Super Bowl 55

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

Quote: billryan

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

I guess Flutie will always have that on him for Patriots QB comparisons.

Quote: Wizard

Smash, 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: billryan

The 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: SOOPOO

I'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: Wizard

Smash, 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?

To 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.

Quote: Wizard

To 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?

Cool chart...surprised 20 is that much more likely than 21

Quote: unJon

Quote: Wizard

To 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.