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Thread: MATH/STATS NERDS!!! Puzzle

01252011, 07:04 PM #1
MATH/STATS NERDS!!! Puzzle
Let me preface this by saying this is not a homework problem. This is a YEAR long debate amongst my friends, my fiance, and myself. It started last year when my fiance, then GF, bought 4 squares on a 10x10 super bowl "board" at work. Along the top numbers 09 were written and team "X". Along the side numbers 09 were written and team "y". At the end of each quarter the teams score would be used to pick a winner on the board. Only the last number of each score would be used. So a score of 77, 177, 2727, or 1717 would all yield the same winner. Also, you can win more than 1 quarter if there is no score change, or a teams score increases by some multiple of 10. Now, assuming that any score is likely, and one isn't more favorable than the other, what are the odds that she would win at least 1 time, by the end of the game?
My rough estimate, of winning one event by the end of the night was 16 in 100. I know to do it properly you actually calculate the odds of NOT winning, then extrapolate that over 4 events. I think this estimate is relatively close though. Essentially, she, and everyone else at the super bowl party, believed that because the winners name is put back in for each quarterly drawing, that your odds stayed the same for the OVERALL likely hood of winning. She stated that they could play 100 quarters of football, have a drawing after each quarter, and the OVERALL likely hood of winning is still 4 in 100. My argument is that more drawings, increases your likely hood, even if the winner of quarter 1 is eligible to win quarter 2.
So who's right? Anyone got any formulas I can use to back up my claim, or am I dead wrong?Last edited by Silver03Termi; 01252011 at 07:15 PM.
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01252011, 07:11 PM #2
E=mc2, that's all you need to know.
"Initial Success or Total Failure"

01252011, 07:17 PM #3
oh, squares were 50$ each, and she won the final quarter. Pay out was:
1st quarter $1000
2nd quarter $1000
3rd quarter $1000
4th quarter $2000

01252011, 07:37 PM #4
Ok so you're saying that you can potentially win all four quarters?
You have 4 combinations out of 100.
There are 4 quarters.
16 combinations out of 400.
Still 4/100 chance.

01252011, 07:47 PM #5
Here is the probability formula, don't have a calculator in front of me so whatever this equals to is your answer...
1((1.01)x(1.01)x(1.01)x(1.01))= ???2004 Mystichrome Cobra, 1 of 515

01252011, 07:52 PM #6

01252011, 07:57 PM #7
No, you take the probability of her winning each quarter, I forgot to mention you multiply the answer by 100 to get the %.
The answer comes to 3.94% so yeah basically 4 out of 100 but with football pools they figure it differently because certain numbers have a greater probability of hitting like 0,3,7, etc. There's a goofy chart if you google "probability of winning a football pool".
Besides that, nice win she had, maybe she'll throw a little sumpin your way for the Cobra, lol.Last edited by SCcobra4me; 01252011 at 08:02 PM.

01252011, 08:00 PM #8

01252011, 08:03 PM #9

01252011, 08:19 PM #10
At the absolute basest form, 4/100 each quarter. For all 4 quarters, multiply both. 16/400. Reduce. There's the odds for winning at least one quarter.
Now to win all 4 quarters you'd need to multiply the percentage just like SCCobra4Me did.
Now this is also not taking into account the odds for each number, as mentioned above, like 0, 3, 7, etc. Those throw an additional layer into the mix that I don't want to think about.
*Edit  I just though of this after rereading your post. If there are only limited number of people playing your odds go significantly up if the rules state that there is always a winner (i.e. If no one picked the winning number it would default to the closest bet). However, if no one wins and there are no rule provisions then the odds are still ~4/100.Last edited by ShadowFist; 01252011 at 08:23 PM.

01252011, 08:21 PM #11
she has a 4 in 100, of winning any of the quarters. You're saying that her odds, are worse over the duration of all 4 quarters? Are you sure you didnt calculate the odds of her winning all 4 drawings? Lets use your formula to calculate the odds of her losing all 4 quarters. Her odds of winning will be whats left over...right?
Last edited by Silver03Termi; 01252011 at 08:24 PM.

01252011, 08:26 PM #12
According to his math (I'm too lazy to check it) there is a 3.94% chance to win each quarter. Since the quarters do not change the variables, each one can be considered independent and the same chance of winning applies for each. Now if you wanted to say (as an example) 'what are the odds of her winning just the FIRST quarter, and none of the rest', then the odds would go down significantly.

01252011, 08:28 PM #13
Last edited by SCcobra4me; 01252011 at 08:31 PM.

01252011, 08:29 PM #14

01252011, 08:31 PM #15

01252011, 08:35 PM #16
Last edited by cobracide; 01252011 at 08:43 PM.
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01252011, 08:36 PM #17
So you knew the answer all along, just wanted us to think is that it??

01252011, 08:53 PM #18

01252011, 08:58 PM #19

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01252011, 09:09 PM #21

01262011, 08:05 AM #22
Edit: Very few seem to be taking into account that certain numbers appear more often in football scores while others appear less. Therefore, if all boxes had an equal chance of winning a straight forward analysis would work. But, you must take into account the fact that different numbers have a different % of winning.
Originally Posted by Albert EinsteinOriginally Posted by FordSVTFan

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01262011, 08:15 AM #24

01262011, 09:44 AM #25
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