Ask the Wizard! (No. 159)

According to my sources in Washington state many casinos waive the 5% commission if the player makes the Fortune side bet. The most common pay table for the Fortune bet in Washington I'm told is 2/3/4/5/25/50/150/400/1000/2000/8000, which has a house edge of 7.83% less 1.07% for each additional player at the table. So the expected loss of a $5 Fortune bet is 39.14 cents less 5.34 cents for each additional player. The following table shows the breakeven point between making and not make the Fortune bet according to the number of players at the table, including yourself.

• 1 player: $27.36
• 2 players: $23.63
• 3 players: $19.90
• 4 players: $16.17
• 5 players: $12.44
• 6 players: $8.71
• 6 players: $4.98

For example, if there are four players (including yourself) you should make the Fortune bet if your pai gow poker bet is $17 or more, and not if it is $16 or less.

As I understand the rules of tennis the winner of a set is the first to win six games, and by a margin of at least two games, except a 6-6 tie will result in a single tie-breaker game. The following table shows the probability of winning a set, given the probability of winning a game.

The formula for any probability of winning a game p, and losing q, is 1*p⁶ + 6*p⁶*q + 21*p⁶*q² + 56*p⁶*q³ + 126*p⁶*q⁴ + 252*p⁷*q⁵ + 504*p⁷*q⁶

I had a number of people ask me to expand on my answer. The solution requires basic matrix algebra.

Start by defining x as the answer, or the average number of flips until the disparity between heads and tails is 3.

Let y be the expected number of flips from a point where one side is up by one flip.

Let z be the expected number of flips from a point where one side is up by two flips.

After the first flip one side will be in the majority by one flip. So x=1+y.

When either side is one flip ahead another flip will result in either the initial tied state, or one side being up by two flips. Both outcomes are equally likely. So y=1+0.5*x + 0.5*z

When either side is two flips ahead another flip will result in either one side being up by one flip, or the end of the game. Again, both outcomes are equally likely. So z=1+0.5*y

So we have three equations and three unknowns:

(1) X= 1+y

(2) Y = 1+ 0.5x + 0.5z

(3) Z = 1+ 0.5y

To solve lets first get rid of the decimals by multiplying the last two equations by 2.

(1) X= 1+y

(2) 2Y = 2+ x + z

(3) 2Z = 2+ y

Let's substitute 1+y, from (1) for x in (2).

2Y = 2 + 1 + y + z

(4) y = 3 + z

No substitute 3+z for y in (3)

2z = 2 + 3 + z

z = 5

Now substitute 5 for z in (4) to get

(5) y = 3+ 5 = 8

No substitute y = 8 in (1) to get

(6) x = 9

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.