Probability of a team wining a best of 3 game

In summary: What happens if N = 100?In summary, the probability of Team A winning the tournament is 60% or higher, as they have a slight edge over Team B with a 60% chance of winning any given game. In a best-of-3 series, there are three possible pathways for Team A to win, and the probability of them winning the tournament is 60% or higher. In a best-of-7 series, there are multiple pathways for Team A to win, and the probability of them winning the tournament is even higher than 60%. This is because as the number of games increases, their chances of winning more games also increases. Therefore, the probability of Team A winning the tournament is greater than 60%.
  • #1
roflmao33
6
0

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.
 
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  • #2
roflmao33 said:

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.
 
  • #3
Dick said:
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?
 
  • #4
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.
 
  • #5
roflmao33 said:
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.
 
  • #6
thx Dick..
 
  • #7
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
 
  • #8
Ryker said:
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.
 
  • #9
Thx for the help guys.. right now I am trying to figure out part (b).Why is it bigger than 60%? Any hints?
 
  • #10
roflmao33 said:
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?
 

Related to Probability of a team wining a best of 3 game

1. What is the probability of a team winning a best of 3 game?

The probability of a team winning a best of 3 game depends on a variety of factors, including the skill level of each team, any potential injuries or absences, and the specific rules and scoring system of the game. It is impossible to accurately determine the exact probability without considering all of these factors.

2. Can the probability of a team winning a best of 3 game be calculated?

While the exact probability cannot be calculated, it is possible to estimate the likelihood of a team winning based on past performance and statistical analysis. However, these estimates are not guaranteed and can change based on the circumstances of the game.

3. Do certain strategies or tactics increase a team's probability of winning a best of 3 game?

There is no definitive answer to this question as different strategies and tactics can work differently for each team and situation. However, some strategies may have a higher success rate than others and can increase a team's probability of winning.

4. How does the probability of a team winning a best of 3 game change if they win the first game?

If a team wins the first game of a best of 3 series, their probability of winning the entire series increases. However, it is important to note that each game is independent of the others and winning the first game does not guarantee a win in the series.

5. Is there a way to improve a team's probability of winning a best of 3 game?

While there is no guaranteed way to improve a team's probability of winning, they can work on improving their skills, team chemistry, and overall strategy. Additionally, analyzing past performances and studying their opponents can also help increase their chances of winning a best of 3 game.

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