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Post by AztecBill on Jan 24, 2020 13:08:52 GMT -8
There are 8 teams that have more Q1 wins than losses. The Aztecs have a record in those game, if every game was a 50/50 shot, that is least likely to happen. The Aztecs 4-0 record would only happen once in 16 tries or 6.25% of the time. | Team | Q1 Wins-Losses | Odds | | San Diego State | 4-0 | 6.25% | | Kansas | 8-3 | 8.06% | | Baylor | 5-1 | 9.37% | | Gonzaga | 4-1 | 15.63% | | Seton Hall | 6-4 | 20.49% | | Oregon | 5-4 | 24.63% | | Dayton | 3-2 | 31.25% | | Florida | 3-2 | 31.25% |
Of course not all Q1 games are equal but in simple terms this is why the Aztecs have the #1 SOR. |
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Post by azdick on Jan 24, 2020 13:46:59 GMT -8
Thanks, Bill. Very interesting. Wuld love to add Kansas to our list!  |
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Post by jcljorgenson on Jan 24, 2020 15:11:17 GMT -8
I understand the 6.25% chance to go 4-0, but how are the losses playing into it? If another team had a 50% chance against all teams, played 8 games and won 50% (went 4-4) - what is their % and what does that mean?
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Post by aztecalbe on Jan 24, 2020 16:46:01 GMT -8
I understand the 6.25% chance to go 4-0, but how are the losses playing into it? If another team had a 50% chance against all teams, played 8 games and won 50% (went 4-4) - what is their % and what does that mean? Not to speak for Bill and perhaps alienating everyone by talking "math" on a sports forum, the calculations were made using something called the binomial distribution formula which uses factorials in its calculation. The equation looks something like this: P(x) = p^x * q^(n-x) * n! / ((n-x)! * x!) where n = the number of trials, x = number of successes, p = probability of success, and q = the probability of failure (i.e. 1- p) If the probability of success and failure are the same, 50%, in the case of Kansas, they had 11 trials (Q1 games) with 8 successes (wins), the probability of that happening is P(x) = 0.5^8 * 0.5^(11-8) * 11! / ((11-8)! * 8!) P(x) = 0.00390625 * 0.125 * 39916800 / (6 * 40320) P(x) = 19490.625 / 241920 = 0.08056640625 P(x) = 8.06% So there was an 8.06% probability that Kansas would win "exactly" 8 of the 11 Q1 games they played given a 50% chance of winning each individual game. In your example, the probably of winning "exactly" 4 games in 8 tries would be 27.43%. Winning "exactly" 5 games, 21.88% (which is the same probability for winning "exactly" 3 games); winning "exactly" 6 games, 10.93% (same for winning "exactly" 2 games); winning "exactly" 7 games, 3.13% (same for winning "exactly" 1 game); winning "exactly" 8 games, 0.39% (same for winning "exactly" 0 games). If I did the math correctly, all possible outcomes should add up to 100%. Apologies to all, I just wanted to use my SDSU math degree for good instead of evil…just once :-) Best regards, Albert |
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Post by tonatiuh on Jan 24, 2020 17:17:28 GMT -8
hxu33jdboy@nguslvg20x7! I lost you at the Drug store! No comprende! But it doesn't matter, we are 20-0!
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Post by jp92grad on Jan 24, 2020 17:56:46 GMT -8
I understand the 6.25% chance to go 4-0, but how are the losses playing into it? If another team had a 50% chance against all teams, played 8 games and won 50% (went 4-4) - what is their % and what does that mean? Not to speak for Bill and perhaps alienating everyone by talking "math" on a sports forum, the calculations were made using something called the binomial distribution formula which uses factorials in its calculation. The equation looks something like this: P(x) = p^x * q^(n-x) * n! / ((n-x)! * x!) where n = the number of trails, x = number of successes, p = probability of success, and q = the probability of failure (i.e. 1- p) If the probability of success and failure are the same, 50%, in the case of Kansas, they had 11 trails (Q1 games) with 8 successes (wins), the probability of that happening is P(x) = 0.5^8 * 0.5^(11-8) * 11! / ((11-8)! * 8!) P(x) = 0.00390625 * 0.125 * 39916800 / (6 * 40320) P(x) = 19490.625 / 241920 = 0.08056640625 P(x) = 8.06% So there was an 8.06% probability that Kansas would win "exactly" 8 of the 11 Q1 games they played given a 50% chance of winning each individual game. In your example, the probably of winning "exactly" 4 games in 8 tries would be 27.43%. Winning "exactly" 5 games, 21.88% (which is the same probability for winning "exactly" 3 games); winning "exactly" 6 games, 10.93% (same for winning "exactly" 2 games); winning "exactly" 7 games, 3.13% (same for winning "exactly" 1 game); winning "exactly" 8 games, 0.39% (same for winning "exactly" 0 games). If I did the math correctly, all possible outcomes should add up to 100%. Apologies to all, I just wanted to use my SDSU math degree for good instead of evil…just once :-) Best regards, Albert WOW, We really took different classes at SDSU. |
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Post by mactec on Jan 24, 2020 18:04:02 GMT -8
I think calculating it as "at least X wins" might show an even better picture. The chance of going 0-4 for example is also 6.25% (using 50/50 odds). Same odds as going 4-0 but obviously less impressive because the chance of going at least 0-4 is 100%. Effectively, using "at least" shows the relative percentile.
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Post by azteclou on Jan 24, 2020 18:08:42 GMT -8
hxu33jdboy@nguslvg20x7! I lost you at the Drug store! No comprende! But it doesn't matter, we are 20-0! 🤯 |
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Post by AztecBill on Jan 24, 2020 19:04:19 GMT -8
I understand the 6.25% chance to go 4-0, but how are the losses playing into it? If another team had a 50% chance against all teams, played 8 games and won 50% (went 4-4) - what is their % and what does that mean? Not to speak for Bill and perhaps alienating everyone by talking "math" on a sports forum, the calculations were made using something called the binomial distribution formula which uses factorials in its calculation. The equation looks something like this: P(x) = p^x * q^(n-x) * n! / ((n-x)! * x!) where n = the number of trails, x = number of successes, p = probability of success, and q = the probability of failure (i.e. 1- p) If the probability of success and failure are the same, 50%, in the case of Kansas, they had 11 trails (Q1 games) with 8 successes (wins), the probability of that happening is P(x) = 0.5^8 * 0.5^(11-8) * 11! / ((11-8)! * 8!) P(x) = 0.00390625 * 0.125 * 39916800 / (6 * 40320) P(x) = 19490.625 / 241920 = 0.08056640625 P(x) = 8.06% So there was an 8.06% probability that Kansas would win "exactly" 8 of the 11 Q1 games they played given a 50% chance of winning each individual game. In your example, the probably of winning "exactly" 4 games in 8 tries would be 27.43%. Winning "exactly" 5 games, 21.88% (which is the same probability for winning "exactly" 3 games); winning "exactly" 6 games, 10.93% (same for winning "exactly" 2 games); winning "exactly" 7 games, 3.13% (same for winning "exactly" 1 game); winning "exactly" 8 games, 0.39% (same for winning "exactly" 0 games). If I did the math correctly, all possible outcomes should add up to 100%. Apologies to all, I just wanted to use my SDSU math degree for good instead of evil…just once :-) Best regards, Albert Good job explaing the general case. It is much more through than I would have done since in this case there is a ton of shortcuts. .5^8 * .5^(11-8) Is simply .5^11 since .5 is used in both. Both the numerator and denominator are both multiplyed by 8! Since 11! = 8! *9 * 10 *11. So it can again be simplified. I am sure you knew those tricks, I just wanted to let others know it isn't that complicated. .5 ^11 * 11*10*9 / 3! (990/2048) / 6 = ~.0806 = 8.06% |
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Post by AztecBill on Jan 24, 2020 19:18:44 GMT -8
I think calculating it as "at least X wins" might show an even better picture. The chance of going 0-4 for example is also 6.25% (using 50/50 odds). Same odds as going 4-0 but obviously less impressive because the chance of going at least 0-4 is 100%. Effectively, using "at least" shows the relative percentile. Since we are only looking at winning records, that is not a problem. At least x wins would make every team except SDSU have a more likely outcome, making the Aztecs look even better. |
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Post by j on Jan 24, 2020 19:19:47 GMT -8
Does aztecmesa have a mathtype add-on?
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Post by azteccc on Jan 24, 2020 20:26:17 GMT -8
Classing this place up a little bit well done everyone
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Post by AztecBill on Jan 24, 2020 21:10:24 GMT -8
I understand the 6.25% chance to go 4-0, but how are the losses playing into it? If another team had a 50% chance against all teams, played 8 games and won 50% (went 4-4) - what is their % and what does that mean? 27.3% which is the most likely occurrence, if all games were 50/50. It means what they did was nothing special in Q1 games meaning they are likely an average Q1-type team. 63.6% of occurrences will be 4 wins or better. |
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Post by 🥸 Hopeless Aztec on Jan 25, 2020 6:30:25 GMT -8
Not to speak for Bill and perhaps alienating everyone by talking "math" on a sports forum, the calculations were made using something called the binomial distribution formula which uses factorials in its calculation. The equation looks something like this: P(x) = p^x * q^(n-x) * n! / ((n-x)! * x!) where n = the number of trails, x = number of successes, p = probability of success, and q = the probability of failure (i.e. 1- p) If the probability of success and failure are the same, 50%, in the case of Kansas, they had 11 trails (Q1 games) with 8 successes (wins), the probability of that happening is P(x) = 0.5^8 * 0.5^(11-8) * 11! / ((11-8)! * 8!) P(x) = 0.00390625 * 0.125 * 39916800 / (6 * 40320) P(x) = 19490.625 / 241920 = 0.08056640625 P(x) = 8.06% So there was an 8.06% probability that Kansas would win "exactly" 8 of the 11 Q1 games they played given a 50% chance of winning each individual game. In your example, the probably of winning "exactly" 4 games in 8 tries would be 27.43%. Winning "exactly" 5 games, 21.88% (which is the same probability for winning "exactly" 3 games); winning "exactly" 6 games, 10.93% (same for winning "exactly" 2 games); winning "exactly" 7 games, 3.13% (same for winning "exactly" 1 game); winning "exactly" 8 games, 0.39% (same for winning "exactly" 0 games). If I did the math correctly, all possible outcomes should add up to 100%. Apologies to all, I just wanted to use my SDSU math degree for good instead of evil…just once :-) Best regards, Albert Good job explaing the general case. It is much more through than I would have done since in this case there is a ton of shortcuts. .5^8 * .5^(11-8) Is simply .5^11 since .5 is used in both. Both the numerator and denominator are both multiplyed by 8! Since 11! = 8! *9 * 10 *11. So it can again be simplified. I am sure you knew those tricks, I just wanted to let others know it isn't that complicated. .5 ^11 * 11*10*9 / 3! (990/2048) / 6 = ~.0806 = 8.06% You two get a private room! 😂 |
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Post by Fishn'Aztec on Jan 25, 2020 7:04:01 GMT -8
Albert, did you mean trials or tries rather than "trails"?
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Post by aztecalbe on Jan 25, 2020 7:42:18 GMT -8
Albert, did you mean trials or tries rather than "trails"? I did mean "trials". Wife was rushing me out the door for a party and I didn't have time to proofread. She gave me the eye roll when I tried to explain what I was doing…so I figured I'd better go :-) |
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Post by Fishn'Aztec on Jan 25, 2020 11:36:00 GMT -8
Albert, did you mean trials or tries rather than "trails"? I did mean "trials". Wife was rushing me out the door for a party and I didn't have time to proofread. She gave me the eye roll when I tried to explain what I was doing…so I figured I'd better go :-) I thought so, spellcheckers suck when you type words like trail/trial which are spelled correctly! |
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Post by 12414 on Jan 25, 2020 12:16:41 GMT -8
To simplify, the three math stats I like are:
The Aztecs are the only undefeated team in the nation.
The Aztecs are the only undefeated team in the nation vs Q1 competition.
The Aztecs are #1 in the NET rankings.
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Post by sdsustoner on Jan 25, 2020 15:29:33 GMT -8
I understand the 6.25% chance to go 4-0, but how are the losses playing into it? If another team had a 50% chance against all teams, played 8 games and won 50% (went 4-4) - what is their % and what does that mean? Not to speak for Bill and perhaps alienating everyone by talking "math" on a sports forum, the calculations were made using something called the binomial distribution formula which uses factorials in its calculation. The equation looks something like this: P(x) = p^x * q^(n-x) * n! / ((n-x)! * x!) where n = the number of trails, x = number of successes, p = probability of success, and q = the probability of failure (i.e. 1- p) If the probability of success and failure are the same, 50%, in the case of Kansas, they had 11 trails (Q1 games) with 8 successes (wins), the probability of that happening is P(x) = 0.5^8 * 0.5^(11-8) * 11! / ((11-8)! * 8!) P(x) = 0.00390625 * 0.125 * 39916800 / (6 * 40320) P(x) = 19490.625 / 241920 = 0.08056640625 P(x) = 8.06% So there was an 8.06% probability that Kansas would win "exactly" 8 of the 11 Q1 games they played given a 50% chance of winning each individual game. In your example, the probably of winning "exactly" 4 games in 8 tries would be 27.43%. Winning "exactly" 5 games, 21.88% (which is the same probability for winning "exactly" 3 games); winning "exactly" 6 games, 10.93% (same for winning "exactly" 2 games); winning "exactly" 7 games, 3.13% (same for winning "exactly" 1 game); winning "exactly" 8 games, 0.39% (same for winning "exactly" 0 games). If I did the math correctly, all possible outcomes should add up to 100%. Apologies to all, I just wanted to use my SDSU math degree for good instead of evil…just once :-) Best regards, Albert Is Einstein your last name? |
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Post by AztecBill on Jan 27, 2020 11:40:09 GMT -8
Baylor now 6-1 which has a 5.67% chance, which is better than the Aztecs and right on queue they are now #1 in NET.
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