Ask the Wizard: Sports - FAQ

$240 bet on Vargas will win $100. A $100 bet on Quartey will win $190. Of course you can bet any amount but the winnings will still be in the same proportion. If you bet $100 on Vargas you will win $100*(100/240)=$41.67. Stay tuned for my new section on sports betting planned for this April.

It doesn’t make any difference who pays. Lets suppose Atlanta is playing Miami and you want to bet on Atlanta and want to win $100. Either way you have to bet $110 to win $100. However it is easier to explain the bet as the loser pays.

I’m quite sure that Nevada is still the only state in which you can bet on all major sports.

Assuming the cells in the grid are chosen at random, which by randomly picking the row and column headings they generally, then the odds of winning any one quarter would be 1/100. If the row and column headings were scrambled for each quarter then the probability of winning all four quarters would be (1/100)^4, or 1 in 100,000,000.

Unfortunately I can’t elaborate much. Sports betting is not my strong point, although I plan to learn more about it when I have the time. I will say it is hard to compare the profitability of sports betting to card counting. Card counting is very technical and by the book. Making money sports betting requires more judgement and is more subject to opinion. There are various strategies one can employ to make money sports betting, for example looking to arbitrage games by taking both sides at different casinos under different point spreads, looking to exploit unusual proposition bets, or going after correlated parlays. I would recommend ’Sharp Sports Betting’ by Stanford Wong for more information on winning at sports betting.

The following table shows the probability of attaining every percentage from 50% to 60%, in increments of 1%, by each number of seasons from 1 to 5. This is based on a 259 game season. I also assume that the overall percentage is rounded down. For example if a handicapper picked 132 out of 259 games, for a percentage of 50.97%, he would only get credit for picking 50%, since he didn’t quite attain 51%. It would not surprise me if these boasting handicappers are rounding in their own favor.

Probability of Handicapping Percentages in the NFL
Ratio	1 Seasons	2 Seasons	3 Seasons	4 Seasons	5 Seasons
0.5	0.5	0.517523	0.5	0.512393	0.5
0.51	0.354641	0.314437	0.282985	0.257059	0.234993
0.52	0.267178	0.178085	0.125486	0.101366	0.074229
0.53	0.160065	0.086589	0.049447	0.025155	0.015098
0.54	0.106982	0.035817	0.013066	0.004959	0.001926
0.55	0.053095	0.012519	0.002569	0.000687	0.000152
0.56	0.023385	0.00282	0.000373	0.000051	0.000007
0.57	0.012645	0.00067	0.000053	0.000003	0
0.58	0.00453	0.000133	0.000004	0	0
0.59	0.00213	0.000022	0	0	0
0.6	0.000617	0.000003	0	0	0

There are 17 regular weeks in the NFL season and 32 teams. The probability of winning any given week is 1 in 32*31*30 = 29760. The expected value of each ticket is $1000*17/29760 = 57.12 cents. So the expected return is 2.28%, or a house edge of 97.72%!

The last couple years I have been devoting a lot of time to sports betting, much more than any other form of gambling. Unless you can play two sport books against each other via different lines, an opportunity which is very rare, there are no guaranteed winning ways to make money sports betting. It is too early to reveal any secrets but I look for any perceived bet with a player advantage and then make my bet. You might like to see some information I did to recently in my sports betting appendix 2.

If we assumed the games were independent then the probability of losing both would be 90%*70%=63%. But since you say the probability of losing both is actually 65% (which is more than the 63%), that means the two events are correlated. If the probability of losing both is 65% and just losing game 2 is 70%, then the probability of winning game 1 and losing game 2 must be 5%. Using the same logic the probability of losing game 1 and winning game 2 must be 25%. That only leaves 5% for winning both games. So the probability of winning exactly once is 25%+5% = 30%.

I do not know the effect of rain on the total. However I do know that sharp handicappers consider the weather very seriously and move the line to reflect the temperature, precipitation, wind speed, and things like that.

First, I’m skeptical that anyone could make 80K with 1K bets and a bankroll of only 250-300K. And don’t even get me started about the word "system." To answer your question, for the most part the best sports bettors may practice openly. Even if a sports book did forbid a professional’s action or 86 them from the property it would be easy to get someone else to do the betting. Then again I once went to a Super Bowl proposition bet seminar by Fezzik, a professional gambler, and he gave his presentation in a Halloween mask.

The office pools I have seen randomized the tables by assigning a random digit to each row and column. However if you can choose the actual terminal digits the following table shows the frequency of each terminal digit for the final score of either team, based on every NFL game from 1983 to 2003.

NFL Terminal Digits per Side
Digit	Frequency	Probability
0	1887	17.75%
1	1097	10.32%
2	348	3.27%
3	1382	13.00%
4	1608	15.13%
5	396	3.73%
6	848	7.98%
7	1945	18.30%
8	631	5.94%
9	488	4.59%
Total	10630	100%

So this table shows 7 is the best choice, followed by 0, 4, and 3.

Some of them I like more than others. My model creates what it estimates to be a fair spread in every game and I list the ones that have a disparity of more than a chosen number of points. Personally I bet more on the ones with the larger disparities. However until I have proven my picks are any good I think it would be pretentious to post degrees of confidence.

Some of them I like more than others. My model creates what it estimates to be a fair spread in every game and I list the ones that have a disparity of more than a chosen number of points. Personally I bet more on the ones with the larger disparities. However until I have proven my picks are any good I think it would be pretentious to post degrees of confidence.

The reason has nothing to do with team loyalty. My program rated them the fifth best team at the close of the 2004 season, and that I carry that power rating over into the 2005 season. However maybe it is too slow too react to recent history. Something for me to think about.

For the 2 to 1 bets, the largest sample size in which this is a good bet is 14 games. Picking randomly you would have a 39.5264% chance of picking 57% or better, or 8 or more right. You need to have better than a 33.3333% probability for it to be a good bet. 21 is close to fair, but the probability is slightly too small at 33.1812%. The larger the number of picks required the lower the required percentage should be. For example, if 1000 picks were required the greatest number required to have more than a 1/3 chance of winning would be 507. The probability of getting 507 or more right in 1000 picks is 34.05%.

The only things that go into the model are scores and home field advantage. Sometimes I seek outside opinions on major injuries, such as Ben Roethlisberger in week 6, or New Orleans playing home games in San Antonio.

Thanks. You probably sent this before week four, which was a bad week for me. Now, after five weeks my record is 17 and 13, for a 56.7% average, which is good but not great. I think that as a bonus to my newsletter readers I’ll give my thoughts on all the games.

First, whoever accepted this bet should honor it, on principle alone. A gentleman honors his debts, especially gambling debts. Second, although I haven’t studied it I think the quarters may actually be negatively correlated. For example if the first quarter has a low total it may be more likely that either team will have good field position at the beginning of the second quarter, and thus likely to make the second quarter high scoring, and vise versa.

I have been thinking a lot about this lately. In my opinion a winning percentage should have an equal weighting per game. You should also have a separate statistic on your overall return of investment, however any statistic should be backed up with a list of the side, date, line source, point spread, and the odds (usually -110). Another issue you don’t bring up is what to do if you have to lay -120 off of a 3 or 7 point spread. It would be easier to attain a good win percentage if you loaded your picks with such bets. So I believe a return on investment figure should be kept even if flat betting. Another thing that bothers me about some other handicappers is they quote lines that are nowhere to be seen. I think it is okay to shop around a bit but quoted lines should be not difficult to find. I admit I don’t do some of these things myself this season, because I didn’t think of these things when I started. Next year, if I do this again, I will document my results as an investment as well.

Regardless of the reason for making the bet, in general it is better to bet underdogs on the money line and favorites against the spread.

Personally I use the NFL Access database from Mr. NFL, which costs $99. If there is anything as good for less I’m not aware of it.

Historically speaking underdogs are a better bet. Here are the results every game played from the start of the 1983 season through week 10 of the 2005 season.

Favorite wins against spread: 2554 games
Underdog wins against spread: 2724 games
Game ends exactly on spread: 150 games

So on resolved bets the underdog have won 51.61% of the time. It is also well known that square bettors prefer to bet favorites, creating value on the underdogs.

The expected return on the straight bets would be (0.5*1 + 0.5*(-1.2))/1.2 = -8.33%. The expected return on the parlay would be 0.25*2.5 + 0.75*-1 = -12.5%. However if I were to only bet two games and want to win or go bust trying then I would go with the parlay. More importantly I would boycott this bookie out of principle, because I’ve never heard of having to lay -120 on straight bets before.

Good point. After giving it some thought I have decided to remove my sports betting appendix 3. It was based on the assumption that the fair line is exactly between the two money lines. For example in the last Super Bowl the money lines were generally Seattle +160 and Pittsburgh -180. My appendix was based on the unrealistic assumption that the fair money lines would be +/- 170. In reality money lines are market driven. Square action on the money line tends to favor the favorite, creating value on the underdog. Assuming Pittsburgh as a 4-point favorite was fair historical data suggests a 4-point favorite has about a 61% chance of winning. That would make the fair money lines +156 on Seattle, and -156 on Pittsburgh. To remind you most casinos had the money lines at +160/-180. Of course Pittsburgh won the game this time, but historically speaking I’m sure you would make out better betting underdogs on the money line over favorites.

For my readers who may not understand the question, a Super Bowl pool has a 10 by 10 grid. Players will buy squares for a set amount each, writing the name of the buyer in each square. After all 100 squares have been purchased the rows and columns should be given headings randomly from 0 to 9. Kind of like a 10 by 10 multiplication table, only with the row and column headings randomly shuffled. Then the terminal digit of each team’s score will used to determine the winner. The reason for the shuffling is that some terminal digits are much more likely than others, as you will see below. For example, whoever ended up with the Seattle 0, Pittsburgh 1 square would have won the pool, because the final score was Seattle 10, Pittsburgh 21.

The following table shows the frequency of each combination in every NFL game from the 1983 to 2005 seasons. It should be noted that the 2-point conversion rule came along about 1998, which would have the effect of smoothing out the distribution a bit.

Terminal Digit in the NFL - Away by Home Total
Away Team	Home Team										Total
	0	1	2	3	4	5	6	7	8	9
0	126	104	34	160	138	37	99	237	64	32	1031
1	73	40	17	41	103	21	36	117	67	31	546
2	25	15	1	20	30	9	13	33	9	12	167
3	194	69	33	66	102	40	102	132	43	40	821
4	122	133	27	78	108	27	48	215	59	35	852
5	32	12	12	21	21	9	13	36	22	5	183
6	91	47	17	75	57	7	28	57	25	39	443
7	217	115	35	135	195	47	65	125	69	47	1050
8	43	59	12	24	41	23	20	38	15	7	282
9	48	28	17	33	40	14	29	33	14	7	263
Total	971	622	205	653	835	234	453	1023	387	255	5638

The next table shows the probability of each combination based on the totals in the table above.

Terminal Digit in the NFL - Away by Home Probability
Away Team	Home Team										Total
	0	1	2	3	4	5	6	7	8	9
0	0.0223	0.0184	0.006	0.0284	0.0245	0.0066	0.0176	0.042	0.0114	0.0057	0.1829
1	0.0129	0.0071	0.003	0.0073	0.0183	0.0037	0.0064	0.0208	0.0119	0.0055	0.0968
2	0.0044	0.0027	0.0002	0.0035	0.0053	0.0016	0.0023	0.0059	0.0016	0.0021	0.0296
3	0.0344	0.0122	0.0059	0.0117	0.0181	0.0071	0.0181	0.0234	0.0076	0.0071	0.1456
4	0.0216	0.0236	0.0048	0.0138	0.0192	0.0048	0.0085	0.0381	0.0105	0.0062	0.1511
5	0.0057	0.0021	0.0021	0.0037	0.0037	0.0016	0.0023	0.0064	0.0039	0.0009	0.0325
6	0.0161	0.0083	0.003	0.0133	0.0101	0.0012	0.005	0.0101	0.0044	0.0069	0.0786
7	0.0385	0.0204	0.0062	0.0239	0.0346	0.0083	0.0115	0.0222	0.0122	0.0083	0.1862
8	0.0076	0.0105	0.0021	0.0043	0.0073	0.0041	0.0035	0.0067	0.0027	0.0012	0.05
9	0.0085	0.005	0.003	0.0059	0.0071	0.0025	0.0051	0.0059	0.0025	0.0012	0.0466
Total	0.1722	0.1103	0.0364	0.1158	0.1481	0.0415	0.0803	0.1814	0.0686	0.0452	1

Although there is a ceremonial home team in the Super Bowl, I think we can ignore that. Let’s also ignore everything about both teams and go strictly off of the historical averages above. Finally, let’s ignore the fact that the Super Bowl can not end in a tie, which would make a match like 4/4 a little less likely to win. So let’s take an average when the terminal digits are different. For example the probability of your Pittsburgh 7, Seattle 4 square would be the average of Away 7, Home 4; and Away 4, Home 7. Doing so results in the following probability for each of your squares.

Pitts 7, Sea 4: (0.0346+0.0381)/2 = 0.0364
Pitts 7, Sea 0: (0.0385+0.0420)/2 = 0.0403
Pitts 4, Sea 4: 0.0192
Pitts 4, Sea 0: (0.0216+0.0245)/2 = 0.0231

So your total probability of winning one of these is 11.90%. Considering you only covered 4% of the squares you made out well.

Although you didn’t specifically ask, here is how often each terminal digit occured. It shows that overall from most the frequent the order is 7043168952.

Terminal Digit in the NFL - Away and Home Totals
Terminal Digit	Away Team	Home Team	Total
0	1031	971	2002
1	546	622	1168
2	167	205	372
3	821	653	1474
4	852	835	1687
5	183	234	417
6	443	453	896
7	1050	1023	2073
8	282	387	669
9	263	255	518

Finally, here is the probability of each terminal digit.

 

People have a misconception that sports betting limits in vegas are really high, which in my experience only applies to football and not always in that case. Could you tell me the approxamite limits in vegas sports books on the 4 major sports? Thank you for your time.
— Betsy from Malibu

I don’t count hockey as a major sport because it gets very little action. I’m told that the Coast casinos have the highest limits. There is no maximum set in stone that I know of but they take large bets on a case by case basis. Here is what I think they would probably take on the average game.

NFL side: $50,000
NFL total: $5000
MLB money line: $10,000
MLB total: $2000
NBA side: $10,000
NBA total: $2,000

 

With many online books offering halftime wagering, I’d love to know your opinion on "dutching". Sometimes it’s obvious, but at what point would you consider betting the opposite of your game bet at halftime? Also, what amount would you consider if you found a favorable situation?
— Anonymous

I assume by "dutching" you mean hedging. The sixth of my ten commandments of gambling is "Thou shalt not hedge thy bets." The only time I would make an exception is when the hedge bet itself has a positive expected value, or life changing amounts of money are at stake.

 

What is the maximum amount of money you can win on a college football wager in a Vegas casino without having to provide your personal information to the casino when you cash out your ticket? I’d like to know the amount I can win (per wager) and still have them pay me cash with no questions asked.
— Harold from Monterey Park, CA

$10,000. A cash transaction of more than $10,000, whether in sports or not, will necessitate a CTR (cash transaction report). You may also ask to be paid in chips, although the same CTR would be generated if and when you cashed them in.

 

Other than the free sites, can you recommend a site where I can get professional sports handicapping for a monthly fee, or percent of win.
— Danny M. from Santa Barbara

No. I do not endorse any touts.

 

You have talked about your betting of the underdogs in the NFL. Did you simply bet the home underdog each time or was it more complicated than that? And most importantly, what line determined if it was a home dog...The early line or the line just before the game. Thanks.
— Steve from Milwaukee

I don’t just blindly bet underdogs, even if I have to lay -105 only, which is the case this season at the Plaza in downtown Las Vegas. If I detect only a small advantage on a sports bet, I won’t bet it. In sports I always assume a certain margin of error, because ultimately people play the game, not statistics. However, if I can find a better than the market line, or the side comes with the recommendation of a trusted handicapper, then I would be happy to make the bet.

 

Thanks for the great website! My father and I are having an argument about hedging bets and could really use your help! The particular situation involves a bet on the Super Bowl. Prior to the start of the season (my father can’t remember the year) my uncle placed a bet that New England would win the Super Bowl. The bet paid 60 to 1. Just before the Super Bowl (in which New England was playing) my uncle hedged his bet (father can’t remember the details) giving up the potential $6,000 pay day but guaranteeing a $3,000 one instead. I’m convinced that this was a sucker bet but my father won’t listen to me. I’m arguing that by hedging in this spot he was giving up expected value and the smart bettor never does this. My father is arguing that it’s fine to give up the expected value because of the money involved and the fact that this bet doesn’t come up often, exactly the same as insuring your house. Of course, I argue that home ownership is inherently different than sports betting in that one can be avoided while the other really can’t. What are your thoughts? Please help us settle this!
— James from Oakville, Ontario

The Seventh of my Ten Commandments of Gambling is, "Thou shalt not hedge thy bets." However, in my remarks I add, "Exceptions can be made for insuring life-changing amounts of money." So if the surrender value of $3,000 is a life-changing amount of money to him, and if the probability of winning were not much more than 50%, then I wouldn’t begrudge his decision. However, unless this was in 2002, the probability of New England winning was much higher than 50%. The other two years they played in the Super Bowl, 2004 and 2005, they were 7-point favorites. I would estimate the probability of winning either year was about 71%. A fair surrender value would have been 0.71 × $6100 (including his original bet back) = $4,331. The house edge on the offer, which was equivalent to an even money bet on the other team, was 29%-71% = 42%. So, if I’m right about the year, he made a very bad decision. He could have gotten much better odds on the open market by betting the other team on the money line. Whoever offered only $3,000 was either very ignorant of the game or took unfair advantage. Interestingly, New England won all three of their recent Super Bowls by three points.

 

What is the probability of an NFL game ending in a tie?
— Lee from Los Angeles

From the 1983 to 2007 seasons, there were 10 ties over 5,901 regular season games played. Rules dictate that a game can not end in a tie in the post-season. So that would make the probability, based on historical games, 0.17%, or about 1 in 590.

 

What is the standard deviation of the difference between the final point differential between the two teams in an NFL game and the pointspread of that game?
— Don from New York

For the benefit of other readers, let’s look at an example from the last Super Bowl, to illustrate what you are asking. The Patriots were a 12-point favorite, but lost by 3. So the game finished 15 points away from the point spread. If I understand the question correctly, you are asking about the standard deviation of this difference. The mean difference was close to zero for each league studied. Here is what I get for the standard deviation:

NFL 13.31 (based on 2000 to 2007 seasons)
College Football 15.72 (based on 1993 to 2007 seasons)
NBA 11.39 (based on 1987 to 2003 seasons)
So, the 2008 Super Bowl finished 15/13.31 = 1.13 standard deviations away from expectations. I’m ignoring the adjustment factor for a discrete distribution, to keep things as simple as possible. The probability of being 1.13 standard deviations or more from expectations, in either direction is 25.85%. This can be found in Excel, using the formula 2 × normsdist(-1.13).

 

As a sports bettor, who likes to play underdogs, and who likes to shop around for lines, I know the value of finding each and every extra half point you can. In your typical NFL or NBA game, what is each half point worth to you? I know that a gambler at -110 on a fair line needs to hit 52.4% to break even. I know lines are dictated by the marketplace, but what would you say each half point is truly worth? If you could get a half point extra on the fair line of each game you bet would that make your break even point truly be 50%. Is there any way to calculate it? Thanks.
— Chris from Chicago

As I show in my page in the NBA, when buying a half point the probability of a win is 51.01%, a loss is 47.01%, and a push is 1.98%, assuming the bettor never buys the half point off of a spread of 0 or -1, which he shouldn’t do. If you only had to lay 110 for the extra half point, the expected return would be (0.5101 - 1.1×.4701)/1.1 = -0.64%. So, a free half point would not be enough to overcome the house edge.

Not that you asked, but if you lay 120, you can buy a half point at most sports books. If you were prepared to bet a game against the spread anyway, is the extra half point a good value? Laying 110, the house edge for a random picker is 4.45%, including ties. Laying 120, the house edge with the half point is 4.50%. So, buying a half point is marginally not worth the price.

The value of buying a half point in football depends a great deal on the point spread, because some margins of victory are much more likely than others. The only time it is worth buying the half point in the NFL is off of a point spread of 3. Unfortunately, the sports books know this too, and won’t let you do it off of 3, most of the time.

 

Knowing that team A scores 1.5 goals per game on average, and team B scores 1.2 goals per game on average, what are the chances that in a game between A and B:

1) A will score more than B
2) B will score more than A
3) Game finishes as a tie.

Is the information provided enough to calculate the probabilities for each outcome?
— Dimitar from Sophia, Bulgaria

That does not take into account that the individual scores should be somewhat negatively correlated, and that the average points each team gives up is just as important as the average points scored. If we can assume that 1.5 and 1.2 are the expected number of points scored in the game, considering both offense and defense, and we ignore the correlation factor, then we can get a decent estimate on your three probabilities. There are lots of Super Bowl props like this, but based on who will score more touchdowns, field goals, interceptions, etc..

The first step is to use the Poisson distribution to estimate the probability of each number of goals for each team. The general formula is the probability that a team has g goals, with a mean of m, is e^-m × m^g/g!. In Excel, you can use the formula poisson(g,m,0). The following table shows the probability for 0 to 10 goals of both teams, using this formula.

Terminal Digit in the NFL - Away and Home Probabilities
Terminal Digit	Away Team	Home Team	Total
0	0.1829	0.1722	0.1775
1	0.0968	0.1103	0.1036
2	0.0296	0.0364	0.033
3	0.1456	0.1158	0.1307
4	0.1511	0.1481	0.1496
5	0.0325	0.0415	0.037
6	0.0786	0.0803	0.0795
7	0.1862	0.1814	0.1838
8	0.05	0.0686	0.0593
9	0.0466	0.0452	0.0459

Probabilities for 0 to 8 Goals for each Team
Goals	Team A	Team B
0	0.223130	0.301194
1	0.334695	0.361433
2	0.251021	0.216860
3	0.125511	0.086744
4	0.047067	0.026023
5	0.014120	0.006246
6	0.003530	0.001249
7	0.000756	0.000214
8	0.000142	0.000032

The next step is rather mundane, but you have to make a matrix of all the 81 possible combinations of 0 to 8 scores for each team. This is done by multiplying the probability of x scores for team A and y scores for team B, from the table above. The following table shows the probability of every score combination from 0-0 to 8-8.

Probabilities Combinations for Both Teams
Goals Team A	Goals Team B
	0	1	2	3	4	5	6	7	8
0	0.067206	0.080647	0.048388	0.019355	0.005807	0.001394	0.000279	0.000048	0.000007
1	0.100808	0.120970	0.072582	0.029033	0.008710	0.00209	0.000418	0.000072	0.000011
2	0.075606	0.090727	0.054436	0.021775	0.006532	0.001568	0.000314	0.000054	0.000008
3	0.037803	0.045364	0.027218	0.010887	0.003266	0.000784	0.000157	0.000027	0.000004
4	0.014176	0.017011	0.010207	0.004083	0.001225	0.000294	0.000059	0.000010	0.000002
5	0.004253	0.005103	0.003062	0.001225	0.000367	0.000088	0.000018	0.000003	0
6	0.001063	0.001276	0.000766	0.000306	0.000092	0.000022	0.000004	0.000001	0
7	0.000228	0.000273	0.000164	0.000066	0.000020	0.000005	0.000001	0	0
8	0.000043	0.000051	0.000031	0.000012	0.000004	0.000001	0	0	0

The next table shows the winner according to each combination of goals, where T represents a tie.

Winner Combinations for Both Teams
Goals Team A	Goals Team B
	0	1	2	3	4	5	6	7	8
0	T	B	B	B	B	B	B	B	B
1	A	T	B	B	B	B	B	B	B
2	A	A	T	B	B	B	B	B	B
3	A	A	A	T	B	B	B	B	B
4	A	A	A	A	T	B	B	B	B
5	A	A	A	A	A	T	B	B	B
6	A	A	A	A	A	A	T	B	B
7	A	A	A	A	A	A	A	T	B
8	A	A	A	A	A	A	A	A	T

Finally, you can use the sumif function in Excel to add the corresponding cells for all three possible outcomes of the bet. In this case the probabilities are:

A wins = 44.14%
B wins = 30.37%
Tie = 25.48%

Appendix C in Sharp Sports Betting by Stanford Wong gives the win/lose/tie probabilities for bets like this. For this case he lists 44%, 30%, and 25%. If anyone knows a simple formula for this kind of problem, I’m all ears.

Follow Up: I received an e-mail from Bob P., who always keeps me on my toes when it comes to math. Here is what he wrote.

Looked up the distrib of the difference between 2 uncorrelated Poissons. It’s a Skellam (new to me).

Anyway, the question can then be posed as P(Z=0), P(Z>0), and P(Z<0) where Z is a Skellam with parameters 1.5 and 1.2.

If you haven’t already done it, you’ll be pleased to know

P(Tie) = P(Z=0) = .254817

P(A beats B) = P(Z>0) = .441465

P(B beats A) = P(Z<0) = 1 - .254817 - .441465 = .303718

almost exactly your answers.

The Wikipedia entry for a Skellam mentioned Bessel functions , which is about the point in calculus where I get scared to go further. So, I’m going to take Bob’s word on this one.

I think these are good bets, but not for the reason you state. A sports team can, in theory, lose endlessly. One reason these are good bets anyway is they are going to be on big underdogs the vast majority of the time, and underdogs are generally stronger bets than favorites. For another reason, the square bettors are generally betting the other way in these situations, creating value your way.

The big question to ask yourself with a prop like this is what is the probability that a given pick will end in a win, loss, or push. From my section on betting the NFL, we can see that 2.8% of games fall right on the line. Let’s just say 3%, to keep it simple. Let’s call p the probability of a win, given that the bet was resolved. For a purely random picker, p would obviously be 50%. It is easy to improve upon that by only picking underdogs. As my previously mentioned page shows, over 25 seasons flat betting underdogs would have resulted in a win rate of 51.5%. It is also easy to improve upon that a bit more by cherry picking the softest lines against the market in general. Between those two, I think it isn’t hard to get to 52%. So I’m going to take it on faith that these guys can at least get as far as 52%.

So, assuming 52% of resolved bets win, the overall probabilities are:

Win: 50.44%
Draw: 3.00%
Loss: 46.56%

Using basic statistics, it is easy to see that the expected win per pick, laying -110, is -0.0078. The standard deviation per pick is 1.0333. The expected win over 70 picks is -0.5432, and the standard deviation is 70^1/2×1.0333 = 8.6452. A win of 8.5 units is 9.0432 units above expectations, or 9.0432/8.6452=1.0460 standard deviations to the right of expectations on the Gaussian Curve. I think we can ignore the adjustment for a discrete distribution because of the pushes, and some games not being -110/-110, will result in a fairly smooth curve down the a factor of 0.05 units.

So, the probability of any one player finishing more than 1.046 standard deviations above expectations is 14.77%. That figure can be found in any table of the Gaussian curve, or with the formula =1-normsdist(1.046) in Excel. The probability of all six players finishing under 1.046 is (1-0.1477)⁶=38.31%. Thus, the probability of at least one player finishing above 1.046 standard deviations up is 61.69%. That makes the over look like a solid bet laying -110. I show it is fair at -161.

The following table shows the probability of the over 8.5 winning given various values of p. Perhaps the person setting the prop was assuming a value closer to 51% for p.

NFL Handicapping Prop
Prob. Correct Pick	Prob. Over Wins
50.0%	41.16%
50.5%	46.18%
51.0%	51.33%
51.5%	56.53%
52.0%	61.69%
52.5%	66.72%
53.0%	71.52%

For the benefit of other readers, this refers to a Nov 15, 2009 game in which the Patriots were up by six points over the Colts. There was 1:57 left in the fourth quarter, it was fourth down and about 1.5 yards, and the ball was on the Patriots’ 28 yard line. Patriots’ coach Bill Belichick made the now controversial decision to seal the victory and go for the first down on fourth and short, rather than kick the can down the road and punt.

There is a good column about this in the Las Vegas Review Journal . It quotes professional gambler and fellow actuary Steve Fezzik as saying that the odds favored going for the first down. I agree completely. In general, I think that other coaches punt too often, and too afraid of taking risks. To argue my position I asked fellow math head and sports bettor Joel B., who is much better than me at analyzing football odds mid-game. He offered the following odds:

• Patriot’s probability of making the first down: 60%
• Patriot’s probability of winning, given they make the first down: 100%
• Patriot’s probability of winning, given they miss the first down: 50%
• Patriot’s probability of winning, given they punt: 75%

So, the probability of winning by going for the first down is 60%×100% + 40%× 50% = 60% + 20% = 80%. That is greater than the 75% by punting.

The Monday morning quarterbacks can vilify Belichick all they want, but I applaud his decision. He shouldn’t be judged by the outcome of the game, but by whether or not the odds favored what he did at the time. I strongly feel that they did. A week later in the Ravens/Steelers game the Ravens went for it on 4th and 5, and made it. Although it was a different kind of situation, I’ve yet to hear anybody second-guess that decision.

In the interests of fairness, I am providing a link to an article taking the opposite point of view, titled Belichick’s fourth-and-reckless by Bill Simmons at ESPN.com

Good question. Straight up, the advantage per wager is 0.55×(10/11) - 0.45 = 0.05. As a parlay, the advantage is (0.55)²×((21/11)²-1)-(1-(0.55)²) = 10.25%. So, it would seem the parlay is the way to go to maximize advantage.

However, the variance is greater as a parlay. If you are following the Kelly Criterion, then you will have to protect your bankroll for the parlay with a smaller wager. In this example, the optimal Kelly bet straight up is 5.48% of bankroll if the two games overlap, 5.50% if you first game ends before you bet on the second game, and 3.88% for the parlay. Multiplying the wager times the advantage, we get 0.00275 straight up (based on a 5.50% advantage) and 0.00397 for the parlay. Thus, the parlay results in the greater profit.

I considered the general case for this kind of question, looking also at 3-team and 4-team parlays and money line wagers. Assuming a small advantage for all bets, as a rule of thumb, if the probability of each event winning is less than 33%, then you should bet straight up. If each probability is between 33% and 52%, then you should do a 2-team parlay. If each probability is between 52% and 64%, then you should do a 3-team parlay. If each probability is greater than 64%, then you should do a 4-team parlay. If you are doing straight up wagers, then you are about equally well off doing 2-team or 3-team parlays, again assuming you have an advantage to begin with.

I should stress that if you are a recreational gambler going against a house edge (what sports bettor will admit to that?), then betting straight up minimizes the house advantage.

From the 2000 to 2009 NFL seasons, not including the 2009 post-season, there were 567 failed 2-point conversion attempts, and 318 successful attempts. Going by that, the probability of success is 318/(567+318) = 35.9%.

That picture was taken at a Leroy’s franchise, which is a small and conservative sports book. They tend to put out the spread first, and later the total and money lines.

It is also unusual to offer money lines on significantly lopsided games, for example the Eastern Michigan vs. Central Michigan with a 23-point spread. It is difficult to set accurate money lines with such games, and the risk of unbalanced action is greater. The best casino for betting money lines on games with big spreads is the Hilton. Finally, if the point spread is very small, like 1 or 1.5, then most sports books don’t bother to post a money line at all, because betting against the spread would likely result in the same outcome.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

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.

1. 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.
   
   
2. Assume the favorite kicks 2 field goals.
3. Asssume the underdog kicks 1 field goal.
4. Subtract the field goal points from each. In the Super Bowl example, this would leave Colts = 25 touchdown points, Saints = 23 touchdown points.
5. Divide touchdown points by 7, to get estimated touchdowns. c=3.57 TD, s=3.29 TD
6. Round the estimated touchdowns to the nearest integer. c=4, s=3.
7. 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 NFL 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 question was raised and discussed in the forum of my companion site Wizard of Vegas .

According to an outstanding article at advancednflstats.com , the historical probability of success is 26%. However, with most onside kicks, the other team is expecting it, which lowers the probability of success. For surprise onside kicks, the probability of success is 60%! The thrust of the article is that smart strategy would be to do surprise onside kicks more often. I agree; it was a brilliant move by the Saints.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

To help answer this question, I graphed average points scored by the week number in the NFL, based on every season from 1983 to 2009. The following graph shows the results.

As you can see, the line goes up and down a lot. The thin black line is a least-squared best fit line, which shows a general trend upward. So as the seaons progresses, and temperatures drop, average points scored increases slightly, but that could easily be random variation.

That is about as far as I can take it. For a general opinion about how weather affects sports betting in general, I turned to my friend Jason Been, who is an expert on the topic. Here is what he said:

In most cases, wind is the predominant aspect of weather that most affects a game; but, it is not the only one. In baseball and other outdoor sports, shadows can have an equal affect, especially during early- and late-season baseball afternoon games. Rain or snow isn’t as big of a factor as most people think in football, as it generally affects both the offense and defense equally. An example would be a defensive back against a wide receiver. The rain and snow would slow them down equally, thus giving neither side an edge. Wind can simply eliminate a passing game in football along with kicking. I have seen games where a passing team has been forced to run the ball almost every play due to a strong crosswind. It doesn’t happen often, but on occasion the wind is the ultimate decider of a game.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

Based on the 2000-2009 NLF seasons, the answer is 57%.

I show after a 21 point loss or more the team will cover the spread 51.66% of the time. However, that is well within the margin of error. The following table shows the outcomes against the spread of the last game, according to the win or loss by the same team the previous game. The results never stray far from 50% and are always within a standard deviation of it. Basically, I find no statistical correlation between the win/loss against the spread and how many points the team won or lost by the previous game.

Win, Loss, or Tie Against Spread According to Margin of Victory or Defeat the Previous Game
Previous Game Outcome	Win Against Spread	Loss Against Spread	Tie Against Spread	Win Ratio	Standard Deviation
Win by 21 or more	233	247	17	48.54%	2.28%
Win by 14 to 20	235	219	11	51.76%	2.35%
Win by 10 to 13	188	180	8	51.09%	2.61%
Win by 7 to 9	198	181	12	52.24%	2.57%
Win by 4 to 6	164	170	12	49.10%	2.74%
Win by 3	202	212	14	48.79%	2.46%
Loss by 2 to win by 2	184	188	14	49.46%	2.59%
Loss by 3	209	207	12	50.24%	2.45%
Loss by 4 to 6	174	163	9	51.63%	2.72%
Loss by 7 to 9	187	195	9	48.95%	2.56%
Loss by 10 to 13	173	189	14	47.79%	2.63%
Loss by 14 to 20	220	232	15	48.67%	2.35%
Loss by 21 or more	249	233	15	51.66%	2.28%

Table based on every NFL game from week 1 of the 2000 season to week 4 of the 2010 season.

Let’s look at the October 25, 2010 Monday Night Football game as an example. The European odds are posted as:

New York Giants 2.750
Dallas Cowboys 1.513

Both figures represent what you will get back for one unit wagered if you win, including your original wager. When the decimal odds are greater than or equal to 2, then the translation is easy: just subtract one, and then multiply by 100. If the odds are less than 2, then (1) subtract 1, (2) take the inverse, and (3) multiply by -100.

For those of you who prefer a formula, if the decimal odds pay x, here is the calculation for the equivalent American odds:

If x>=2: 100*(x-1)
If x<2: -100/(x-1)

In the example above, the lines in the American format are:

New York Giants: 100*(2.750-1) = +175
Dallas Cowboys: -100/(1.513-1) = -195

You can also automatically convert all the lines by selecting "American Odds" in the pulldown menu in the upper left of Pinnacle’s web site, above the logo.

Outstanding bet, like taking candy from a baby. It’s like getting 2 to 1 on a coin flip. Even better, since you would win on a tie.

At the time of this writing, 95 games have been played out of 256 in the regular season. Using the binomial distribution, I show your probability of winning is 99.87%. A fair settlement price would be $998.02.

This question was raised and discussed in the forum of my companion site Wizard of Vegas .

I would lay 10 to 1 you’ll get paid, although they may have a few words with you first. The reason I have some doubt is a sports book can rescind a wager that was obviously made in error, with permission of the Gaming Control Board.

"A book may not unilaterally rescind any wager without the prior written approval of the chairman." -- Nevada Regulation 22.115 (PDF).

While this right exists, based on anecdotal evidence, it is very seldom invoked. Let me know what happens.

p.s. I was notified later by the questioner that he was paid without much trouble.

I like creative bets like this. It can be found at the Boyd casinos, as well as the Palms, El Cortez, and South Point. To answer the question I looked at another set of bets on the specific player to score the first touchdown. Those odds are shown in the second column of the table below. For the sake of simplicity, I’m ignoring the field at 5-1, and no touchdown scored at 100-1. I then converted those wins to the &lquo;fair probability” in the third column, meaning the probability of winning you would need for the bet to be exactly fair. These probabilities are inflated, due to depressing what each win pays, which is why the sum is 166%. The “adjusted probability” in the fourth column shows the fair odds divided by 1.660842, so that the total probability is 100%. The fifth column shows the number of Scrabble points in each player’s name. The “expected Scrabble points” in the sixth column is the product of the probability and Scrabble points. The lower right cell shows the mean Scrabble points is 14.18521.

Going by the mean, the over looks to be the right side. Going player by player, the probability of 11 or more Scrabble points is 0.641894, which converts to a fair line of -179. So that makes the over at -115 an outstanding bet. Laying 115, the player has a 20% advantage on the over.

Unfortunately, by the time I went to bet it the line had moved to -180.

Scrabble Points in First Player to Score a Touchdown in Super Bowl
Name	Posted Odds	Fair Probability	Adjusted Probability	Total Scrabble Points	Expected Scrabble Points
Mendenhall	4	0.200000	0.120421	20	2.408416
Jennings	4.5	0.181818	0.109473	22	2.408416
Starks	5	0.166667	0.100351	10	1.003507
Wallace	7	0.125000	0.075263	15	1.128945
Ward	8	0.111111	0.066900	8	0.535204
Rodgers	8	0.111111	0.066900	10	0.669005
Nelson	8	0.111111	0.066900	9	0.602104
Miller	10	0.090909	0.054737	11	0.602104
Driver	10	0.090909	0.054737	11	0.602104
Jones	12	0.076923	0.046316	15	0.694735
Roethlisberger	12	0.076923	0.046316	22	1.018945
Sanders	15	0.062500	0.037632	9	0.338684
Brown	18	0.052632	0.031690	12	0.380276
Redman	18	0.052632	0.031690	11	0.348587
Quarless	20	0.047619	0.028672	19	0.544761
Kuhn	25	0.038462	0.023158	12	0.277894
Jackson	30	0.032258	0.019423	24	0.466145
Moore	30	0.032258	0.019423	8	0.155382
Totals		1.660842	1.000000		14.185214

Unfortunately, by the time I went back to the casino to bet it the line had moved to -180.

P.S. Hours before the game I got in a bet at -170. Unfortunately, it lost. The first player to score a touchdown was Jordy Nelson. Nelson has 9 Scrabble points.

In the 267 games played, the favorite beat the spread 128 times, the underdog won 133 times, and 6 games landed on a push. Of bets resolved, the underdog won 51.0%.

What is more interesting is the overs and unders. The over won 148, the under 114, and 5 landed on a push. Of bets resolved, the over won 56.5% of the time. The probability of the under winning 114 games or less out of 262 resolved is 2.1%.

The probability of at least one safety per game is 5.77%, based on historical experience.

The expected number of safeties per game would be -ln(1-0.0577) = 0.0594.

The expected number per quarter per team would be 0.0594/8 = 0.0074.

The probability of exactly two safeties by the same team in a single quarter would be e^-0.0074×0.0074²/fact(2) = 1 in 36,505.

In an NFL season there are 267 games, and 267×8=2,136 team quarters. So, according to my estimate, this will occur on average once every 36,505/2,136 = 17.1 years.

This should be considered as just a rough guess. There are factors to the game that I'm not taking into account, in the interest of simplicity.