Probability of a team wining a best of 3 game

[roflmao33]

Homework Statement

Two teams, teams A and B, are competing in a tournament. Suppose the tournament is a ‘best-of-3’ series; in other words, the first team to win 2 games wins the tournament. Let’s suppose that team A has a slight edge on team B, such that the probability that team A wins any given game is 60%, and the outcome of any previous match does not influence team A’s probability of winning in any future match.
a. What is the probability that team A wins this tournament?
b. Explain why the probability that team A wins the tournament is greater than 60%.
c. If this had been a ‘best-of-7’ series (first team to win 4 games), would the probability that team A wins the tournament be greater or less than what you calculated in a)?
d. Calculate the probability team A wins a ‘best-of-7’ series.

The Attempt at a Solution

There should be 3 ways that team A can win the tournament (3 choose 2=3). Since the probability of team A winning a game is 0.6. The probability of team A winning the tournament will be 3C2*(0.6)^2*(0.4)=0.432?

What am I doing wrong. I can't get a probability of greater than 60% as of question (b).

Thx in advance.

 

[Dick]

Homework Statement

Two teams, teams A and B, are competing in a tournament. Suppose the tournament is a ‘best-of-3’ series; in other words, the first team to win 2 games wins the tournament. Let’s suppose that team A has a slight edge on team B, such that the probability that team A wins any given game is 60%, and the outcome of any previous match does not influence team A’s probability of winning in any future match.
a. What is the probability that team A wins this tournament?
b. Explain why the probability that team A wins the tournament is greater than 60%.
c. If this had been a ‘best-of-7’ series (first team to win 4 games), would the probability that team A wins the tournament be greater or less than what you calculated in a)?
d. Calculate the probability team A wins a ‘best-of-7’ series.

The Attempt at a Solution

There should be 3 ways that team A can win the tournament (3 choose 2=3). Since the probability of team A winning a game is 0.6. The probability of team A winning the tournament will be 3C2*(0.6)^2*(0.4)=0.432?

What am I doing wrong. I can't get a probability of greater than 60% as of question (b).

Thx in advance.

They don't have to win just two games. They could also win all three games.

 

[roflmao33]

They don't have to win just two games. They could also win all three games.

Okay. So they still play the last game even though team A won the first 2? That will give me (0.6)^3 +0.432=0.648.Am i right?

 

[Ryker]

No, if it's a best-of-three, then they can't win three games. I'm getting the same result as you just got, but I don't think you went about it the proper way in your last post. You were on the right track when you said at the beginning that there are three possible pathways to team A winning, though. Can you explain what those three paths are? If you can, then maybe you will see where you have gone wrong in your original calculations.

Hint: Those three pathways don't all have the same probability of occurring.

 

[Dick]

Okay. So they still play the last game even though team A won the first 2? That will give me (0.6)^3 +0.432=0.648.Am i right?

Sure, you calculated the probability they win EXACTLY two games. So add the probability they win 3. It doesn't really matter if they stop the tournament when team A has won 2. If they do and you want to count the actual cases, then A could win the first two, or lose the first or second and win the third. If you add that up, you'll get the same number.

 

[roflmao33]

thx Dick..

 

[Ray Vickson]

In the best two of three scenario, A wins the tournament if the game winners are: AA*, ABA and BAA.

In the best of 7, look at what happens if: (1) A wins the first game; or (2) B wins the first game. In case (1) we are left with up to 6 games in the tournament, and A must win 3 (equivalently, B cannot win 4). In case (2) we are essentially swapping A and B.

RGV

 

[Dick]

No, if it's a best-of-three, then they can't win three games. I'm getting the same result as you just got, but I don't think you went about it the proper way in your last post. You were on the right track when you said at the beginning that there are three possible pathways to team A winning, though. Can you explain what those three paths are? If you can, then maybe you will see where you have gone wrong in your original calculations.

Hint: Those three pathways don't all have the same probability of occurring.

If you think about it you'll realize that it doesn't really matter if they play out the tournament or not. How can it? The cases just get more complicated if you start trying to figure in the possibility of skipping games.

 

[roflmao33]

Thx for the help guys.. right now I am trying to figure out part (b).Why is it bigger than 60%? Any hints?

 

[Dick]

Thx for the help guys.. right now I am trying to figure out part (b).Why is it bigger than 60%? Any hints?

If you have a 60% probability of winning in any individual game, can you explain in words why if you play more games that your probability of winning over 1/2 of the games gets better and better as the number of games increases? If you play N games, how many do you expect to win, on average?