Packers versus Bears Math problem

[pbmax]

http://sports.yahoo.com/blogs/nfl-shutdown-corner/chicago-bears-cornerback-charles-tillman-does-not-pro-163415977--nfl.html;_ylt=AjnFUMvEuDXsdxcZF3iwm1lDubYF

Via PFT and Deadspin

http://l2.yimg.com/bt/api/res/1.2/ulL4_vvDNO4BPlwkbXB9eA--/YXBwaWQ9eW5ld3M7cT04NTt3PTYzMA--/http://media.zenfs.com/en/blogs/sptusnflexperts/yahoo_tillmannote.jpg

[smuggler]

"My kid will take the zero if you feel like getting cute about my team again! We only lost those last 4 games because..."

[Smidgeon]

So Mr. Tillman thinks that his celebrity status gives him the ability to dictate what the teacher teaches? Did I get that right?

[hoosier]

Poor Charles Tillman, he gets elected to one pro bowl and all of a sudden he thinks he's an "all pro corner." And that's a tricky question cuz it cannot be answered as stated. The answer is either 20% (if "winning each game" means winning all four of them) or just under 60% (if "each" refers to any specific game).

[Tony Oday]

THE BEARS STILL SUCK!

[Bossman641]

Poor Charles Tillman, he gets elected to one pro bowl and all of a sudden he thinks he's an "all pro corner." And that's a tricky question cuz it cannot be answered as stated. The answer is either 20% (if "winning each game" means winning all four of them) or just under 60% (if "each" refers to any specific game).

The probability the Bears win all 4 games is 1/625 (1/5 * 1/5 * 1/5 * 1/5)

The probability the Bears win at least one game 4/5 (1/5 + 1/5 + 1/5 + 1/5)

[hoosier]

The probability the Bears win at least one game 4/5 (1/5 + 1/5 + 1/5 + 1/5)

If that were true then the chances of them winning one of six games would be greater than 100%! :-)

Real probability is 100% - 0.84%

[MadtownPacker]

You gotta love the "bear down" trolling P.S. at the end! :lol:

[Fritz]

The math teacher missed a comma in the problem, too.

But really, ya gotta like Tillman's response.

[MJZiggy]

The math teacher missed a comma in the problem, too.


You're right. I must be slipping. Either that or I've had a glass of wine and just don't care. You choose which it is.

[Gunakor]

You gotta love the "bear down" trolling P.S. at the end! :lol:


I love the Geaux Bears even more. Schizophrenia?

[Fosco33]

I love the Geaux Bears even more. Schizophrenia?

Geaux = Go

He must be from Louisiana...

[Iron Mike]

Packers are 29-16 vs. the Bears since the 1990s. My math-addled brain interprets that as a 75% probability the Pack will beat the Bears in any game.

[Gunakor]

Geaux = Go

He must be from Louisiana...

Geaux = Saints

[Upnorth]

Packers are 29-16 vs. the Bears since the 1990s. My math-addled brain interprets that as a 75% probability the Pack will beat the Bears in any game.

Maybe this is why Tillman is pissed. If your kids teacher screws up they should be held accountable :?:

[sharpe1027]

If that were true then the chances of them winning one of six games would be greater than 100%! :-)

Real probability is 100% - 0.84%

That would be 99.5904%. Way to high. I think it is actually: 100 - 100*(.84) = 59.04% ;)

[pbmax]

Its (4/5)^4 or 40.96% chance they win one game of four.

Its not 1- or 100- since there is no replacement here. Its the same odds, same conditions each time. There is no equivalent to taking away a card from the deck.

[sharpe1027]

Its (4/5)^4 or 40.96% chance they win one game of four.

Its not 1- or 100- since there is no replacement here. Its the same odds, same conditions each time. There is no equivalent to taking away a card from the deck.

Isn't 40.96% the chance that the Packers win all 4? 4/5 for the first, 4/5 for the second, 4/5 for the third, 4/5 for the fourth -> odds that the Packers win all 4 is 4/5*4/5*4/5*4/5...the chance that the Bears win at least one is the inverse of that.

If you were doing a card from the deck it would be 4/5 * 3/4 * 2/3 *1/2, right?

[Freak Out]

LOL.

[hoosier]

Same odds each time, yes, but you have to look at the problem (what are the odds of Chicago winning at least one) as the reverse probability of GB winning all four. The probabilty of that happening is 80% for the first game, and for the second game it's 80% of the 80% in which they won the first game, and for the third it's 80% of the 64% of the time they won the first two, and so on, which finally makes 40.96%. If that is the probability of GB winning all four, the probabiliy of Chi taking at least one is 100-40.96.

[smuggler]

http://www.youtube.com/watch?v=dwjQaJ5xt1o

[Fritz]

Wow, Smuggler. Usually I watch youtube to see 16 year old boys lighting their crotches on fire, or to see slinky women in high heels and short tight dresses.

I'm feeling all edumucated and righteous after watching that youtube video you posted. Thanks!

[Iron Mike]

Wow, Smuggler. Usually I watch youtube to see 16 year old boys lighting their crotches on fire, or to see slinky women in high heels and short tight dresses.

I'm feeling all edumucated and righteous after watching that youtube video you posted. Thanks!


http://www.youtube.com/watch?v=zU0R5Y7t80Y

[Fritz]

Oh hell yeah!

But wait...what's the probability that Drew Barrymore will do that with me?

Hmm....P(A) + P(B) = 1. Thus, let's see...plug in .0000413....times that by a negative integer.....subtract from the right side.....mmm.....wait, almost done........then determine the co-efficient of her boobs......hmmmm......let "B" stand for flight times from Detroit to LA......hmmmm......okay......

Aha. The probability is 0%.

[MJZiggy]

You missed the whole thread I put in the Romper Room about her getting married...The probability is 0%.

[LP]

You missed the whole thread I put in the Romper Room about her getting married...The probability is 0%.

Yeah but...she's a model and an actress. The divorce papers are probably drawn up, just waiting for dates and signatures.

[sharpe1027]

Oh hell yeah!

But wait...what's the probability that Drew Barrymore will do that with me?

Hmm....P(A) + P(B) = 1. Thus, let's see...plug in .0000413....times that by a negative integer.....subtract from the right side.....mmm.....wait, almost done........then determine the co-efficient of her boobs......hmmmm......let "B" stand for flight times from Detroit to LA......hmmmm......okay......

Aha. The probability is 0%.

You just need to get cast in a leading role with her. How good is your acting?

[ThunderDan]

You just need to get cast in a leading role with her. How good is your acting?

Fritz could star in Porkies but that's about all!

[SkinBasket]

God damn we got some math retards around here.

[sharpe1027]

God damn we got some math retards around here.

Why? What do you think the probability is that Drew Barrymore will do that with Fritz?

[MJZiggy]

Yeah but...she's a model and an actress. The divorce papers are probably drawn up, just waiting for dates and signatures.
And she's like 7 months pregnant.

[LP]

And she's like 7 months pregnant.

So they have to add custody documents and visitation schedules. What else is new in the world of "it's all about me"?

[woodbuck27]

God damn we got some math retards around here.

No....they can't be serious. If you read the entire thread the correct response is in here.

a) Chance the BEARS win just one of the four games is the same as it is to win any game 1-4.

b) Chance the BEARS win 'all four' games ? Take the correct response to a) and multiply that by itself three more times.

[smuggler]

:whaa:

Haha, woody.