 8 games, and B has won 4 games. who is now most likely to win the championship?
sorry for my english ^ _ ^
On
The problem can be modeled through a Markov chain:
Red arrows represent transitions with probability $\frac{1}{3}$ and green arrows represent transitions with probability $\frac{2}{3}$. The transition matrix is $$ P=\left(\begin{smallmatrix}0 & \frac{1}{3} & \frac{2}{3} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{2}{3} & 0 & 0 & 0 & 0 & 0 & \frac{1}{3} \\ 0 & 0 & 0 & \frac{1}{3} & \frac{2}{3} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{2}{3} & 0 & 0 & 0 & \frac{1}{3} \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{3} & \frac{2}{3} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{2}{3} & 0 & \frac{1}{3}\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{2}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{2}{3} & \frac{1}{3} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{smallmatrix}\right)$$
and the probability that $B$ wins is $$\left(1,0,0,0,0,0,0,0,0,0\right)\,P^5 \left(0,0,0,0,0,0,0,0,0,1\right)^T = \color{red}{\frac{131}{243}}.$$
What we have here is a negative binomial.
$A$ must do one of the following to win:
Win $4$ consecutive games
Win $3$ of the next $4$ games and then win the $5th$ game.
Otherwise, $B$ would have won $2$ games and the game would be over.
We have $n$ trials given $k$ successes where
$${n-1\choose{k-1}}p^k(1-p)^{n-k}$$
Using our numbers,
$${3\choose{3}}{\frac{2}{3}}^4\frac{1}{3}^{0}+{4\choose{3}}{\frac{2}{3}}^4\frac{1}{3}^{1}\approx 0.4609$$
Thus $B$ has a greater chance of winning.
Just to check:
The probability that $B$ wins would be
$${1\choose{1}}{\frac{1}{3}}^2\frac{2}{3}^{0}+{2\choose{1}}{\frac{1}{3}}^2\frac{2}{3}^{1}+{3\choose{1}}{\frac{1}{3}}^2\frac{2}{3}^{2}+{4\choose{1}}{\frac{1}{3}}^2\frac{2}{3}^{3}\approx 0.5391$$
which is the compliment of the probability that $A$ wins, as expected.