On your web page, you state that continuous shuffling machines reduce the house edge in Blackjack. Although I have no doubt that you have proven this mathematically, this result appears to be in conflict with the "Floating Advantage" concept described in Chapter 6 of Don Schlesinger's Blackjack Attack. Basically, the "Floating Advantage" concept appears to indicate that the house edge decreases as more decks are removed from play (regardless of the true count). For me, it is difficult to reconcile the fact that the house edge can decrease when cards are immediately replenished (when CSMs are used) and also when cards are utilized and not replenished (when shoes are used). Can you comment on this observation?

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.

Thanks for teaching me Pai Gow Tiles. I had a most profitable time on my last Vegas trip. The question is that at many casinos the limit is \$25. When winning, the house vig at some casinos is \$1 on \$25 bet equaling 4% not the 5% usually as they don’t have quarters. Does this affect the house’s edge if you do not vary the amount bet or stick to odd amounts bet?

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 Player 0.023896 0.020811 -0.003085 Banker 0.007377 0.004207 -0.00317

Some of the online casinos such as Bodog pay 9 to 1 for the tie bet in baccarat. What is the house edge for the tie bet with the 9 to 1 payout?

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%.

Do you have a good rule for getting up from the blackjack table for a non-card counter playing basic strategy? Obviously we’d all like to quit while we’re ahead, but how far ahead. And is there a time to quit when you’re behind?

Scott from Chicago

For recreational gambling, my rule is to get up when you’re not having fun any longer.

Two questions, please: 1) What is the probability of rolling 6,6,6,6,6,6 @ one time, with (6)6-sided die? 2)What is the probability of rolling 1,2,3,4,5,6 @ one time with (6) 6-sided die? Thanks! It’s killing me!

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

Spirit Mountain Casino in Grand Ronde Oregon added a side bet in the last 24 hours called "Field Gold 21." It resolves before the rest of the hand begins and concerns the first two cards dealt to a player. The side bet can be between 1 and 25 dollars. The pay table follows.

• 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

 Event Pays Permutations Probability Return Ace/jack suited 25 144 0.002968 0.074202 Two aces 10 276 0.005689 0.056888 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 All other -1 33012 0.680435 -0.680435 Total 48516 1 -0.056641

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.

I am a pit supervisor at a local casino and recently had a dealer deal two players two seven of clubs each and give himeself the last seven of clubs as his upcard on a five-deck shoe. What are the odds of five of the same card coming out of a five-deck shoe in order?

Jesse from Scottsdale

That probability would be 52/combin(260,5) = 5/9525431552 = 1 in 1,905,086,310.

I am going to Las Vegas next weekend and like to play \$5 blackjack & Let It Ride on the tables. Will I be able to find an open \$5 table on a weekend or should I plan on bringing more funds than usual with me? If \$5 tables are few and far between, where might I find them?

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.

Pennsylvania recently legalized "slot parlors." They are advertising electronic blackjack and baccarat. Do you know if these electronic versions of table games have the same odds and payout as real live games based on truly random deals? Or are they set with specific payout percentage, a la slot machines?

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.

I recently acquired a carnival wheel that belonged to my great uncle, it’s about a hundred years old and I’m trying to develop a game around it. It’s numbered 1-60 in random order and it alternates black and red with a green star every fifteenth mark, could you help me outline how much the payoffs should be for each spin?

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%.

Playing blackjack on a continuous shuffling 5-deck system, are the odds of winning different than playing the dealer with 1 deck or 2 decks?

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.