Tennis matches

[danique]

I am not sure how to continue on a problem I have with the following question:
Two tennis players are playing a couple of matches. Every match player1 has probability p to win and player2 has probability q=1-p to win. They will stay playing till one of them have won two matches in a row. For example: “2-0, 3-1”
A) Find P(X=k) for k=0,1,2,…..
B) What is EX and for what value of p is this the biggest?

This are my solutions:
A) I thought about question A, that player1 wins if he has 2 more matches won than player2. The total of matches are x. To get the total matches won by player1 is (x/2)+1 and lost (x/2)-1. This got me the following formula for the probability:

\(\displaystyle \dfrac{x!}{\left(\dfrac{x}{2}\, +\, 1\right)!\,\left(\dfrac{x}{2}\, -\, 1\right)!}\, \left(p^{\frac{x}{2}+1}\right)\, \left(q^{\frac{x}{2}-1}\right)\)

B) I have no idea how to do B. If I have to say something I would say (1/p)*2, but I know this wrong.

I hope that someone can help me with this questions

 

[Ishuda]

I am not sure how to continue on a problem I have with the following question:
Two tennis players are playing a couple of matches. Every match player1 has probability p to win and player2 has probability q=1-p to win. They will stay playing till one of them have won two matches in a row. For example: “2-0, 3-1”
A) Find P(X=k) for k=0,1,2,…..
B) What is EX and for what value of p is this the biggest?

This are my solutions:
A) I thought about question A, that player1 wins if he has 2 more matches won than player2. The total of matches are x. To get the total matches won by player1 is (x/2)+1 and lost (x/2)-1. This got me the following formula for the probability:

\(\displaystyle \dfrac{x!}{\left(\dfrac{x}{2}\, +\, 1\right)!\,\left(\dfrac{x}{2}\, -\, 1\right)!}\, \left(p^{\frac{x}{2}+1}\right)\, \left(q^{\frac{x}{2}-1}\right)\)

B) I have no idea how to do B. If I have to say something I would say (1/p)*2, but I know this wrong.

I hope that someone can help me with this questions

For (A), what you have is then way I would approach this
(a)Player 1 wins first and wins after 2n+2 matches pⁿ⁺²qⁿ
(b)Player 1 wins first and Player 2 wins after 2n+3 matches pⁿ⁺¹qⁿ⁺²
(c)Player 2 wins first and wins after 2n+2 matches pⁿqⁿ⁺²
(d)Player 2 wins first and Player 1 wins after 2n+3 matches pⁿ⁺²qⁿ⁺¹
where n = 0, 1, 2, 3, ... The combination of (a) and (d) would almost be equivalent to what you have. Notice that if the player who doesn't win first wins, then there must be an odd number of matches. If the player who does win first wins, then there must be an even number of matches. Also the minimum number of matches is 2 for an even number of games, and 3 for an odd number of games.

For (B), I'm not sure what you mean. Do you mean the expected value of k?

 

[danique]

Thank you for helping me.

I do not fully understand why +3, because let's say you have an odd number of matches. This mean that you could have 6+3, this means that we have a difference of 3 games, but I only want to calculate the probability to win 2 matches in a row. Please correct me if I'm wrong.

For B, I want indeed the expected value of k, but I have no idea how I can do this.

For (A), what you have is then way I would approach this
(a)Player 1 wins first and wins after 2n+2 matches pⁿ⁺²qⁿ
(b)Player 1 wins first and Player 2 wins after 2n+3 matches pⁿ⁺¹qⁿ⁺²
(c)Player 2 wins first and wins after 2n+2 matches pⁿqⁿ⁺²
(d)Player 2 wins first and Player 1 wins after 2n+3 matches pⁿ⁺²qⁿ⁺¹
where n = 0, 1, 2, 3, ... The combination of (a) and (d) would almost be equivalent to what you have. Notice that if the player who doesn't win first wins, then there must be an odd number of matches. If the player who does win first wins, then there must be an even number of matches. Also the minimum number of matches is 2 for an even number of games, and 3 for an odd number of games.

For (B), I'm not sure what you mean. Do you mean the expected value of k?

 

[Ishuda]

Thank you for helping me.

I do not fully understand why +3, because let's say you have an odd number of matches. This mean that you could have 6+3, this means that we have a difference of 3 games, but I only want to calculate the probability to win 2 matches in a row. Please correct me if I'm wrong.

For B, I want indeed the expected value of k, but I have no idea how I can do this.

The 2n+3 is just so I can start n at 0. If I had started n at 1, it would be 2n for the even and 2n+1 for the odd. Let take case (b) Player 1 wins the first but Player 2 wins the whole thing. The situation is then we have pqpqpqpq....pqpqpqq which is an odd number of matches, i.e pqq, pqpqq, pqpqpqq, ... However if Player 1 wins first and wins the whole thing [case (a)] it is pp, pqpp, pqpqpp, ..., i.e. an even number of matches.

For (B), use the definition of Expected Value?

 

[Steven G]

I am not sure how to continue on a problem I have with the following question:
Two tennis players are playing a couple of matches. Every match player1 has probability p to win and player2 has probability q=1-p to win. They will stay playing till one of them have won two matches in a row. For example: “2-0, 3-1”
A) Find P(X=k) for k=0,1,2,…..
B) What is EX and for what value of p is this the biggest?

This are my solutions:
A) I thought about question A, that player1 wins if he has 2 more matches won than player2. The total of matches are x. To get the total matches won by player1 is (x/2)+1 and lost (x/2)-1. This got me the following formula for the probability:

\(\displaystyle \dfrac{x!}{\left(\dfrac{x}{2}\, +\, 1\right)!\,\left(\dfrac{x}{2}\, -\, 1\right)!}\, \left(p^{\frac{x}{2}+1}\right)\, \left(q^{\frac{x}{2}-1}\right)\)

B) I have no idea how to do B. If I have to say something I would say (1/p)*2, but I know this wrong.

I hope that someone can help me with this questions

1st you really really need to define X. I can only assume that P(X=k) means the prob of winning after k games. Here is how I remember doing/thinking about this problem when I 1st saw it in the dark ages.
Suppose payer 1 wins the first game.
X=2 means 11 (payer 1 won then player 1 won again) So P(X=2)=pp=p^2
X=3 means 122. P(X=3)=p*q^2
X=4 means 1211. P(X=4)=p^3*q
X=5 means 12122. P(X=5)=p^2*q^3
X=6 means 121211. P(X=6)=p^4*q^2
X=7 means 1212122. P(x=7)=p^3*q^4
Find a rule for P(X=2n) and a rule for P(X=2n+1).
How should the answer be modified since player could have won 1st?