Math Question

[Mikie2times]

I am afraid of math, but was wondering this:

 

If each NFL game is a 50/50 Win Loss proposition, what is the statistical probibility that Denver, Baltimore and Jax lose at least one of their final two games?

 

or, in other words...

 

It's like 4 people flipping a coin twice. What's the chances that all four people get a Heads at least once?

 

Can anyone solve that equation? I would love to know.

 

Thanks!

168163[/snapback]

 

I just made a post about this with my own calculations. Here is a simple way to figure out the odds of those teams winning out.

 

Take your odds for each game. For example in my post I have the Jaguars at 6/10 vs Houston. In other words for every 10 games they play against Houston in Jacksonville they will win 6 of them. Then you take the odds of your second game, in Jacksonville's case I have them at 8/10 vs Oakland. Turn those fractions into percentages and multiply them together. So its . 60*.80= 48%.

 

So within my estimates the Jaguars have a 48% chance at winning out.

[eSJayDee]

f each NFL game is a 50/50 Win Loss proposition, what is the statistical probibility that Denver, Baltimore and Jax lose at least one of their final two games?

 

I think you asked 2 questions so, I'll answer this one.

The easiest way to visualize this is to look at the chance it doesn't occur and subtract that from 100%.

If a team fails to lose at least one game, that means they win both. The odds of that (independent) are 25%. So the odds of them losing at least 1 is 75%.

You want to know the odds of all happening together, that's .75*.75*.75 =27/64=~42.8%.

 

I was a math major, BTW (concentrating in calculus, though).

 

Also, give SDS the gold star. He got the answer 1st. I figured it was easiest to figure the answer myself than to go over which previous posts were right & which ones were wrong.

[Guest cal t]

I am afraid of math, but was wondering this:

 

If each NFL game is a 50/50 Win Loss proposition, what is the statistical probibility that Denver, Baltimore and Jax lose at least one of their final two games?

 

or, in other words...

 

It's like 4 people flipping a coin twice. What's the chances that all four people get a Heads at least once?

 

Can anyone solve that equation? I would love to know.

 

Thanks!

168163[/snapback]

Let me correct the Math here.If each game is a 50/ 50 chance.The each individual team has a 25% chance of losing or winning both games. Each team has a 50% chance of winning one/losing one.If this is the case...you multiple .50 x .50 x .50 and you get .125 or 12.5 percent.-BUT--there is also a .25x.25 x.25 chance of allthese teams losing both games(if they dont play eachother) or about 2 percent(less actually) BUTthere is a chance that one team lose both & the others lose 1--.25 x .5 x .5=.625 or approx 6 percent. BUT--2 teams losing both plus one winning 1---.25x.25 x.5=3.125..So add 12.5 plus 1.5625plus6.25plus3.125...soooo

There is a 23.4375 % chance that the Bills will make the playoffs ASSUMING 2 Bills wins--using the 50 percent scenario that you chose.

[Fan in San Diego]

Here is my entry in the stats derby

 

4 events of 50/50

 

2 * 2 * 2 *2

 

=16

 

1 in 16

 

or 6.25 % chance of 4 heads in a row.

[Guest cal t]

I think you asked 2 questions so, I'll answer this one.

The easiest way to visualize this is to look at the chance it doesn't occur and subtract that from 100%.

If a team fails to lose at least one game, that means they win both.  The odds of that (independent) are 25%. So the odds of them losing at least 1 is 75%.

You want to know the odds of all happening together, that's .75*.75*.75 =27/64=~42.8%.

 

I was a math major, BTW (concentrating in calculus, though).

 

Also, give SDS the gold star.  He got the answer 1st.  I figured it was easiest to figure the answer myself than to go over which previous posts were right & which ones were wrong.

168319[/snapback]

Sorry you were both wrong.

sincerely, cal t ...statistics expert. You have to get the odds of all possibilities putting the bills in the playoffs.Those scenarios are all 3 teams losing both. all 3 teams losing once, 1 losing both 2 winning 1,2 losing both 1 winning 1.then you add these......and you ASSUME the Bills win their 2(for true odds--you'd need to add that in--but we are ASSUMING 2 Bills wins)...and you get the 24 + % figure.

Thank you,

You've been a lovely audience.