• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Anyone know how to do this?

A

alexjohnson16

Platinum Member
Q: The World Series of baseball is an example of a best-of-seven series. The team that wins the majority of the seven games (4, 5, 6 or 7 games assuming all seven games are played no matter what) that are played is the winner of the series. Suppose two teams, the Sanborn Rockets and the Sheldon Mavericks are competing in a best-of-seven series in which all seven games are played no matter what. Research throughout the season has shown that the Rockets have a 58% chance of winning any game they play. Given this information, how likely is it the Rockets will win the best-of-seven series against the Mavericks (that is, how likely is it the Rockets will win 4, 5, 6 or 7 games)? Show your work to support your answer.

Do I do a probability distribution and add 4, 5, 6, and 7? It seems way too low to be the correct answer (like a 22% chance, I don't buy it).
 
What would be the assumptions?

I know the first would be that all seven games are played.

Would you have to assume that the opposing team doesn't have a higher rate of success?
 
Grrr... I can't think of the word that goes in front of probability... bi-something or other.. (shame on me!)
i.e. if the probability of winning a game is 2/3, then the probability of winning 5 out of 7 games would be:
7combination5 * (2/3)^5 *(1/3)^2 = .307

Of course, you're going to want the sum of winning 4 out of 4,
4 out of 5
4 out of 6
and 4 out of 7

But, I'm thinking (for a moment) that because of the 7combination4, that isn't quite right... because the only way to win 4 out of 7 games in the world series is if you win 3 ouf of the first 6, plus the 7th.

Thus, the probability of winning in the 6th game of the world series would be calculate from
the probability of winning 3 out of the first 5, then the 6th...

Or, 5combination3 * .58^3 * .42^2 *.58 = .19962

Is this enough to help?




 
uhhh, missed this part in the original problem:
(4, 5, 6 or 7 games assuming all seven games are played no matter what)
Click to expand...

So, it's going to be 7combination4 * .58^4 *.42^3
plus
7combination5 *.58^5 * .42^2
plus
7combination6 *.58^6 *.42^1
plus
7combination7 *.58^7 *.42^0

 
Binomial.

nvm, drpizza's correct. Didn't see the all 7 games part.



below is the probability of the realistic winning 4 games of a best of seven.

5 choose 4 (combination of 4 wins from 5 games)*.58^4*(1-.58)
+
6 choose 4 *.58^4*(1-.58)^2
+
7 choose 4 *.58^4*(1-.58)^3
+
4 choose 4 * .58^4

forgot about a sweep :X
 
to explain one of the lines in my previous post
7combination5 *.58^5 *.42^2

The probability of winning exactly 5 games out of 7...
They could WWWWWLL
or they could WWLWWLW
or they could WWLLWWW
etc.
There are 7combination5 ways to win 5 out of 7 games.
The probability of winning any particular 5 out of 7 games is .58*.58*.58*.58*.58*.42*.42
Thus, all these possibilities add up to
7combination5 * .58^5 * .42^2
(.42 is the probability of losing (or a tie, I suppose))
 
Uncle Wai, you're overlooking one thing with the 7 choose 4...
The only realistic way they're going to win 4 out of 7 is if the 7th game is a win...
Thus, that one would be 6 choose 3, and then the probability of winning the 7th.
(perhaps I'm wrong, but I don't think so)
 
win 0 Game
1 way to win zero games
1 * .42^7 = 0.00230539333248
~0.23%

win 1 Game
7!/(6!*1!) -> 7 ways to win one game
7 * .58 * .42^6 = 0.02228546888064
~2.22%

win 2 Games
7!/(5!*2!) -> 21 ways to win two games
21 * .58^2 * .42^5 = 0.09232551393408
~9.23%

win 3 Games
7!/(4!*3!) -> 35 ways to win three games
.58^3 * .42^4
= 0.2124952304832
~21.25%

win 4 Games
7!/(3!*4!) -> 35 ways to win four games
35 * .58^4 * .42^3 = 0.2934457944768
~29.34%

win 5 Games
7!/(2!*5!) -> 21 ways to win five games
21 * .58^5 * .42^2 = 0.24314080113792
~24.31%

win 6 Games
7!/(1!*6!) -> 7 ways to win six games
7 * .58^6 * .42 = 0.11192195607936
~11.19%

win 7 Games
7!/(1!*6!) -> 1 way to win seven games
1 * .58^7 = 0.02207984167552
~02.21%



I get 0.6705883933696 or ~67.06%

Notice that when you add up all the percentages for winning 0,1,2,3,4,5,6,7 games you get 100%. That's one way to double check that the calculations at least make sense.
 
Originally posted by: DrPizza
Uncle Wai, you're overlooking one thing with the 7 choose 4...
The only realistic way they're going to win 4 out of 7 is if the 7th game is a win...
Thus, that one would be 6 choose 3, and then the probability of winning the 7th.
(perhaps I'm wrong, but I don't think so)
Click to expand...

Yes, you are right. Good call.
 
Originally posted by: DrPizza
Uncle Wai, you're overlooking one thing with the 7 choose 4...
The only realistic way they're going to win 4 out of 7 is if the 7th game is a win...
Thus, that one would be 6 choose 3, and then the probability of winning the 7th.
(perhaps I'm wrong, but I don't think so)
Click to expand...
No, you're right. He's doing a lot of double counting with that calculation. It gives an answer of 94.4% when the actual answer is closer to 67.1%. 4 choose 4 * .58^4 counts all cases where they win the first 4 games, but 7 choose 4 *.58^4*(1-.58)^3 also includes that case so it has been double counted.
 
Thanks, Uncle Wai, Kyteland... I was pretty sure, but it's a relief to know I was thinking correctly 🙂
 
Originally posted by: DrPizza
Thanks, Uncle Wai, Kyteland... I was pretty sure, but it's a relief to know I was thinking correctly 🙂
Click to expand...
It's easy to tell when you're correct when working with probability. Everything adds up to 1. 😎
 
Originally posted by: SampSon
I don't like this example, once you win 4 games the series is over!
Click to expand...
It doesn't change the answer if you stop after 4 wins. They phrased that way to try and make the steps to solve the problem more obvious.
 
Originally posted by: alexjohnson16
Doh. Wish I would've read this before I turned it in.

I went with the second post to the thread.
Click to expand...
When doing probability it's easy to see when your answer is wrong. The probability of something happening is always 100% so your numbers must sum to 1.

If (.58^4 + .58^5 + .58^6 + .58^7) is the probability of of them winning the series then (.58^0 + .58^1 + .58^2 + .58^3 + .58^4 + .58^5 + .58^6 + .58^7) must be the probability of either team winning the series. But that adds up to 235% which raises a big red flag.
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top