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Ask the Wizard #247
FinsRule
I think the odds are better than that with a good educated guess. Here is my basic strategy for picking the exact outcome of any NFL game.
 Using the total and spread, estimate the total points of each team. For example, if we use a total of 57 and a spread of 5 for the Super Bowl, letting c=Colts points, and s=Saints points then...
(1) c+s=57
(2) c5=s
Substituting equation (2) in equation (1):
c+(c5)=57
2c5=57
2c=62
c=31
s=315=26
The problem with stopping here is sometimes you get values that are unlikely to be scored by a single team. For example, the probability of a singleteam total of 24 is 6.5%, but the probability of 25 is only 0.9%. The table below shows the singleteam total probability, based on the 20002009 seasons. So we’re going to estimate the total points for each team based on realistic combinations of field goals and touchdowns.
 Assume the favorite kicks 2 field goals.
 Asssume the underdog kicks 1 field goal.
 Subtract the field goal points from each. In the Super Bowl example, this would leave Colts = 25 touchdown points, Saints = 23 touchdown points.
 Divide touchdown points by 7, to get estimated touchdowns. c=3.57 TD, s=3.29 TD
 Round the estimated touchdowns to the nearest integer. c=4, s=3.
 Following this method, we get for total points c=(4×7)+(2×3)=34, s=(3×7)+(1×3)=24.
Using this method on all 6,707 games from the 1983 through the 2009 seasons would have resulted in 69 correct picks, for a success rate of 1.03%. The last time it would have been right was the Titans/Colts game in week 13 of 2009. That game had a spread of Colts 6.5, and a total of 46. The score was Titans 17, Colts 27.
One critic thought a better and simpler strategy would be to pick the nearest significant oneteam total for both teams. Using such a method resulted in 51 wins only, for a win rate of 0.76%. In my opinion, splitting the field goals 2 and 1 between the stronger and weaker teams is important.SingleTeam Totals in the NFL20002009 Seasons
OneTeam Total  Total in Sample  Probability 
0  93  1.75% 
1  0  0.00% 
2  0  0.00% 
3  148  2.79% 
4  0  0.00% 
5  2  0.04% 
6  114  2.15% 
7  210  3.96% 
8  9  0.17% 
9  76  1.43% 
10  316  5.96% 
11  9  0.17% 
12  49  0.92% 
13  289  5.45% 
14  238  4.49% 
15  55  1.04% 
16  170  3.21% 
17  373  7.03% 
18  33  0.62% 
19  92  1.73% 
20  368  6.94% 
21  234  4.41% 
22  64  1.21% 
23  218  4.11% 
24  347  6.54% 
25  47  0.89% 
26  103  1.94% 
27  282  5.32% 
28  159  3.00% 
29  52  0.98% 
30  127  2.39% 
31  242  4.56% 
32  23  0.43% 
33  57  1.07% 
34  164  3.09% 
35  76  1.43% 
36  27  0.51% 
37  68  1.28% 
38  108  2.04% 
39  11  0.21% 
40  21  0.40% 
41  62  1.17% 
42  31  0.58% 
43  6  0.11% 
44  24  0.45% 
45  33  0.62% 
46  1  0.02% 
47  7  0.13% 
48  28  0.53% 
49  15  0.28% 
50  1  0.02% 
51  5  0.09% 
52  7  0.13% 
53  0  0.00% 
54  2  0.04% 
55  1  0.02% 
56  4  0.08% 
57  1  0.02% 
58  1  0.02% 
59  1  0.02% 
Total  5304  100.00% 
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
Dice Test Data  
Dice Total  Observations 
2  6 
3  12 
4  14 
5  18 
6  23 
7  50 
8  36 
9  37 
10  27 
11  14 
12  7 
Total  244 
C. from Las Vegas
7.7%.
The chisquared test is perfectly suited to this kind of question. To use the test, take (ae)^{2}/e for each category, where a is the actual outcome, and e is the expected outcome. For example, the expected number of rolls totaling 2 in 244 throws is 244×(1/36) = 6.777778. If you don’t understand why the probability of rolling a 2 is 1/36, then please read my page on dice probability basics. For the chisquared value for a total of 2, a=6 and e=6.777778, so (ae)^{2}/e = (66.777778)^{2}/6.777778 = 0.089253802.
ChiSquared Results
Dice Total  Observations  Expected  ChiSquared 
2  6  6.777778  0.089253 
3  12  13.555556  0.178506 
4  14  20.333333  1.972678 
5  18  27.111111  3.061931 
6  23  33.888889  3.498725 
7  50  40.666667  2.142077 
8  36  33.888889  0.131512 
9  37  27.111111  3.607013 
10  27  20.333333  2.185792 
11  14  13.555556  0.014572 
12  7  6.777778  0.007286 
Total  244  244  16.889344 
Then take the sum of the chisquared column. In this example, the sum is 16.889344. That is called the chisquared statistic. The number of "degrees of freedom" is one less than the number of categories in the data, in this case 111=10. Finally, either look up a chisquared statistic of 10.52 and 10 degrees of freedom in a statistics table, or use the formula =chidist(16.889344,10) in Excel. Either will give you a result of 7.7%. That means that the probability fair dice would produce results this skewed or more is 7.7%. The bottom line is while these results are more skewed than would be expected, they are not skewed enough to raise any eyebrows. If you continue this test, I would suggest collecting the individual outcome of each die, rather than the sum. It should also be noted that the chisquared test is not appropriate if the expected number of outcomes of a category is low. A minimum expectation of 5 is a figure commonly bandied about.
Lisa from Las Vegas
Pai gow consists of 16 pairs of tiles. There are combin(16,2)=120 ways to choose 2 pairs out of 16. Once the two pairs are chosen, there is one way only to choose the specific tiles. There are combin(32,4)=35,960 ways to choose four tiles out 32. So the probability of two pairs is 120/35960 = 0.33%, or 1 in 300.
Not that you asked, but the probability of one pair is 16×combin(15,2)×2^{2}/combin(32,4)=18.69%.
Pictures taken from my companion site Wizard of Vegas.
blackchipjim
Based on six decks, I get a house edge of 3.40%. I show all my math in my blackjack appendix 8. An extremely high or low count would indicate the ranks in the remaining cards are clumped together, which would lower the house edge, but I don’t think it would be enough to warrant bothering with.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.