# 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) c-5=s

Substituting equation (2) in equation (1):

c+(c-5)=57

2c-5=57

2c=62

c=31

s=31-5=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 single-team total of 24 is 6.5%, but the probability of 25 is only 0.9%. The table below shows the single-team total probability, based on the 2000-2009 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 one-team 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.### Single-Team Totals in the NFL2000-2009 Seasons

One-Team 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 chi-squared test is perfectly suited to this kind of question. To use the test, take (a-e)^{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 chi-squared value for a total of 2, a=6 and e=6.777778, so (a-e)^{2}/e = (6-6.777778)^{2}/6.777778 = 0.089253802.

### Chi-Squared Results

Dice Total | Observations | Expected | Chi-Squared |

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 chi-squared column. In this example, the sum is 16.889344. That is called the chi-squared statistic. The number of "degrees of freedom" is one less than the number of categories in the data, in this case 11-1=10. Finally, either look up a chi-squared 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 chi-squared 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.