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~

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- Aug 30th 2009, 06:01 PMuglingprobability 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~

- Aug 30th 2009, 07:48 PMmatheagle
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.

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

$\displaystyle P({\rm win\; in\; 5})=4P(LWWWW)=4qp^4$

since the loss can come in any game EXCEPT the fifth game (geometric)

$\displaystyle P({\rm win\; in\; 6})=10P(LLWWWW)=10q^2p^4$

since the loss can come in any game EXCEPT the sixth game (neg binomial)

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

$\displaystyle P({\rm win\; in\; 7})=20P(LLLWWWW)=20q^3p^4$

since the loss can come in any game EXCEPT the sixth game (neg binomial)

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

Now, add these, since they are mutually exclusive. - Aug 30th 2009, 08:54 PMuglingthank you~
I think your answer is the right one~(Clapping)

- Aug 30th 2009, 09:11 PMmatheagle
was there a contest here?