Ask the Wizard: Sports - Other Games

1-(1+(d/100))*(1-(100/f))/(2+(d/100)-(100/f))

The amount you must bet to get back one unit is 1/[(d/100))*(1-(100/f))/(2+(d/100)-(100/f))].

For example with money lines of +130 and -150 the house edge on both bets would be 3.3613% and the expected return on a bet of 1.034783 units would be 1 unit.

At a land casino I would assume the fair set of money lines to be +140 and -140 in this example, resulting in a house edge of 2.78% on the favorite and 4.17% on the dog. All other things being equal this would suggest that Pinnacle is a good place to bet on underdogs.

100*p - 160*(1-p) = 0
260p = 160
p = 160/260 = 8/13 = 61.54%.

So the expected return on a $145 bet at a -145 line would be (8/13)*100 + (5/13)*-145 = 75/13 = $5.77. So the player advantage would be $5.77/$145 = 3.98%.

Let’s define t as the true money line with no house edge and a as the actual money line. Following are the formulas for the player’s expected return:

A is negative, t is negative: (100*(t-a) / (a*(100-t))
A is positive, t is positive: (a-t)/(100+t)
A is positive, t is negative: (a*t + 10000)/((t-100)*100)

So in your case your expected return is 100*(-160 -(-145))/(-145*(100-(-160))) = 3.98%.

To answer your question, to beat the 6-point teaser you would need an overall probability of winning of greater than 50%, so per pick the probability of covering the tease would need to be the square root of 0.5, which equals 70.71%. To beat the 6.5-point teaser the probability per bet would be 52.38%, or 0.5238^0.5=72.37% per pick.

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

A reader wrote in to express his comments about the assumptions I made in my answer. Here is a link to his commentary.

Interesting question. I looked up a few bits of data on the internet, and it looks like the ball will go further on a humid day than on a dry day, everything else being equal. The two factors that are most relevant are: 1) the air density; and 2) air viscosity.

1) Air Density

Contrary to popular belief, humid air is lighter than dry air. This is because the water molecules take up the same space but weigh less than the O2/N2 mixture. Lighter air results in less buoyant force on the football because the football is displacing less mass. However, the density of dry air at 20 C and 700 kPa(*) is 8.33 kg/m3, and with 42.1% relative humidity at the same temperature and pressure the density is 8.32 kg/m3 according to the sources listed, a difference of about 1/10th of 1%. So this isn’t going to effect the distance much.

(*) - 700 kPa is a high pressure, but it’s the only data I could find. However, in engineering terms it’s not much different from normal atmospheric pressure so I believe the properties listed in the data will be applicable to the situation at normal atmospheric pressure (101.325 kPa).

2) Air viscosity

Viscosity is the force that contributes to skin drag on the football. A lower viscosity will contribute less to drag, resulting in a longer flight. For dry air at 20 C and 700 kPa, the dynamic viscosity is 18.3 Pa*s, while for the air with 42.1% humidity the viscosity is only 17.8 Pa*s. This is a difference of about 3%, again not much but a little more significant than the effect of air density. However, humid air will still contribute to a slightly longer football flight.

To see if this makes sense in the real world, I found a golf website that has some data on golfball flight distance in dry and humid conditions:

As you can see, in humid air the golfball goes further, but only by a yard or two at most. So humid air definitely results in a longer projectile (golfball or football) flight, but the effect is very slight.

Andrew N

Data culled from:

wipos.p.lodz.pl/HighTech/example1.html (data on humid air at 20 C and 700 kPa)
physics.holsoft.nl/physics/ocmain.htm (calculators for humid air properties)

Wizard’s comments: To add to the first point, Boyle’s Law says that given the same temperature, the volume of gas is inversely proportional to the pressure. So given the same temperature and pressure the volume of gas will be constant, in other words the same number of molecules per unit area. The atomic weight of oxygen is 16, nitrogen is 14, and hydrogen is 2. So a water molecule (H2O) has an atomic weight of 18, while O2 and N2 are much heavier at 32 and 28 respectively. So when it is humid the lighter water molecules push the heavier O2 and N2 molecules out of the way, causing for less resistance for the football to cut through the air.

Major League Baseball

Indeed, the American League West has a big advantage, and the National League Central has a big disadvantage, all other things being equal. I do not know of any compensating rules. Nor do I know the reason for this imbalance. Before 1998 there were only two divisions. In 1998, Major League Baseball added two new teams, the Tampa Bay Devil Rays and Arizona Diamondbacks. They also increased the number of divisions from four to six, and added the Wild Card rule. However, why they didn’t balance the leagues, I have no idea. The best solution to this inequity, in my opinion, would be to move the Houston Astros to the American League West. Some may say Houston is not west enough, but the Texas Rangers are also in that division.

I posted the answer and solution to the probability question at my companion site, mathproblems.info , as problem number 200.

p.s. Since I posted this column, one reader wrote that the reason for the imbalance was to keep the number of teams in each league an even number. This allows every team to play on a given day, and keep the play within the division. However, this I don’t buy as an excuse. In 2008 the regular season consisted of 162 games per team, played over 185 days (not counting the all-star game day, and a day on each side). So, each team played 0.8757 games per day. Of those 162 games, 18 are played against teams in the opposite division, and 144 in the same division. I suggest that with balanced divisions of 15 teams each, on any given day 12 teams play within their own league. Over 185 days, that will accomplish 185 × (12/15) = 148 games. In the other 37 days, schedule 14 interleague games, for a total of 162 games. So, the only change will be a reduction in the number of interleague games per team from 18 to 14. It seems to me, most fans oppose interleague games to begin with, including me.

p.p.s. Another reader wrote to say that my system would not accommodate the baseball traditions of keeping every team playing on Saturdays and Sundays, and designating interleague play to only certain times of the seasons. Okay, fair points. However, if tradition is so important in baseball, why introduce interleague games at all? Personally, I value fairness over tradition. Put me in charge of baseball scheduling, and I’ll no only balance the leagues, but keep every team playing on weekends. However, it would come at the expense of the days off being clumped together. Maybe the easier thing to do would be to add two more teams. My hometown of Las Vegas will be the first to volunteer to be one of them.

1. Take out the track cut from the total pool of place bets. For the sake of argument, let’s use the usual American take-out of 17%.
2. Divide the rest into three pools.
3. Pay the winners on each dog a pro-rata basis according to the size of the pool and the amount bet on the dog. If the amount bet on the dog exceeds its share of the pool, then bettors will get a refund.

Let’s look at an example. Suppose $100,000 is bet on place bets in an 8-dog race. Assume bets on the winning dogs total $5,000 on dog A, $10,000 on dog B, and $15,000 on dog C. First, the 17% take-out would be deducted, leaving $83,000. That would be divided by 3, leaving $27,667 to pay the winners of each dog. Winning bets on dog A would be paid $27,667/$5,000 = 5.53 for 1, before any rounding (I’m not sure how they round down under). Likewise, winning bets on dog B would be paid 27667/10000=2.77 for 1 and winning bets on dog C would be paid 27667/15000 = 1.84 for 1.

The bettor in this case exploited the rules by betting such huge amounts that he pretty much controlled the odds. For the sake of simplicity, let’s assume he was the only bettor. The article said he bet $350,000 on the two favorites and $5,000 on each additional dog. With six underdogs (pun intended), that resulted in a total pool of 2?$350,000 + 6?$5,000 = $730,000. After the take-out and split, there was $201,997 to the winners of each dog. The rule about getting at least a push resulted in bets on the two favorites being refunded, because $350,000 > $201,997. However, the share of the pool on the third dog was huge compared to wagers on it. The winning odds would have been 201,997/5000 = 40.4 to 1. So, the profit on the third dog was $5,000 ? 39.4 = $197,000. He actually only won $170,000, probably because of other bets on the third dog.

This technique would not work in the U.S., by the way, because in the U.S., we deduct the original wagers made on each winning dog from the total show pool and then add them back in after dividing by 3. This deduction would have caused the pools on the two favorites to be negative, resulting in just small winnings of the minimum $0.10 per $2 bet.