## Wizard Recommends

# Ask the Wizard #159

Michael from Marysville

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.

Mike from Perth

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.

### Probabilities in Tennis

Probability Game Win |
Probability Set Win |

0.05 | 0.000003 |

0.1 | 0.000189 |

0.15 | 0.001899 |

0.2 | 0.009117 |

0.25 | 0.028853 |

0.3 | 0.06958 |

0.35 | 0.138203 |

0.4 | 0.23687 |

0.45 | 0.361085 |

0.5 | 0.5 |

0.55 | 0.638915 |

0.6 | 0.76313 |

0.65 | 0.861797 |

0.7 | 0.93042 |

0.75 | 0.971147 |

0.8 | 0.990883 |

0.85 | 0.998101 |

0.9 | 0.999811 |

0.95 | 0.999997 |

The formula for any probability of winning a game p, and losing q, is 1*p^{6} + 6*p^{6}*q + 21*p^{6}*q^{2} + 56*p^{6}*q^{3} + 126*p^{6}*q^{4} + 252*p^{7}*q^{5} + 504*p^{7}*q^{6}

Scott from Pittsburgh

Thanks for your vote of confidence. According to my count I only received four such questions in the first 134 columns. However starting in August 2005 they flooded in, partly because my webmaster, Michael Bluejay got us the #1 rank in Google for a search on "Is my boyfriend cheating on me". We have since dropped to #2. To answer your question, yes, there is a connection. I take a cold calculating look at both casino games and life. When I give advice in either venue it is based on what I think will make the asker happier in the long run.

Wayne from Honolulu

The Casino Royale, which offers 100x odds. For complete playing conditions in Vegas please see my new Vegas craps directory.

Al from Calgary

Yes. Assuming the commission is paid only on a win then it should be applied to the bet amount, not the win amount.

Amanda from Woonsocket

This is not a good time to be turning a small problem into a big one. Your evidence that there is anything going on between him and this other girl is pretty light. If you want him to be around for you and the baby then be nicer to him. All this nagging about an address in his wallet is only going to drive him away, probably to Puerto Rico.

Jim from Atlantic City, NJ

My webmaster, Michael Bluejay, loves the dives too. He never misses a chance to visit the Western, although he laments, "That place has really gone uphill." Unlike the other low-end casinos of downtown, mainly the El Cortez and the Gold Spike, the air is not quite as thick with cigarette smoke at the Western. Closer to your friends I truly like both downtown Henderson casinos, the Rainbow Club and the El Dorado. Both are old fashioned and cater the low rolling locals. However both are clearly worlds above the Western. Here is Bluejay’s rundown of skanky Vegas hotels.

Mahalo

You are subject to tax for any gambling winnings. However table games players are basically on the honor system. An exception that a W2G form is generated if a win is 300 for 1 or more odds and is over $600. That is usually only an issue with progressive jackpots. Also, if there is a cash transaction of $10,000 or over the casino is obligated to fill out a CTR, which stands for Cash Transaction Report. Yet these are nothing to worry about, and I think many big bettors are overly paranoid about them.

David from Chicago

That is what can happen if you rush a relationship. It takes a while to truly get to know somebody. Next time take it slow and don’t fall in love until you truly know the other person well. An iPod, even crammed with love songs, is only going to get you so many points. My advice is to agree to a cooling off period of a month or two where you don’t see each other. Then after it is over, and it is mutually agreeable, try it again, but be more patient.

Rick from Covington, LA

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

"Anonymous" .

Physics is not my strong subject so I asked two physics experts, my father and Andrew N., this question. Both agree the ball will go further if the humidity is high. Here is how Andrew N explained why.

*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). Link no wonder works.

physics.holsoft.nl/physics/ocmain.htm (calculators for humid air properties, Link no wonder works)

**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.