Ask the Wizard #368
The following image is of a "money wheel" game I have seen at charity casinos. There are six bets, one to six. After betting is closed, the dealer spins the wheel. The player wins according to how many times the number he bet on appears on the slice where the wheel stops, as follows:
- 3 times pays 3 to 1
- 2 times pays 2 to 1
- 1 time pays 1 to 1
- 0 times loses
What is the house edge of this game?
The following table shows how often any given single bet will have 0 to 3 matches, what it pays, the probability, and contribution to the return. The lower right cell shows a house edge of 21.43%.
I know a dice influencer who claims to have recorded the following rolls in craps. The roller claims he objective was to hit the inside numbers (4, 5, 6, 8, 9, and 10). Can you analyze his results?
First, let's add a column to the table to show the expected tally of each total, assuming a fully random roll.
Craps Data with Expectations
You didn't ask me how to analyze the data, so I'll do it a few different ways.
A chi-squared test has a chi-squared statistic of 21.43009, with 10 degrees of freedom. The probability of data this skewed, or more, is 1.83%.
Looking at just the inside numbers, which you mentioned was the goal, the total achieved was 12,802, while the expected total would be 25,227 × (2/3) = 12613.5. This excess of inside numbers is 2.52 standard deviations above expectations. The probability of such an excess number, or more, is 0.59%.
I couldn't help but notice the lack of sevens. In 25,227 rolls, the expected sevens are 25,227 × (1/6) = 4204.5. The shooter had 3,963. That is 4.08 standard deviations away from expectations. The probability of such a shortfall is 0.0000225, or one in 44,392.
However, I must say it's usually easy to look at historic data and find something fishy about it. Then again, avoiding sevens is an intrinsic goal to the dice influencer.
The scientific way to test whether the shooter can influence the dice is to state the goal BEFORE data is collected.
In Las Vegas, what is the lowest house edge bet in a table game that does not require skill or a secondary bet?
Since you rule out games of skill, we can't look at blackjack, which would otherwise be the correct answer, assuming the best of rules.
Since you rule out bets that require a secondary bet, we must rule out the odds bet in craps.
That said, I claim the answer to be the buy bet on the 2 or 12 in Crapless Craps, where the casino charges the 5% commission on wins only. It is my understanding the MGM/Mirage casinos as well as Resorts World charge the commission on wins only. That said, the win is 5.95 to 1 and the probability of winning is 1/7. That calculates to a house edge of 0.714%. I'm not aware of anything better that meets your requirements.
At Resorts World, they allow "even money" in blackjack, including tables that pay 6 to 5 on a winning blackjack. How much does this lower the house edge?
For my answer, I shall assume six decks of cards.
If the dealer pays 6 to 5 only on a winning blackjack, then "even money" is usually not allowed. However, I'll take your word for it that they do offer it at Resorts World.
At a win of 6 to 5, a player blackjack against an ace is worth 83% of the bet amount. So, to get 100% is a great deal. This situation will happen with probability 0.352%. Overall, the is worth 0.00352 × (1 - 0.83) = 0.0006 to the player. In other words, it decreases the house edge by 0.06%.
I must remind my readers that if blackjack pays 3 to 2, the player should decline it. In that case, a blackjack against an ace is worth 1.037 times the bet amount, so to accept only one unit would be a bad decision.