100% Welcome Bonus
$11000 Welcome Bonus
$3000 Welcome Bonus
Ask the Wizard #182
Mark from Mississauga
I hope you’re happy; I spent two days running simulations to answer this one. For the benefit of others, let me briefly summarize both arguments. The Cut-Card Effect states the house edge is less in a continuously shuffled game, as opposed to a cut-card game, all other things being equal. The "Floating Advantage" concept states that for the card counter the odds get better the deeper into the deck or shoe the dealer gets. According to Stanford Wong, "...after we have counted the deck down to where n decks are left, the edge with a count of zero is about the same as if we had started with n decks." - Blackjack Attack, third edition, page 71. So, for example, the house edge in a six-deck with one deck left, at a count of zero, has about the same house edge as a single-deck game with the same rules.
Unfortunately the Floating Advantage does not benefit non-counters. While they will benefit, unknowingly, at close to neutral true counts, they will do worse at extremely high and low counts. According to Schlesinger, "Seems that, at the one-deck level, extremely high counts produce less edge than expected for the basic strategist (many pushes) and the extreme negative counts were found to be even more unfavorable than previously thought (doubles, splits, and stands tend to be disastrous)." - Blackjack Attack, third edition, page 70.
As I understand it, the reason the counter benefits from the Floating Advantage, even if he may not know about it, is that he bets more when the Floating Advantage works to his advantage, and less when it doesn’t. For the non-counter, who doesn’t know when the Floating Advantage is strongest, the pros and cons exactly cancel each other out.
In conclusion, both the Cut-Card Effect and Floating Advantage are distinct topics and do not contradict each other. To compare them is to compare apples and oranges. For more information please read chapter 6 in Blackjack Attack by Don Schlesinger.
John from Raleigh, NC
You’re welcome. That was a difficult game to explain. The following table shows the house edge both ways, and the difference, assuming both player and dealer use the same house way.
House Edge in Pai Gow
|Event||5% Comm.||4% Comm.||Difference|
Bryan from Mill Valley
Yes, Bodog does indeed pay 9 to 1 on the tie. Assuming eight decks, that lowers the house edge from 14.360% to 4.844%.
Scott from Chicago
For recreational gambling, my rule is to get up when you’re not having fun any longer.
Heather from Petaluma
The probability of six sixes is (1/6)6 = 1 in 46656. The probability of rolling 1,2,3,4,5,6 with six dice is 6!/66 = 1 in 64.8
- Ace, Jack Suited = 25 - 1
- 2 Aces = 10 - 1
- 3 or 4 Total = 3 - 1
- 9 or 10 Total = 2 - 1
- 11 or 12 Total = 1 - 1
- Any Blackjack = 3 - 2
Aces always count as 1 and 10’s and faces count as 10. What is the house advantage? If I keep an Aces and Fives count is there a positive count where the possible remaining aces make the bet a positive proposition? Would counting remaining aces divided by remaining decks be better?
Stan from Beaverton, Oregon
You didn’t tell me the number of decks, but assuming six the house edge is 5.66%. Here is the return table.
Field of Gold — Six Decks
|3 or 4 total||3||1428||0.029434||0.088301|
|9 or 10 total||2||4884||0.100668||0.201336|
|Any other blackjack||1.5||2160||0.044521||0.066782|
|11 to 12 total||1||6612||0.136285||0.136285|
Just eyeballing it, I would say aces would be the best card to track, betting into an ace-rich deck. My advice would be to count aces as −12 and all other cards as +1.
Jesse from Scottsdale
That probability would be 52/combin(260,5) = 5/9525431552 = 1 in 1,905,086,310.
Kathy from Hitchcock
It will be difficult finding $5 blackjack on the Strip on a weekend. You’ll probably have to settle for a low-roller casino like the Riviera, Sahara, Frontier, or Circus Circus. It will be a lot easier downtown. Let It Ride is slowly fading away, but if you find it the minimum unit is usually $5.
Kevin from Philadelphia
Here in Nevada there is a state law that an electronic representation of playing cards must have the same probabilities as if a human being were dealing the game. To do business in Nevada a game maker must abide by this law in every machine it places anywhere in the world. So if they use major U.S. brands like IGT or Bally I’m sure the games are fair. However if the games are low-budget imports then I can make no assurances. As with a live game, check the rules before you play. Most importantly, avoid games that pay even money on blackjack.
Mike from Olympia
So there are 30 black, 30 red, and 4 green numbers. That would make the probability of black 30/64, red 30/64 and green 4/64. If the probability of an event is p then the fair odds are (1-p)/p to 1. So fair odds for any red would be (34/64)/(30/64) = 34 to 30 = 17 to 15. Same for black. The fair odds on green are (60/64)/(4/64) = 60 to 4 = 15 to 1. For a specific number the fair odds are (63/64)/(1/64) to 63 to 1.
I suggest paying 1 to 1 on red and black, 14 to 1 on green, and 60 to 1 on any individual number. One formula for the house edge is (t-a)/(t+1), where t is the true odds, and a is the actual odds. In this case the house edge on the red or black bet is (63-60)/(63+1) = 3/64 = 4.69%. On the green bet the house edge is (15-14)/(15+1) = 1/16 = 6.25%. On individual numbers the house edge is (63-60)/(63+1) = 3/64 = 4.69%.
Tom from Aurora, CO
For the beneit of other readers, my blackjack appendix 10 explains, the house edge in a five-deck game is 0.028% less if a continuous shuffler is used, as opposed to a hand shuffle. The difference between five decks and two decks, all other rules being equal, is 0.18%. So the two-deck game without a shuffler would be much better. Let’s compare a 5-deck continuous shuffler game to a 4-deck hand shuffled game. As my blackjack calculator show difference in house edge between four decks and five decks is 0.0329%. So the benefit of a continuous shuffler is worth less than removing a single deck.