# probability of winning a no-ad tennis game

• August 30th 2009, 06:01 PM
ugling
probability of winning a no-ad tennis game
if the probability of winning a point is p, what is the probability of winning a tennis game under the "no-ad" scoring? (The first player who wins four points wins the game.) Thanks~
• August 30th 2009, 07:48 PM
matheagle
Does that mean you can win the game 4-3?
I once solved a math monthly that asked to maximize teams winning a 7 games series.
This is the same idea. First to four wins.
Then I did calculus to show a max wrt p.
You just need to go through the possibilities.

$P({\rm win\; in\; 4})=P(WWWW)=p^4$

$P({\rm win\; in\; 5})=4P(LWWWW)=4qp^4$
since the loss can come in any game EXCEPT the fifth game (geometric)

$P({\rm win\; in\; 6})=10P(LLWWWW)=10q^2p^4$
since the loss can come in any game EXCEPT the sixth game (neg binomial)

we need that $10={5\choose 3}$ here

$P({\rm win\; in\; 7})=20P(LLLWWWW)=20q^3p^4$
since the loss can come in any game EXCEPT the sixth game (neg binomial)

we need that $20={6\choose 3}$ as a coefficient in this one.

Now, add these, since they are mutually exclusive.
• August 30th 2009, 08:54 PM
ugling
thank you~