Ask the Wizard #385
I hear quadruple-zero roulette is now out there. What is the house advantage?
Yes, such a wheel is marketed by TCS John Huxley. I am not aware of any placements yet.
The formula for the house edge in standard roulette is z/(36+z), where z = number of zeroes. In this case, the house edge is 4/(36+4) = 4/40 = 1/10 = 10%.
At what value is x1/x a maximum?
The answer is e =~ 2.71828182845905.
There, the value of e^(1/e) =~ 1.44466786100977.
Here is my solution (PDF).
I have seen rolling-chip commissions as high as 2.4% in Asian casinos. What is the house edge at that commission?
As a reminder, the way these programs work is the player buys non-negotiable chips with cash. These are the use-until-you-lose type of chips, which are usually called "dead chips" in Macau. Wins are paid in cashable chips. After all the non-negotiable chips have been played through, the player will be given a commission, based on the original buy in. I assume the commission itself is paid in non-negotiable chips as well. The commission can also be paid up-front, for which the math is still the same.
To answer your question, let's review baccarat probabilities. Here is the probability of each outcome.
- Banker wins = 0.458597423
- Player wins = 0.446246609
- Tie wins = 0.095155968
Let's look at the Banker bet. The number of times the player must wager on the Banker bet, on average, before a loss is 1/0.446246609 = 2.240913385 bets.
The expected win the Banker bet is 0.95*0.458597423 - 0.446246609 = -0.01057900.
The expected cost to play off a non-negotiable chip is 0.01057900 × 2.240913385 = 0.02370675 units.
Assuming the player gets an extra 2.4%, the value of the chip is (1+0.024) × (1-0.02370675) = 0.02343104.
Overall, the house edge to the player is the cost to play less the expected value of the promotion. This is 0.02370675 - 0.02343104 = 0.00012304.
So, the house edge is 0.01%.
Using the same logic, the house edge on the Player bet with a 2.4% commission is 0.164089%.
The break-even commission on the Banker bet is 2.4282409%.
This question was asked and discussed in my forum at Wizard of Vegas.
In this YouTube video, David Blaine walks up to a double-zero roulette game and correctly predicts the color of the next five spins. How did he do it? Assuming he was randomly picking, how many times would have to do this before being successful, on average?
I think he did this over and over until it worked. I'm not sure if he used different players each time or those women were audience plants who pretended it never failed before.
The probability of winning each bet is 18/38. So, the probability of winning five in a row is (18/38)5 = 1889568/79235168 = 2.3848%.