# Thread: Probability of wining a series

1. ## Probability of wining a series

A and B play a series of game, the probability of A wining is $\displaystyle p$ and the probability of B wining is $\displaystyle 1-p$. The winner of the series is the player that wins 2 more games.

A) Find the probability that A is the final winner.
B) Compute the expectation of games played.

A) in order for A to be the winner, one of these outcomes would have to occur:

$\displaystyle AA = p^2$
$\displaystyle ABA = (p^2)(1-p)$
$\displaystyle BAA = (p^2)(1-p)$

I think I would have to sum up the values, which would give:

$\displaystyle p^2+2(p^2)(1-p)= p^2(3-2p)$

B) I would think that would be around 3, considering that there are a total of 6 possible outcomes, I would think it would look something like this:

$\displaystyle E[N] = E[N=2|Y=1]P(Y=1)+E[N=3|Y=1]P(Y=1)$+$\displaystyle E[N=2|Y=0]P(Y=0)+E[N=3|Y=0]P(Y=0)$

$\displaystyle E[N] = E[N=2|Y=1]p^2+E[N=3|Y=1](p^2(3-p)$+$\displaystyle E[N=2|Y=0](1-p)^2+E[N=3|Y=0](1-p)^2(1+2p)$

2. making some modifications to part B), would it be:

$\displaystyle Y = \left\{ \begin{array}{rcl} 1 & \mbox{if} & \mbox{A wins the series} \\ 0 & \mbox{if} & \mbox{B wins the series} \end{array}\right.$

so then I would have:

$\displaystyle E[N] = E[N|Y=1]P(Y=1)+E[N|Y=0]P(Y=0)$

$\displaystyle E[N] = E[N|Y=0](p^2(3-2p))+E[N|Y=1](1-p^2)(1+2p)$

at which point I'm stuck.

3. Originally Posted by lllll
A and B play a series of game, the probability of A wining is $\displaystyle p$ and the probability of B wining is $\displaystyle 1-p$. The winner of the series is the player that wins 2 more games.

A) Find the probability that A is the final winner.
B) Compute the expectation of games played.

A) in order for A to be the winner, one of these outcomes would have to occur:

$\displaystyle AA = p^2$
$\displaystyle ABA = (p^2)(1-p)$
$\displaystyle BAA = (p^2)(1-p)$

I think I would have to sum up the values, which would give:

$\displaystyle p^2+2(p^2)(1-p)= p^2(3-2p)$

B) I would think that would be around 3, considering that there are a total of 6 possible outcomes, I would think it would look something like this:

$\displaystyle E[N] = E[N=2|Y=1]P(Y=1)+E[N=3|Y=1]P(Y=1)$+$\displaystyle E[N=2|Y=0]P(Y=0)+E[N=3|Y=0]P(Y=0)$

$\displaystyle E[N] = E[N=2|Y=1]p^2+E[N=3|Y=1](p^2(3-p)$+$\displaystyle E[N=2|Y=0](1-p)^2+E[N=3|Y=0](1-p)^2(1+2p)$
Do you mean 'The winner of the series is the player that first wins 2 games' ....? That's what your answer implies .... Except for AA, none of the others have A winning two more games than B ..... ABA and BAA both have A winning only one more game than B.

4. it's 2 more games then the other

5. Originally Posted by lllll
it's 2 more games then the other
Then for reasons I've implied in my earlier post, your calculation is not correct. Consider the following cases:

Case 1:

AA
ABAA
ABABAA
ABABABAA
.
.
.

Case 2:

BAAA
BABAAA
BABABAAA
.
.
.

And there are obviously more cases .... I don't have time now to go into it further.