# Ask the Wizard #288

Who offers the best futures odds in Vegas?

Anonymous

According to futures bets on the 2015 Super Bowl, here is the average house edge at various Vegas sports book groups.

### House Edge in Sports Futures

Sports Book | House Edge |
---|---|

CG Technology | 21.90% |

William Hill | 26.63% |

Wynn | 27.96% |

Caesars | 35.49% |

Stations/El Cortez | 38.33% |

Golden Nugget | 39.75% |

MGM | 40.88% |

Boyd/Coast | 49.35% |

TI | 57.93% |

To calculate the average house edge on any set of futures bets, please use my Sports Futures Calculator.

I hear that Jerry's Nugget no longer offers the juicy odds for NFL teasers. Is this true?

Anonymous

Sadly, it is. Jerry's Nugget was the last place to offer the liberal odds of -110 on a 2-leg 6-point NFL teaser, +180 for three legs, and +300 for four legs. By doing Wong teasers (crossing the 3- and 7-point margins of victory), this was a solid advantage play.

You can find all the current parlay and teasers odds around Vegas at my sports book survey at Wizard of Vegas.com.

For more on football teasers in general, please see my page on Teaser Bets in the NFL.

The Hustling the House show on the Discovery Channel had a long segment on the best way to turn $30 into $1,000. It featured Andy Bloch saying, "If you have $30 in your pocket and you want to turn it into $1,000, then roulette is your only game." Andy went on to explain why betting the whole $30 on a single number was better than parlaying even money bets five times.

Is Andy correct in that the best way to turn $30 into $1,000 is to put the whole $30 on a single number in roulette?

Anonymous

No, he is not correct. The probability of Andy's single bet strategy is 1/38 = 2.6316%.

After much trial and error, I devised my "Hail Mary" roulette strategy, which will increase the odds of turning $30 into $1,000 to 2.8074%.

Wizard's "Hail Mary"strategy for roulette:

This strategy assumes that bets must be in increments of $1. In all bet calculations, round down.

Let:

b = Your bankroll

g = Your goal

- If 2*b >=g, then bet (g-b) on any even money bet.
- Otherwise, if 3*b >=g, then bet (g-b)/2 on any column.
- Otherwise, if 6*b >=g, then bet (g-b)/5 on any six line (six numbers).
- Otherwise, if 9*b >=g, then bet (g-b)/8 on any corner (four numbers).
- Otherwise, if 12*b >=g, then bet (g-b)/11 on any street (three numbers).
- Otherwise, if 18*b >=g, then bet (g-b)/17 on any split (two numbers).
- Otherwise, bet (g-b)/35 on any single number.

In other words, always try to reach the goal, with just one bet, if you can, without exceeding the goal. If there are multiple ways to accomplish this, then go with the one with the greatest probability of winning.

What about other games, you might ask? According to the Discovery Channel voice-over guy, "Everyone agrees that roulette is the best get rich quick scheme in the casino." Well, I don't. Even limiting ourselves to common games and rules, I find craps to be better. In particular, betting the don't pass and laying the odds.

Following my Hail Mary strategy for craps (explained below), the probability of turning $30 into $1,000 is 2.9244%. This assumes the player may lay 6x odds, regardless of the point (which is the case when 3x-4x-5x odds are allowed taking the odds). This probability of success is 0.117% higher than my Hail Mary strategy for roulette, and 0.2928% higher than the Andy Bloch strategy.

Andy might argue that my argument above relies on an assumption of a minimum bet of $1, which is hard to find in Vegas on a live dealer game. Expecting somebody might say that, I ran through both games under an assumption of a $5 minimum and betting in increments of $5. In that case, the probability of success using my Hail Mary strategy is 2.753% in roulette and is 2.891% in craps. In both cases, greater than the 2.632% under the Andy Bloch strategy.

In all fairness, the Discovery Channel would have never put the insane rant above on the air and was surely looking for something simple that the masses would understand. Andy was surely giving them something they wanted to hear. The basic premise of his advice is that if you want to reach a certain goal, then a hit-and-run strategy is much better than letting the house edge grind you down with multiple bets. That is definitely true and something I've been preaching for 17 years.

Wizard's "Hail Mary" strategy for craps.

This strategy assumes that bets must be in increments of $1 and wins will be rounded down to the nearest dollar. In calculating bets, never bet so much that you overshoot the goal. Also, never make a bet amount that will cause you to get rounded down.

Let:

b = Your bankroll

g = Your goal

- Bet max($1,min(b/7,(g-b)/6)) on the don't pass.
- If a point is rolled, and you have enough for a full odds bet, then lay the full odds. Otherwise, lay whatever you can.

So, I hope Andy and the Discovery Channel are happy. I've spent days running simulations to prove them wrong.

This question was raised and discussed on my forum at Wizard of Vegas.

Suppose you were offered the opportunity to play a coin flipping game. If the first flip is heads, then you get back $2 and the game is over. Otherwise, you flip again. If the second flip is heads, then you get back $4. If the second flip is also tails, you keep flipping until you do get heads. For each flip, the prize doubles. In other words, you get back 2^n, where n is the number of flips (including the ending flip on heads). How much would you pay to play this game? I hear the mathematical answer is an infinite amount of money, but this doesn't make sense, because you must win a finite amount of money at some point.

Omaha

This is known as the Saint Petersburg Paradox.

It is indeed true that the expected win of the game is ∞, while at the same time the probability is that the coin will eventually land on tails, leading to a finite amount of money. The calculation of the expected win is:

Expected win = pr(1 flip)×2 +
pr(2 flips)×4 +
pr(3 flips)×8 +
pr(4 flips)×16 +
pr(5 flips)×32 +
pr(6 flips)×64 + ... =

(1/2)^{1} × 2^{1} +
(1/2)^{2} × 2^{2} +
(1/2)^{3} × 2^{3} +
(1/2)^{4} × 2^{4} +
(1/2)^{5} × 2^{5} +
(1/2)^{6} × 2^{6} + ...

= ((1/2)*(2/1)) ^{1} +
((1/2)*(2/1)) ^{2} +
((1/2)*(2/1)) ^{3} +
((1/2)*(2/1)) ^{4} +
((1/2)*(2/1)) ^{5} +
((1/2)*(2/1)) ^{6} + ...

= 1^{1} + 1^{2} +
1^{3} +
1^{4} +
1^{5} +
1^{6} + ...

= 1 + 1 + 1 + 1 + 1 + 1 + ...
= ∞

Where this is paradoxical is the player must win a finite amount of money, but the expected win is infinite. How can that be?

This is probably not a very satisfying answer, but there are lots of paradoxes when it comes to ∞. This may cause me to get some angry emails, but what lets me sleep at night, despite such infinity paradoxes, is that I believe that ∞ is a mathematical or philosophical concept that is unproven to exist in the real physical universe. This concept or theory of infinity carries with it built-in paradoxes.

For those who disagree with this, please tell me anything that is proven to have infinite quantity or measurement. Please don't say a black hole has infinite density unless you have evidence of its size.

To answer the initial question of how much should one pay to play this game, we should keep in mind that happiness is not proportional to the amount of money. Personally, I was taught in economics classes, and believe, that the utility, or happiness, from money is proportional to the logarithm of the amount of the money.
Under this assumption, if you increase or decrease any two people's wealth by the same percentage, other than an initial wealth of zero, then they both experience the same change in happiness. For example, if Jim's wealth suddenly increases from $1,000 to $1,100 and John's wealth suddenly increases from $10,000,000 to $11,000,000 they both experience the same increase in happiness, because in both cases their wealth increased by 10%.
Assuming that the happiness from money is indeed proportional to the log of the amount, then the following table shows the most somebody should be willing to pay according to his wealth before paying to play.

### Indifference Amount to Play

Wealth | Indifference Amount |
---|---|

$ 10 | $ 4.97 |

$ 100 | $ 7.79 |

$ 1,000 | $ 10.96 |

$ 10,000 | $ 14.26 |

$ 100,000 | $ 17.78 |

$ 1,000,000 | $ 20.88 |

$ 10,000,000 | $ 24.19 |

$ 100,000,000 | $ 27.51 |

$ 1,000,000,000 | $ 30.84 |

As you can see, under realistic conditions, the amount you should pay is much less than $∞. For example, if your wealth is one million dollars, then you should be indifferent to playing at a cost of $20.88.

This question is raised and discussed on my forum at Wizard of Vegas.