## Poll

5 votes (26.31%) | |||

1 vote (5.26%) | |||

2 votes (10.52%) | |||

4 votes (21.05%) | |||

3 votes (15.78%) | |||

1 vote (5.26%) | |||

No votes (0%) | |||

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1 vote (5.26%) | |||

3 votes (15.78%) |

**19 members have voted**

I guess Flutie will always have that on him for Patriots QB comparisons.Quote:billryanThe only thing Tom Brady has not accomplished in his storied career is a drop kick.

Quote:WizardSmash, I did. Fair odds on the yes are 92 to 1.

I'm going to put this out there - if I'm mistaken in any way I apologize - but I don't think I'm mistaken

if the fair odds are 92/1 and the book offers -1400

and a player wins 92 times betting $1,400 he will win $9,200 - and if he loses one time he will lose $1,400 - his net profit is $7,800

his total action (93 * $1,400) is $130,200

his edge is 5.99% ($7,800 divided by $130,200)

the edge is still not large

for comparison purposes - a winning player who can hit 57% on traditional against the spread bets in the NFL has an edge of 8.8%

on the scorigami bet:

if he bets $10,000 his dollar expectation is to average $599 profit on the bet in the long run

if he bets $1,400 his dollar expectation is to average $83.86 profit on the bet in the long run

if he bets $1,000 his dollar expectation is to average $59.90 profit on the bet in the long run

it's a lot of money to risk for a comparatively small edge and small profit

I still don't think it's a great bet - but that comes down to a matter of opinion - obviously others could have a different opinion

*

Quote:billryanThe only thing Tom Brady has not accomplished in his storied career is a drop kick.

That is exactly why Doug Flutie is a better NFL player than Brady.

Quote:SOOPOOI'll take 75-1 for as much money as you will let me bet on the yes! It is not implausible for the loser to have the highest score ever by a losing team.... a scorigami!

the highest score ever by a losing team is 51 - not impossible but I would say very, very unlikely for the loser to score more than that

more important - as you can see from the image - the highest Super Bowl scores have been way, way less than the highest playoff and highest regular season scores, and that might be what Mr. Wizard is looking at

the obvious reason for this is the much larger sample size - but it also is likely to have to do with the increased pressure in the Super Bowl - which IMO is much more likely to affect the offense than the defense

https://en.wikipedia.org/wiki/List_of_highest-scoring_NFL_games#:~:text=The%20Washington%20Redskins%20and%20the,outscored%20the%20Giants%2072%E2%80%9341.

Quote:WizardSmash, I did. I'm in a rush to get out the door, but I'll write up my method later today or tomorrow.

Bottom line is my probability of a scorigami is 0.010736. Fair odds on the yes are 92 to 1.

I think I didn't correctly account for the fact that a non-tied score can happen two ways. For example, there has never been a score of 5 to 6 in an NFL game. However, that could happen two ways in the Super Bowl:

TB 5 -- KC 6

TB 6 -- KC 5

Looking at games from 1994 (when the two-point conversion rule began) to 2018, the probability of a given side having a total of five points is 0.000396762. The probability of a given side having a total of six points is 0.021187113.

Since there are two combinations for a 5-6 score, I contend a decent estimate of the probability is 2*0.000396762*0.021187113 = 0.00001681.

Doing this for every set of scores that hasn't happened yet, I get a probability of 0.017925, or 1 in 55.

Any comments?

Total | Total | Combined Probability |
---|---|---|

0 | 170 | 0.013490 |

1 | 0 | 0.000000 |

2 | 2 | 0.000159 |

3 | 303 | 0.024044 |

4 | 0 | 0.000000 |

5 | 5 | 0.000397 |

6 | 267 | 0.021187 |

7 | 420 | 0.033328 |

8 | 29 | 0.002301 |

9 | 188 | 0.014918 |

10 | 706 | 0.056023 |

11 | 32 | 0.002539 |

12 | 123 | 0.009760 |

13 | 646 | 0.051262 |

14 | 530 | 0.042057 |

15 | 128 | 0.010157 |

16 | 434 | 0.034439 |

17 | 892 | 0.070782 |

18 | 91 | 0.007221 |

19 | 282 | 0.022377 |

20 | 860 | 0.068243 |

21 | 511 | 0.040549 |

22 | 189 | 0.014998 |

23 | 548 | 0.043485 |

24 | 821 | 0.065148 |

25 | 118 | 0.009364 |

26 | 267 | 0.021187 |

27 | 673 | 0.053404 |

28 | 382 | 0.030313 |

29 | 131 | 0.010395 |

30 | 336 | 0.026662 |

31 | 578 | 0.045866 |

32 | 61 | 0.004841 |

33 | 146 | 0.011585 |

34 | 394 | 0.031265 |

35 | 200 | 0.015870 |

36 | 71 | 0.005634 |

37 | 163 | 0.012934 |

38 | 265 | 0.021028 |

39 | 30 | 0.002381 |

40 | 50 | 0.003968 |

41 | 146 | 0.011585 |

42 | 78 | 0.006189 |

43 | 25 | 0.001984 |

44 | 58 | 0.004602 |

45 | 85 | 0.006745 |

46 | 7 | 0.000555 |

47 | 16 | 0.001270 |

48 | 47 | 0.003730 |

49 | 35 | 0.002777 |

50 | 5 | 0.000397 |

51 | 15 | 0.001190 |

52 | 14 | 0.001111 |

53 | 1 | 0.000079 |

54 | 4 | 0.000317 |

55 | 6 | 0.000476 |

56 | 6 | 0.000476 |

57 | 2 | 0.000159 |

58 | 3 | 0.000238 |

59 | 5 | 0.000397 |

60 | 0 | 0.000000 |

61 | 0 | 0.000000 |

62 | 2 | 0.000159 |

Total | 12602 | 1.000000 |

For any given score combination that has never happened, I take 2*probability(score 1)*probability(score 2). Sum that up for every unseen score and you get my answer in the previous post.

I'll save the forum some trouble and offer my own criticism:

1. This does not take into account the probability a team scores a total of one point. Yes, the NFL has such a thing as a one-point safety. Since no team has ever had a total of one at the end of the game, my method assigns a probability of zero to this. To end the game with one point, would take scoring ONLY a one-point safety in the entire game.

2. This assumes the two individual team scores are independent of each other. In reality, I think there is a negative correlation. If one team scores a LOT of points, like over 50, it is probably at the expense of the other team scoring very little.

My defense is factoring in these criticisms would have had a very marginal effect, but significantly increased the difficulty and complexity of the analysis.

Quote:WizardTo expand, this table shows the count, from 1994 to 2018, of individual team scores and the probability.

Total Total Combined Probability 0 170 0.013490 1 0 0.000000 2 2 0.000159 3 303 0.024044 4 0 0.000000 5 5 0.000397 6 267 0.021187 7 420 0.033328 8 29 0.002301 9 188 0.014918 10 706 0.056023 11 32 0.002539 12 123 0.009760 13 646 0.051262 14 530 0.042057 15 128 0.010157 16 434 0.034439 17 892 0.070782 18 91 0.007221 19 282 0.022377 20 860 0.068243 21 511 0.040549 22 189 0.014998 23 548 0.043485 24 821 0.065148 25 118 0.009364 26 267 0.021187 27 673 0.053404 28 382 0.030313 29 131 0.010395 30 336 0.026662 31 578 0.045866 32 61 0.004841 33 146 0.011585 34 394 0.031265 35 200 0.015870 36 71 0.005634 37 163 0.012934 38 265 0.021028 39 30 0.002381 40 50 0.003968 41 146 0.011585 42 78 0.006189 43 25 0.001984 44 58 0.004602 45 85 0.006745 46 7 0.000555 47 16 0.001270 48 47 0.003730 49 35 0.002777 50 5 0.000397 51 15 0.001190 52 14 0.001111 53 1 0.000079 54 4 0.000317 55 6 0.000476 56 6 0.000476 57 2 0.000159 58 3 0.000238 59 5 0.000397 60 0 0.000000 61 0 0.000000 62 2 0.000159 Total 12602 1.000000

For any given score combination that has never happened, I take 2*probability(score 1)*probability(score 2). Sum that up for every unseen score and you get my answer in the previous post.

I'll save the forum some trouble and offer my own criticism:

1. This does not take into account the probability a team scores a total of one point. Yes, the NFL has such a thing as a one-point safety. Since no team has ever had a total of one at the end of the game, my method assigns a probability of zero to this. To end the game with one point, would take scoring ONLY a one-point safety in the entire game.

2. This assumes the two individual team scores are independent of each other. In reality, I think there is a negative correlation. If one team scores a LOT of points, like over 50, it is probably at the expense of the other team scoring very little.

My defense is factoring in these criticisms would have had a very marginal effect, but significantly increased the difficulty and complexity of the analysis.

Have you thought about testing this distribution against the actual final scores over the same period to check the goodness of fit?

Quote:unJonQuote:WizardTo expand, this table shows the count, from 1994 to 2018, of individual team scores and the probability.

Total Total Combined Probability 0 170 0.013490 1 0 0.000000 2 2 0.000159 3 303 0.024044 4 0 0.000000 5 5 0.000397 6 267 0.021187 7 420 0.033328 8 29 0.002301 9 188 0.014918 10 706 0.056023 11 32 0.002539 12 123 0.009760 13 646 0.051262 14 530 0.042057 15 128 0.010157 16 434 0.034439 17 892 0.070782 18 91 0.007221 19 282 0.022377 20 860 0.068243 21 511 0.040549 22 189 0.014998 23 548 0.043485 24 821 0.065148 25 118 0.009364 26 267 0.021187 27 673 0.053404 28 382 0.030313 29 131 0.010395 30 336 0.026662 31 578 0.045866 32 61 0.004841 33 146 0.011585 34 394 0.031265 35 200 0.015870 36 71 0.005634 37 163 0.012934 38 265 0.021028 39 30 0.002381 40 50 0.003968 41 146 0.011585 42 78 0.006189 43 25 0.001984 44 58 0.004602 45 85 0.006745 46 7 0.000555 47 16 0.001270 48 47 0.003730 49 35 0.002777 50 5 0.000397 51 15 0.001190 52 14 0.001111 53 1 0.000079 54 4 0.000317 55 6 0.000476 56 6 0.000476 57 2 0.000159 58 3 0.000238 59 5 0.000397 60 0 0.000000 61 0 0.000000 62 2 0.000159 Total 12602 1.000000

For any given score combination that has never happened, I take 2*probability(score 1)*probability(score 2). Sum that up for every unseen score and you get my answer in the previous post.

I'll save the forum some trouble and offer my own criticism:

1. This does not take into account the probability a team scores a total of one point. Yes, the NFL has such a thing as a one-point safety. Since no team has ever had a total of one at the end of the game, my method assigns a probability of zero to this. To end the game with one point, would take scoring ONLY a one-point safety in the entire game.

2. This assumes the two individual team scores are independent of each other. In reality, I think there is a negative correlation. If one team scores a LOT of points, like over 50, it is probably at the expense of the other team scoring very little.

My defense is factoring in these criticisms would have had a very marginal effect, but significantly increased the difficulty and complexity of the analysis.

Have you thought about testing this distribution against the actual final scores over the same period to check the goodness of fit?

Oh never mind. It looks like you are using the historical data to generate this.