SUPER BOWL

Jonny_B said:
i find any sport/game can be interesting to watch once you know the rules and begin to understand the strategy. i can even watch nascar and regular season baseball.
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likewise, i even like cricket. but i don't compare sports saying "this one is for people who are harder" because im not a cock.

fed up trying to point out the difference in rugby and american footy. suffice to say any particularly bad injuries i got was from playing american football, not rugby, and i've played both a hell of a lot
 
Jonny_B said:
if you generate a random string of 1's and 0's you'll see lots of instances where you get repetitive 1's or repetitive 0's.

if you calculate the odds of getting 12 consecutive wins out of 43 trials you'll find it's very unlikely, but their odds of winning next year remain 50/50 since a coin toss is an independent test, so previous outcomes do not affect the future outcome.

edit: that's still freaky though. but don't believe anyone who tells you that the AFC is "due"
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But out of a set of 43 - what are the chances of that???
 
fly said:
But out of a set of 43 - what are the chances of that???
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oh christ, man.... um.... i'm not sure that i care enough to calculate it.

hold on, did the NFC always call heads and it was heds 12 times, or did they just win 12 times in a row? i think the odds are the same, just curious.
 
And to think, I almost spent a rediculous amount of Marklar on the under, but didn't after reviewing the final tallys of both teams games throughout the season and realizing that 47 pts wasn't really that much.
 
let's say they have a 50/50 chance of winning every time. 43 attempts.

total number of possible outcomes over 43 trials is 2^43 = 8,796,093,022,208

number of outcomes with 12 wins is
(2^31)*32 = 68,719,476,736

68,719,476,736/8,796,093,022,208 = 0.0078125.

or a 1 in 128 chance of it happening. hmmm, that seems wrong. mikey?
 
Jonny_B said:
let's say they have a 50/50 chance of winning every time. 43 attempts.

total number of possible outcomes over 43 trials is 2^43 = 8,796,093,022,208

number of outcomes with 12 wins is
(2^31)*32 = 68,719,476,736

68,719,476,736/8,796,093,022,208 = 0.0078125.

or a 1 in 128 chance of it happening. hmmm, that seems wrong. mikey?
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I hate statistics but I don't think it is 2^43. It's more like 43 factoral / 43 factoral - 2 factoral :lol: I missing a multiplier in that denominator and I can't think of it offhand.
 
thrawn said:
I hate statistics but I don't think it is 2^43. It's more like 43 factoral / 43 factoral - 2 factoral :lol: I missing a multiplier in that denominator and I can't think of it offhand.
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the outcome is either 1 (win) or 0 (loss) in each of 43 places. 2 outcomes for each place, 43 places, 2^43. just like the number of difference license plates. (3 letters 3 numbers = 26^3 * 10^3)

so there are 2^43 different strings of outcomes.

put 1 in the first 12 places and there are 31 remaining so 2^31 different outcomes that start with 12 1's.

put 1 in the 2nd to 13th places and it is 2 * 2^30 or 2^31 outcomes with 1's there.

continue moving the block of 12 ones over and you find there are 32 difference places for the block with the total number for each location being 2^31, so 2^31*32.

you have to do the factorial stuff when dealing with "bag of 10 red balls and 4 blue balls, what are the odds of drawing exactly 2 red balls" questions. with the coin tossing each trial is independant.

(though i just realised you have to assume the places next to the block are 0. i'll have to recalculate)
 
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2^30 + 2^30 + 2^29*30 = 1073741824 + 1073741824 + 16106127360 = 18,253,611,008

18,253,611,008 / 8,796,093,022,208 = 0.0020751953125

or 1 in 482. That's better.

keep in mind that means a total of 43 * 482 coin tosses or 20,726 coin tosses to make getting 12 in a row likely (without allowing overlap between groups of 43)
 
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blah, i grabbed a book. i didn't read your post correctly. the number of ways you can get heads on 43 flips is 43!/(2!(43−2)!)
 
Jonny_B said:
2^30 + 2^30 + 2^29*30 = 1073741824 + 1073741824 + 16106127360 = 18,253,611,008

18,253,611,008 / 8,796,093,022,208 = 0.0020751953125

or 1 in 482. That's better.

keep in mind that means a total of 43 * 482 coin tosses or 20,726 coin tosses to make getting 12 in a row likely (without allowing overlap between groups of 43)
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1231311709384987.jpg
 
thrawn said:
blah, i grabbed a book. i didn't read your post correctly. the number of ways you can get heads on 43 flips is 43!/(2!(43−2)!)
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right, but we're not trying to figure that out. we're trying to figure out the number of ways to get 12 heads in a row. for that you just calculate the number of outcomes with 12 heads and divide it by the number of total outcomes. plus, order matters so you can't use combinations.
 
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Jonny_B said:
right, but we're not trying to figure that out. we're trying to figure out the number of ways to get 12 heads in a row. for that you just calculate the number of outcomes with 12 heads and divide it by the number of total outcomes. plus, order matters so you can't use combinations.
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There are only 32 ways to get 12 heads in a row.
Flips 1-12, 2-13, 3-14, etc... up to 32-43
 
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tre said:
There are only 32 ways to get 12 heads in a row.
Flips 1-12, 2-13, 3-14, etc... up to 32-43
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ways to get 3 in a row out of six tries:

111000
111001
111010
111011

011100
011101

001110
101110

000111
100111
010111
110111

that's 12, not 1-3, 2-4, 3-6 (3)

or 2^2 + 2^2 + 2^1*2
 
Jonny_B said:
right, but we're not trying to figure that out. we're trying to figure out the number of ways to get 12 heads in a row. for that you just calculate the number of outcomes with 12 heads and divide it by the number of total outcomes. plus, order matters so you can't use combinations.
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I know but you have to figure out the number of possible heads out of a given number of flips to figure out the probability of getting heads each time.

And for arguments sake I am just going with 12 heads in a row. Guessing the correct coin toss 12 tiimes in a row is a bit more complicated.
 
thrawn said:
I know but you have to figure out the number of possible heads out of a given number of flips to figure out the probability of getting heads each time.

And for arguments sake I am just going with 12 heads in a row. Guessing the correct coin toss 12 tiimes in a row is a bit more complicated.
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it's actually the same. you have a 50/50 chance of guessing correctly, which is the same as the odds of getting heads.
 
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