series calculations

raycab · 2012-08-16 00:22:29

Assuming one side of a 2 way proposition is 52.5% (obv making the other side 47.5%), what are the chances of the 52.5% proposition winning a race to 5,10,15,20,30,50 games etc. vs the 47.5% side? Though it is obvious to see logically that the longer the race goes the better the odds will be for the 52.5% side to win the race, I don't actually know how to figure it out mathematically. Any help here would be greatly appreciated. Thanks in advance.

bobbym · 2012-08-16 01:44:05

Hi raycab;

Do you mean getting to 5, 10, 15,... first? Or do you mean 3 to 2 for 5? 6 to 4 or 7 to 3 etc for 10?

If each race is a separate race then these are the percentages you want.

That is the probability the stronger side will be ahead after some number of games. Notice the chart is counter intuitive, do you see why?

raycab · 2012-08-16 03:43:10

Sorry for the confusion. What I mean is a race to 5 or best of 9, race to 10 or best of 19, race to 50 or best of 99 etc. Therefore it has to be some type of glide up from lowest amount of games in a race to highest amount of games in a race being that each individual game has a 52.5% chance of winning. So the higher the race goes the better the chances of the 52.5% side winning the series. Again thanks for your help.

raycab · 2012-08-16 03:45:38

Not sure if this helps but I have an NBA series calculator that calculates best of 7 series or races to 4 and the 52.5% side would win 55.455% of the time in a race to 4.

raycab · 2012-08-16 03:50:34

I see what you did. Your answer for (5) is actually the correct answer for a race to 3 or best of 5. Your answer for (15) I'm sure is correct for a race to 8 and your answer for (10) is taking 5-5's (ties) into account therefore the percentage being lower. Long story short what ever formula you used is 100% correct. I just didn't explain correctly what I was looking for in the original post with the amount of games. Again thanks a ton.

raycab · 2012-08-16 04:04:06

And to answer the question, yes because the even numbered samples account for ties and yes each race is a separate race.

bobbym · 2012-08-16 08:12:06

Hi raycab;

Then it appears we are done with this except for one more thing. Welcome to the forum! Or do you need the table to be more filled in?

raycab · 2012-08-16 09:27:19

bobbym wrote:

Hi raycab;
Then it appears we are done with this except for one more thing. Welcome to the forum! Or do you need the table to be more filled in?

That or the formula to figure it out would be great. Thanks again.

bobbym · 2012-08-16 10:26:57

Hi;

In probability the wording of a problem is extremely important. Sometimes making a small change in how a problem is phrased can turn an easy one into one that is very, very difficult.

We have been using two different phrasings. In one sense we talk about a 9 game match as always playing to 9 games and seeing who has the most games. This is what my table has solved. The other phrasing suggests a best 5 out of 9 games. Notice that in this structure a match can be over in 5 games. No one ever has more than 5, whereas in the first structure the favorite could win by 9 to 0 or 8 to 1 or 7 to 2...

So, what is your problem really like the first structure or the second?

raycab · 2012-08-16 11:14:59

bobbym wrote:

Hi;
In probability the wording of a problem is extremely important. Sometimes making a small change in how a problem is phrased can turn an easy one into one that is very, very difficult.
We have been using two different phrasings. In one sense we talk about a 9 game match as always playing to 9 games and seeing who has the most games. This is what my table has solved. The other phrasing suggests a best 5 out of 9 games. Notice that in this structure a match can be over in 5 games. No one ever has more than 5, whereas in the first structure the favorite could win by 9 to 0 or 8 to 1 or 7 to 2...
So, what is your problem really like the first structure or the second?

The 2nd structure best describes what I'm looking for. When I say a race to 5 games(meaning the winner is the 1st to reach 5 games) to me that is the same as saying a best 5 out of 9 games. Another words, which ever side gets to 5 games 1st whether it takes 5 games or 9 games. In a perfect world the table would give me the percentage outcome of a best 5 out of 9,10 out of 19,20 out of 39,30 out of 59, 50 out of 99, and 100 out of 199. I'm very sorry for the confusion and again I thank you very much for your help.

bobbym · 2012-08-16 16:51:30

Hi;

NBA series calculator that calculates best of 7 series or races to 4

Can you provide the page for that calculator?

You can use the following sum to get the answers you require.

Where r is the number of games like 4 out of 7.

p is the probability of the favorite team winning a single game.

length is the length of the playoff like 7 games.

So to know the probability that team A with a winning percentage of .525 will win a 7 game series we have:

r = 4 ( 4 out of 7 needed )

p =.525

length=7

Math Is Fun Forum

#1 2012-08-16 00:22:29

series calculations

#2 2012-08-16 01:44:05

Re: series calculations

#3 2012-08-16 03:43:10

Re: series calculations

#4 2012-08-16 03:45:38

Re: series calculations

#5 2012-08-16 03:50:34

Re: series calculations

#6 2012-08-16 04:04:06

Re: series calculations

#7 2012-08-16 08:12:06

Re: series calculations

#8 2012-08-16 09:27:19

Re: series calculations

#9 2012-08-16 10:26:57

Re: series calculations

#10 2012-08-16 11:14:59

Re: series calculations

#11 2012-08-16 16:51:30

Re: series calculations

Board footer