WOO logo

Ask The Wizard #309

I encountered specific rules on a blackjack machine located at Binion's in Las Vegas: it utilized a single deck, the dealer must stand on soft 17, blackjacks pay out at 1:1, both players and dealers are required to stand after reaching six cards, and doubling or splitting is not permitted. Can you tell me what the house edge is with these poor rules?

Charles from Winnipeg

Ouch! According to Norm Wattenberger, who designed the Casino Vérité blackjack software, the house advantage with these conditions stands at 4.30%. This figure does not take into consideration the rule regarding standing after six cards, which I believe balances out for both players and the house. Sadly, other bar-top games tend to follow this same unsatisfactory set of regulations.

This is a topic that's frequently discussed in my forum at Wizard of Vegas .

What would be the solution to the dice scenario in Ask the Wizard column #179 if the players alternate rolling the dice and only the player actively rolling can progress based on the value rolled?

anonymous

The initial question posed in column #179 was: When two dice are repeatedly rolled until one of two specific outcomes occurs, which outcome is more likely to happen first:

  • Rolling a total of six and eight in any order, allowing for duplicates.
  • A total of seven is rolled twice.

Your variation is that a single roll cannot benefit both players. Instead, they take turns, and only the player rolling can make use of their roll.

The outcome relies on who goes first. If the player needing a six and eight rolls first, they have a winning probability of 57.487294%. Conversely, if the first roll belongs to the player needing two sevens, the probability of the player needing six and eight winning drops to 52.671614%. I solved this using a straightforward Markov Chain process.

This is a topic that's frequently discussed in my forum at Wizard of Vegas .

That game Flip It At the Rio, is it feasible to keep track? Do you have any tips on which bet appears to be the most vulnerable?

anonymous

The bets on red and black seem to be the weakest. I'd suggest conducting a basic red/black count as follows:

  • Let C represent count (where red cards are marked as +1 and black cards as -1)
  • J = count of jokers left in the shoe
  • if J-C < 0 then bet on black
  • If J+C < 0 then bet on red

For example, if the cards played after a certain time are:

red = 100
black = 75
jokers = 10

The remaining cards would be:

red = 108
black =133
jokers = 14

The count would yield +25. The calculation for remaining jokers would then be - C = 14-25 = -9. Since this value is negative, it would be wiser to place a bet on black, as there are more favorable cards left (133) compared to unfavorable ones (122) in the black category.

Is there any discernible pattern in the arrangement of numbers on the roulette wheel?

anonymous

The expectation is for the wheel to be balanced. Essentially, if you were to average a section of consecutive numbers, the result should be around the average number of 18.5. I thought of a method to clarify how numbers are spread on a double-zero wheel, as follows.

  • Set the 00 at the top of the wheel and the 0 at the bottom.
  • The numbers situated at the top of the wheel are 13, 1, 00, 27, 10*, 25. This sequence should just be memorized.
  • All low odd numbers (ranging from 1 to 17) are placed on the left side, while the higher odd numbers (19 to 35) are on the right.
  • Starting at 1 and moving to 13, move four places counter-clockwise to find the next odd number, ensuring you don’t pass the zero.
  • As an exception to rule 4, avoid positioning the 19 four spaces counter-clockwise from the 17, as 19 is a high odd number and belongs on the right side. Instead, place 11, the only low odd number not previously covered under rule 4.
  • All even numbers are located directly across from the preceding odd numbers, 180 degrees apart.

I'm uncertain how to explain the arrangement of numbers on a single-zero wheel at the moment. However, if you position the zero at the top:

  • On the left, all low red numbers and high black numbers will be found.
  • On the right, you will see all high red numbers paired with low black numbers.

This topic is brought up and debated in my forum at Wizard of Vegas .

As many are aware, the 16-seed UMBC team managed to defeat the 1-seed University of Virginia during the NCAA basketball championship tournament. This has marked a historical first in the 34-year existence of March Madness under its current format. What implications does this hold for the likelihood of creating a flawless bracket?

anonymous

After modifying my model based on that match and the entirety of the 2018 season, I determined the chances of achieving a perfect bracket to stand at 1 in 42,743,890,552. Details of my strategy and the calculations behind that figure can be found on my webpage Probability of a Perfect Bracket .