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bluepenguin
03-20-2008, 01:54 PM
ok math majors, this is freaky.

what are the odds that the losing team in all three of the first games played had the same half time score - and that two of the games would have identical scores?

Kansas 49 Portland St. 26
Georgia 35 Xavier 26
MSU 35 Temple 26

And all three losing teams would also end up with the same total number of points.

Kansas 85 Portland St. 61
Xavier 73 Georgia 61
MSU 72 Temple 61

ok, i probably have too much time on my hands.

sagegrouse
03-20-2008, 02:06 PM
The probability that two teams would identical halftime scores is on the order of 0.10 (ten percent). The prob. that two teams would have identical endgame scores is lower, say, 0.05 (five percent), due to greater dispersion of scores.

Then the compound prob. is 0.1*0.1*0.05*0.05 = 0.000025, or 25 chances out of one million.

OTOH, there are other potential patterns that could have occurred (equal winning scores, equal point margins), so the odds of something this strange happening are a bit less.

sagegrouse
'Or, at least, that is what I remember from Math 135-136 many decades ago'

Jaymf7
03-20-2008, 03:22 PM
It depends a lot on the score, doesn't it? There are only a limited number of scores, and the incidence of some is much greater than others.

If the halftime and final scores we are talking about are 55 and 98, respectively, that is much less likely than 35 and 70, right? (or 19 and 88).

26 and 61 seem a bit low, but this is NCAA play and such games might historically feature slightly lower scores.

Pretty remarkable in any event.

Jaymf7
03-20-2008, 03:29 PM
After posting that, I figured out that you probably mean the aggregate of all possible scores. Interesting question. Beyond my simple math skills, but if anyone can explain a solution that would be neat.

jacone21
03-20-2008, 03:52 PM
Assume (big assumption) that a given team will score between 20 and 130 inclusive in a game

With three teams, the number of combinations of scores at the end of the game is

111 choose 3 or 234136


The number of combinations where the numbers are the same (ex/ 61,61,61) is 111, obviously.


So, the odds of it happening in the range of 20 to 130 is 111/234136 or about .0004741

Of course that assumes that the probability of a team scoring 20 is the same as the probability of a team scoring 80. Another big assumption. That's a variable I'm not ready to take on.

shadowfax336
03-20-2008, 03:55 PM
of course this is all a bit innacurate because Xavier was one of the losing scores at halftime and went on to win...

phaedrus
03-20-2008, 05:55 PM
sagegrouse
'Or, at least, that is what I remember from Math 135-136 many decades ago'

I took both those classes (Probability and Statistics, respectively) in the last 4 years. You must be a Duke grad or something.

sagegrouse
03-20-2008, 10:40 PM
I took both those classes (Probability and Statistics, respectively) in the last 4 years. You must be a Duke grad or something.

Yup. However, I took the courses in 1962-1963.

sagegrouse