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Conditional Probabilities in Tennis

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Homework Statement


This isn't really homework, just reviewing for a test. This is problem 3.17 in 'A modern introduction to probability and statistics: understanding why and how' Dekking.

But since it can be seen as a HW problem, might as well post here.

Question:

You and I play a tennis match. It is deuce, which means if you win the next two rallies, you win the game; if I win both rallies, I win the game; if we each win one rally, it is deuce again. Suppose the outcome of a rally is independent of other rallies, and you win a rally with probability p. Let W be the event you win the game, G the game ends after the next two rallies, and D it becomes deuce again.

(a) Determine P(W|G).
(b) Show that P(W) = p^2 + 2p(1 - p)P(W|D) and use P(W) = P(W|D) (why is this so?)
to determine P(W).
(c) Explain why the answers are the same.



The attempt at a solution
(a) P(W|G) = p^2
Since you have a probability p of winning each rally (and they are independent).

(b) P(W) = P(W ∩ G) + P(W ∩ D) , since D and G are mutually exclusive exhaustive events
P(W) = P(W|G)P(G) + P(W|D)P(D) = p^2 * P(G) + p(1-p) P(W|D)

I know P(W|G) = p^2 from (a), and P(D) = p(1-p) since we would each have to win one rally for it go to deuce.

I am not sure how to compute P(G), I thought it should P(G) = p^2 + (1-p)^2, either you win both or I win both; however, this doesn't give the desired result what we were supposed to show.

----------------
P(W) = P(W|D) : I suppose because if it's a Deuce we "reset" and have an equal chance of winning again.

Taking what we had to show for granted I can solve for P(W)

P(W) = p^2 + 2p(1-p)P(W|D)
P(W) - 2p(1-p)P(W) = p^2
P(W)(1 - 2p(1-p)) = p^2
P(W) = p^2 / (1 - 2p + p^2)

(c) What two answers is the question referencing to being the same?
Was I supposed to get the same answer for P(W) and P(W|G)?
 

Answers and Replies

  • #2
haruspex
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.., since D and G are mutually exclusive exhaustive events

.., and P(D) = p(1-p) .

I am not sure how to compute P(G),
Ponder those three statements.
 
  • #3
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Ponder those three statements.
G: Game ends in the next two rallies
D: Game is deuce again

So P(G U D) = 1 , they are exhaustive. Meaning either game ends or it doesn't, and if it doesn't then it must be deuce.

Similarly, it is the case that either the game ends or it's deuce, and certainly can't be both, so they do seem disjoint.

P(D) = 2 * p (1 - p), I did overlook a factor of 2, since there are two ways to arrange it p * (1 - p) or (1 - p) * p.

If everything else in my computation is right, it's saying P(G) must be 1, but that doesn't make much sense. I still think P(G) = p^2 + (1 - p)^2 (i.e. I win twice or you win twice, since rallies are independent, just multiply).
 
  • #4
haruspex
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So P(G U D) = 1 , they are exhaustive. Meaning either game ends or it doesn't, and if it doesn't then it must be deuce.

Similarly, it is the case that either the game ends or it's deuce, and certainly can't be both, so they do seem disjoint.

P(D) = 2 * p (1 - p), I did overlook a factor of 2, since there are two ways to arrange it p * (1 - p) or (1 - p) * p.
Well spotted
If everything else in my computation is right, it's saying P(G) must be 1,
No, if G and D are mutually exclusive and between them cover all eventualities then the sum of their probabilities is 1
. I still think P(G) = p^2 + (1 - p)^2 (i.e. I win twice or you win twice, since rallies are independent, just multiply).
Yes, and that's the same as 1 - P(D).
 
  • #5
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Well spotted

No, if G and D are mutually exclusive and between them cover all eventualities then the sum of their probabilities is 1
Yes, and that's the same as 1 - P(D).
Thanks, I'm still having trouble with what is required to show (unless this is an error in the book).

I want to show: P(W) = p^2 + 2p(1 - p)P(W|D)

I have

P(W) = P(W ∩ G) + P(W ∩ D) = P(W|G)P(G) + P(W|D)P(D) = p^2 * P(G) +2p(1-p) P(W|D)

Where I have P(W|G) = p^2, P(D) = 2p(1 - p) and if I substitute P(G) = p^2 + (1 - p)^2 or 1 - 2p(1 - p) I do not get what I want to show.
 
  • #6
haruspex
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I just noticed your answer to a is wrong. Surely P(W&G) is p squared?
 
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  • #7
Ray Vickson
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Homework Statement


This isn't really homework, just reviewing for a test. This is problem 3.17 in 'A modern introduction to probability and statistics: understanding why and how' Dekking.

But since it can be seen as a HW problem, might as well post here.

Question:

You and I play a tennis match. It is deuce, which means if you win the next two rallies, you win the game; if I win both rallies, I win the game; if we each win one rally, it is deuce again. Suppose the outcome of a rally is independent of other rallies, and you win a rally with probability p. Let W be the event you win the game, G the game ends after the next two rallies, and D it becomes deuce again.

(a) Determine P(W|G).
(b) Show that P(W) = p^2 + 2p(1 - p)P(W|D) and use P(W) = P(W|D) (why is this so?)
to determine P(W).
(c) Explain why the answers are the same.



The attempt at a solution
(a) P(W|G) = p^2
Since you have a probability p of winning each rally (and they are independent).

(b) P(W) = P(W ∩ G) + P(W ∩ D) , since D and G are mutually exclusive exhaustive events
P(W) = P(W|G)P(G) + P(W|D)P(D) = p^2 * P(G) + p(1-p) P(W|D)

I know P(W|G) = p^2 from (a), and P(D) = p(1-p) since we would each have to win one rally for it go to deuce.

I am not sure how to compute P(G), I thought it should P(G) = p^2 + (1-p)^2, either you win both or I win both; however, this doesn't give the desired result what we were supposed to show.

----------------
P(W) = P(W|D) : I suppose because if it's a Deuce we "reset" and have an equal chance of winning again.

Taking what we had to show for granted I can solve for P(W)

P(W) = p^2 + 2p(1-p)P(W|D)
P(W) - 2p(1-p)P(W) = p^2
P(W)(1 - 2p(1-p)) = p^2
P(W) = p^2 / (1 - 2p + p^2)

(c) What two answers is the question referencing to being the same?
Was I supposed to get the same answer for P(W) and P(W|G)?
Sometimes it helps in such problems to look at some more details---in particular, the nature of the "sample space" and the events therein.

To change notation, let the two players be called A and B, and let T denote a "tie" in one round (what you call a deuce). (Here, a round equals two rallies). The sample space S is the set of outcomes
S ={A,B,TA,TB,TTA,TTB,TTTA,TTTB, ....}, where these show the outcomes of the successive rounds.

The event that A wins the game is {A wins} = {A,TA,TTA,TTTA, ... }. If a = P{A} = p^2, b = P{B} = (1-p)^2 and t = P{T} = 1-a-b = 2*p*(1-p), we can get P{A wins} as a convergent infinite series involving t and a.

In this notation, your event G = {A,B} and your event D = {T,TA,TB,TTA,TTB, ....}, which are all the sample points starting with 'T'.
 
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  • #8
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I just noticed your answer to a is wrong. Surely P(W&G) is p squared?
Oh.
a) P(W|G) = P(W & G) / P(G) = p^2 / (p^2 + (1-p)^2) = p^2 / (2p^2 + 1 -2p)
b) Follows trivially from the rest.
P(W)(1 - 2p(1-p)) = p^2
P(W) = p^2 / (1 + 2p^2 - 2p)
c) P(W|G) = P(W) = P(W|D)
Which means winning is independent of whether or not the game ends in two rallies or goes to deuce.

Sometimes it helps in such problems to look at some more details---in particular, the nature of the "sample space" and the events therein.

To change notation, let the two players be called A and B, and let T denote a "tie" in one round (what you call a deuce). (Here, a round equals two rallies). The sample space S is the set of outcomes
S ={A,B,TA,TB,TTA,TTB,TTTA,TTTB, ....}, where these show the outcomes of the successive rounds.

The event that A wins the game is {A wins} = {A,TA,TTA,TTTA, ... }. If a = P{A} = p^2, b = P{B} = (1-p)^2 and t = P{T} = 1-a-b = 2*p*(1-p), we can get P{A wins} as a convergent infinite series involving t and a.

In this notation, your event G = {A,B} and your event D = {T,TA,TB,TTA,TTB, ....}, which are all the sample points starting with 'T'.
Thanks! That does provide a good conceptual map to understand the underlying outcomes.
 

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