Ask The Wizard #247

I believe the chances could be improved with an informed guess. Here’s my straightforward approach to determining the exact scores in any NFL matchup.

1. By examining the total and point spread, we can forecast the total points for each team. For instance, using a total of 57 with a spread of -5 for the Super Bowl, we define c as the Colts' score and s as the Saints' score.
   
   (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 issue with stopping our calculation at this point is that it may yield improbable scores for one team. For example, the odds of a team scoring exactly 24 points is 6.5%, while for 25 points, it drops to just 0.9%. The accompanying table illustrates the single-team scores from the 2000-2009 seasons. Therefore, we will estimate the scoring based on credible combinations of touchdowns and field goals.
   
   
2. Let’s assume the favored team scores 2 field goals.
3. Now, let's assume the underdog secures 1 field goal.
4. Next, we subtract the points from the field goals. In our Super Bowl scenario, this results in the Colts scoring 25 points and the Saints scoring 23 points.
5. To calculate the estimated touchdowns, we divide the touchdown points by 7. Thus, we have c=3.57 TD and s=3.29 TD.
6. Now we'll round the estimated touchdowns to whole numbers. This gives us c=4 and s=3.
7. Using this formula, we calculate for total points as follows: c=(4×7)+(2×3)=34, s=(3×7)+(1×3)=24.

Employing this method across all 6,707 NFL games played from 1983 to 2009 resulted in 69 accurate predictions, yielding a success rate of 1.03%. The last accurate prediction was during the Titans versus Colts match in week 13 of 2009, which had a spread of Colts -6.5 and a total of 46. The final score was Titans 17, Colts 27.

One observer suggested a simpler approach might be to select the nearest significant score for both teams. However, this method only resulted in 51 correct predictions, giving a win percentage of 0.76%. Personally, I think it’s important to distribute the field goals as 2 for the stronger team and 1 for the weaker team.

NFL Single-Team Scoring Totals for the 2000-2009 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 topic was discussed in the forum attached to my affiliated website. Wizard of Vegas .

7.7%.

The chi-squared test This kind of question is ideally suited for statistical analysis. To apply the test, you'll need to calculate (a-e)/e for each outcome category, where 'a' represents the actual results and 'e' the expected results. As an example, the anticipated number of rolls equaling 2 in 244 acts is 244×(1/36) = 6.777778. If you're unsure why rolling a 2 has a probability of 1/36, I recommend checking my dedicated page on the subject.²For the chi-squared calculation related to rolling a 2, set a=6 and e=6.777778, leading to (a-e). Next, sum up the results from the chi-squared column, yielding a total of 16.889344. This number is referred to as the chi-squared statistic. The degrees of freedom equate to one less than the total number of categories, which here is 11-1=10. You can then look up the chi-squared statistic of 10.52 with 10 degrees of freedom in a statistics table, or utilize Excel with the formula =chidist(16.889344,10). Both will indicate a probability of 7.7%. This suggests that while these outcomes are more skewed than anticipated, they aren't strangely skewed enough to raise concerns. For further testing, I advise recording each die's individual outcomes instead of their sums. It's important to note that when expected outcomes for a category are low, the chi-squared test may not be suitable; a common threshold is a minimum expected value of 5. dice probability basics What are the chances of being dealt two pairs in pai gow?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

Pai gow consists of 16 pairs of tiles, and there are combin(16,2)=120 combinations for selecting 2 pairs out of 16. Once the specific pairs have been chosen, there is only one way to select which tiles to use. Additionally, there are combin(32,4)=35,960 methods to pick four tiles from a total of 32 tiles. Thus, the likelihood of ending up with two pairs is 120/35960, which amounts to 0.33%, equivalent to about 1 in 300.

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I encountered a new side bet in a blackjack game and I'm curious about the house edge associated with it. Alongside the house edge, what would be the best strategy to mitigate that edge and gain an advantage, if it's available? The player wins if the rank of the dealer's up card lies between the ranks of the player's first two cards. The smaller the rank gap between the player's two cards, the more lucrative the payout. A win with a one-rank gap pays 12 to 1, with two ranks it pays 6 to 1, and three ranks result in a 4 to 1 payout, while four ranks or more pays even money. Additionally, a three of a kind would pay 30 to 1. Any guidance you can offer would be greatly appreciated.²/combin(32,4)=18.69%.

From my analysis of six decks, I calculated the house edge to be 3.40%. I have documented all my calculations in my report. A very high or low count can suggest that the remaining ranks in the cards are clustered, potentially reducing the house edge, although I don’t believe it would make a significant difference worth pursuing. Wizard of Vegas .

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This topic was discussed in the forum attached to my affiliated website. Wizard of Vegas .