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        riddles >> medium >> Super Bowl
        
(Message started by: SWF on Jan 20^th, 2003, 6:40am)

Title: Super Bowl
Post by SWF on Jan 20^th, 2003, 6:40am

A statistician has a mathematical model for projecting the scoring in the Super Bowl: a football game between the Buccaneers and the Raiders.

For each inifinitesmal time increment, each team is assumed to have a certain probability of scoring. For a given team this probability remains constant throughout the game, but the constants are different for the two teams. The probabilities are such that the Buccaneers will score on average 6.2 times per game, and the Raiders will score on average 4.4 times per game. However, only 20% of the the Buccaneers' scores are touchdowns (the other 80% are field goals), while 59% of the Raiders' scores are touchdowns (41% are field goals). It is assumed that there is no chance of any safeties, missed extra points, or two point conversions during the game, i.e. touchdowns are worth 7 points each and field goals are worth 3 points each. Based on this model, which team is most likely to win.

Title: Re: Super Bowl
Post by redPEPPER on Jan 20^th, 2003, 7:48am

How do you play American football in the first place? :P

From what I understand, I calculate an average score of 23.56 for the Buccaneers vs. 23.584 for the Raiders. Is that correct?

Title: Re: Super Bowl
Post by towr on Jan 20^th, 2003, 10:22am

the average scores are correct.. But that's not an answer to the question, since the one with the highest avarage could still loose most of the time..

Image for instance a game where there's a way to score 100 points and a way to score 1 point. one team has 5% chance of scoring 100, and 0% of scoring 1. The other has 0% chance of scoring 100 points, but 50% of scoring 1 point. avarages: team 1 gets 5, team 2 gets 0.5 points, yet team 1 only wins 5% of the time, and team 2 wins slightly less than half of the time.. (rest are draws)

anyway..
hmm.. I don't really know what to do, don't I need the variation/standard deviation for the number of times they score, and for the touchdowns and field goals.. Or, not assuming a normal distribution, their distribution?
One of the teams might once in every hundred year have a very good game where they score a two thousand touchdowns and eight thousand field goals, and loose pretty much every other game..
The model seems incomplete to me..

Title: Re: Super Bowl
Post by James Fingas on Jan 20^th, 2003, 11:05am

towr,

I agree with you. What I believe we need to do is to take the Buccaneers' possible scores and their probabilities of happening, take the negative of the Raiders' possible scores and their probabilities of happening, and then convolve the two. Although it is true that the mean of such a convolution will be slightly negative, likely we'll find that the median is positive (I'm just saying that because there's probably a trick in the question).

This is a brute-force approach, but I'm not sure there's anything else to do ... maybe we can simplify the convolution somehow.

To find their possible scores, I would suggest considering the touchdowns separately from the field goals. I don't remember the name of the discrete probability distribution you'll have to use (or its formula!), but then you can convolve the touchdowns with the field goals to get the overall distribution of points for each team.

NOTE:
For those people who haven't done convolutions, the idea is pretty simple. Let's say we want to convolve the discrete functions y₁ and y₂, to get y₃ = y₁*y₂ (the asterisk indicates convolution here):

x = 0 1 2 3 4 5 6 7 . . .
y₁ = 0 1 3 0 0 0 0 0 . . .
y₂ = 0 1 0 0 2 0 0 0 . . .

y₃ = 0 1 3 0 2 6 0 0 . . .

You just replace every '1' in y₂ with a copy of y₁, and replace every '2' with twice y₁, etc.. Of course, when we deal with probabilities, we'll be working with numbers less than 1, but the same idea applies.

The 'x' row is just my lame-ass way of trying to make a number line.

Title: Re: Super Bowl
Post by Chronos on Jan 20^th, 2003, 2:02pm

Quote:

I don't remember the name of the discrete probability distribution you'll have to use (or its formula!)

Would that be the Poisson distribution?

Title: Re: Super Bowl
Post by SWF on Jan 20^th, 2003, 7:49pm

I was trying to explain briefly and in a way that did not require knowledge of the rules of American football, but it may not have been completely clear.

The winner of the Super Bowl is the team with the most points after 60 minutes of play. If tied at that point, play continues with the first team to score declared the winner.

With this model, the probability of a given team scoring remains constant throughout the game. i.e. for any period of time with duration T, the probability of the Buccaneers scoring during that time interval is the same as for any other time segment of the game with duration T. When the Buccaneers do score, there is a 20% chance of it being worth 7 points and an 80% chance of it being worth 3 points.

Title: Re: Super Bowl
Post by Cyrus on Jan 23^rd, 2003, 1:54pm

I'll just throw out an answer saying that the Raiders are more likely to win even though the 2 AVERAGE scores are 23.584 for the Raiders and 23.56 for the Bucs, the Raiders also have a MEDIAN score of 23.2 while the Bucs only have a median score of 21.8.

Not sure if that's a smart way to look at it, but I figured in this case the median score was more important than the average/mean.

Title: Re: Super Bowl
Post by towr on Jan 23^rd, 2003, 2:15pm

What's the difference between mean and average (arithmetic mean)?

Title: Re: Super Bowl
Post by william wu on Jan 23^rd, 2003, 4:02pm

they're supposed to be the same

Title: Re: Super Bowl
Post by Chronos on Jan 23^rd, 2003, 7:25pm

What we need to look at is the median of the difference of the scores (is this the same thing as the difference of the medians?). The median is the quantity such that the result is equally likly to be above or below that number.

Title: Re: Super Bowl
Post by towr on Jan 23^rd, 2003, 11:40pm

I suggest we simulate a million plays between them, and use that to judge who will win more often..

Title: Re: Super Bowl
Post by Cyrus on Jan 24^th, 2003, 8:16am

God, I'm an idiot. I say mean but I MEANT median.

What I did was I assumed the Bucs will score 6 times in 80% of their games and 7 times in 20% of their games, giving an average of 6.2 times per game.

Then I calculated that if the Bucs score 6 times what is the probability it will be (6 Field Goals=18 Pts) or (5 FG + 1 TD=22 Pts) or (4 FG + 2 TD) . . . etc. I then multiplied these probabilities by 80% because the Bucs will score 6 times in 80% of their games.

And then did the same for if they scored 7 times. Multiplying the probabilities by 20%.

It gives a table like this (rounded to 1 decimal):
Tampa Bay
18 - 21.0 %
21 - 04.2 %
22 - 31.5 %
25 - 7.34 %
26 - 19.7 %
29 - 05.5 %
30 - 06.6 %
33 - 02.3 %
34 - 01.2 %
37 - 00.6 %
38 - 00.1 %
41 - 00.1 %
42 - 00.0 %
45 - 00.0 %
49 - 00.0 %

Using this table I just did a really quick estimate of the median score for Tampa Bay and came up with 21.8 points per game.

I did exactly the same thing for the Raiders and came up with 23.2 Points per game

There's probably a better way to come up with the median but I don't know it. You must remember I'm a rather major moron with only grade 12 math behind me, and that was 5 years ago.

Title: Re: Super Bowl
Post by Cyrus on Jan 24^th, 2003, 8:20am

Of course I know what someone is going to say:

"There's also a possibility that the Bucs will score only 5 times, or 4 or 3 or not at all...."

That's true, but I figured the likelyhood of that is the same as it is of them scoring 8, 9, 10, 11 or even 12 times in a game, so I just went with what I knew.

Again, there's probably some complex formula that I don't know of that can solve this question in 30 seconds.

Title: Re: Super Bowl
Post by James Fingas on Jan 24^th, 2003, 9:54am

Cyrus,

The Poisson, as Chronos pointed out, calculates exactly this scenario. It gives the probabilities of them getting 0,1,2,3,... goals, based on a single constant, v, which just happens to be equal to the average number of goals.

P(k) = v^ke^-v/k!

mean = v

I think it probably is important to use the actual distribution in this case, rather than making assumptions (but that's just a gut feeling).

As I said before, I think you should calculate field goals and touchdowns separately.

Title: Re: Super Bowl
Post by Kozo Morimoto on Jan 25^th, 2003, 2:45am

Can this problem be solved using a binomial option pricing setup? So you would end up with 6 normal branches with uneven up/down probabilities and the last branch would be the same but the scores would be x0.2 to represent a partial branch?