1. ## Simple probability

Hey guys, newbie here. I'm in trouble with this question, and I would really appreciate help.

In a game of tennis, each point is won by one of the two players A and B. The usual rules of scoring for tennis apply. i.e. the winner of the game is the player who first scores four points, unless each player has won three points, when deuce is called and play proceeds until one player is two points ahead of the other and hence wins the game.

A is serving and has probability of winning any point of $\frac{2}{3}$. The result of each point is assumed to be independent of every other point.

a) Show that the probability of A winning the game without deuce being called is $\frac{496}{729}$ - I can do this so don't worry about this one.

b) Find the probability of deuce being called.

c) If deuce is called, show that A's subsequent probability of winning the game is $\frac{4}{5}$

d) Hence determine A's overall chance of winning the game.

For b) I did:
$9 ( \frac{2}{3} . \frac{2}{3} . \frac{2}{3} . \frac{1}{3} . \frac{1}{3} . \frac{1}{3} ) = \frac{8}{81}$
I multiplied it by 9 is because, unless I'm mistaken, there are 9 different, acceptable combination it occuring. Of course, you can't let player B win 3 points in a row, or else the game will be over. Likewise with player A. Is this right?

For c) I think it involves the sum of an infinite geometric series. But I don't have any idea.

2. Because I do not fully understand the scoring of tennis this may be wrong.
But as you describe a deuce would be any arrangement of the string “AAABBB”.
If that is correct, then there are $\frac{{6!}}{{\left( {3!} \right)^2 }} = 20$ ways to rearrange that string.
So $\frac{{6!}}{{\left( {3!} \right)^2 }}\left( {\frac{2}{3}} \right)^3 \left( {\frac{1}{3}} \right)^3 = {\rm{0}}{\rm{.219}}$.
Is that correct?

3. Yes, I think so. I guess they can win 3 points in a row.

Any ideas on the geometric one (if it is a geometric series question)?

4. Don't worry, I've worked it out.

Thanks for the help Plato.