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The index number of 16 against a 10 in most blackjack counting systems is zero. So if the deck were completely neutral you should stand, because you stand if the count is equal or exceeds the index number. Yet the basic strategy tables tell us to hit. This seems to be a contradiction.

"Anonymous" .

In all honesty this is an old question but I got a better answer from Chris F.. He correctly says the reason is that when the basic strategy charts are created they assume the player’s first two cards and the dealer’s up card have already been removed from the deck. A good example of this is that in single deck the correct play is to stand on 7,7 against a 10, because half of the sevens in the deck are already gone, and that is what you need to beat a dealer 20 with 3 cards.

In the case of 16 against a 10 the player’s hand is either composed of a 10 and 6 or a 9 and 7. Either way two cards that would bust the player by hitting have been removed. So the deck is slightly rich in small cards that will not bust the player, giving the player an incentive to hit. While this is true I was skeptical because in an infinite deck game the odds still favor hitting. However except in a few Internet casinos an infinite deck is just an abstraction. I was curious what would be the best play in an 8-deck game if the dealer just said without dealing a single card "You have a 16 and I have a 10, but don’t have a blackjack." Using the blackjack analyzer at gamblingtools.net (site no longer exists), I entered eight decks and then carefully depleted the deck of 1 of every card, except only no sixes, and 2 tens. Then I gave the dealer a 10 and myself a 10 and 6. So the player was playing this hand against a neutral deck with 31 of each card A-9 and 124 tens. Here is the expected values:

### 10+6 vs 10 — Eight Decks

Play Expected Value
Stand -0.5399
Hit -0.5399

Although the expected value numbers are the same the applet highlights standing as the better play, presumably because it is higher beyond four decimal places. It is the same if I remove the following: A,2,3,4,5,6,8,10,10,10 to simulate 9,7 vs 10, because the player is going against the exact same neutral shoe.

It just goes to show how powerful the effect of removal is, even when just three cards in an eight-deck game. Getting back the original question, a zero count reflects a totally neutral deck after the player’s two cards and dealer’s up card have been accounted for. So as I just showed going into a neutral deck the odds favor standing. The reason hitting is correct in an infinite deck is because there is no effect of removal. If you accidentally hit a 16 vs 10 in a neutral shoe, and got a low card, then the dealer would have a better chance of getting a 10 in the hole. This fact is reflected in the higher expected value for standing in an 8-deck game, but would not matter in an infinite deck. For the record, here are the expected values in an infinite deck game:

### 10+6 vs 10 — Infinite Decks

Play Expected Value
Stand -0.5404
Hit -0.5398

I just read your answer about baccarat as played in 007 movies, and I would like to let you know that in South America with a point of 5 the player can choose taking a hit or not. As this option should be made prior the bank showing his card, only a fool would take a hit, since in this situation there are 4 cards that favors the player and 5 that hurts him. Best regards from your loyal fan

Marcio

Thanks for your comments. I just watched the scene in question from For Your Eyes Only several more times and am still not sure what is going on. It doesn’t help that the dealer giving the running commentary is doing so in French. It also doesn’t help that the table is mostly plain, like a poker table, unlike an American table where you can tell a bet by its location.

We see Bond dealing the cards but an unseen dealer is paying players. Bond is apparently betting the opposite of what the only other bettor at the table is doing. In the first hand the other character turns over a 2-card natural 8, Bond turns over a 2-card 5, and Bond wins the hand. This would imply that the other player bet on the banker hand, and thus Bond on the player hand. In the second hand the other bettor increases his bet from half a million to on million, at the goading of his wife. After receiving his first two cards he requests a third. Bond turns over his two cards, revealing a face card and a 5, and gives the other bettor a third card. The other bettor’s cards are not turned over yet but he seems pleased with his hand. Then a third character, who just walked up, comments to Bond, "The odds favor standing pat." However Bond takes a card anyway, which is a 4, for a total of 9. The other player storms off without turning over his cards.

This is consistent with what you said, except Bond is acting last, or as the banker. I tend to think the American makers of the movie didn’t understand European baccarat rules and incorrectly gave the banker the free will take card a card, as opposed to the player. It certainly wouldn’t be the first time a gambling scene was depicted incorrectly in the movies. I have seen numerous card counting scenes in the movies and television, and yet to find anything close to being realistic.

I agree that if given the choice the odds favor standing on 5 as the player. Assuming the banker rules are the same either way then if the player stands on a 5 the following is the house edge per bet, based on an 8-deck game.

### Player Hits 5

 Bet House Edge Banker 0.79% Player 1.52% Tie 17.27%.

So if the player consistently hits on 5 the house edge goes up by 0.29% on the player bet. The player will get a 5, while the dealer does not have a natural, 9.86% of the time, for a cost per 5 of 2.94%.

I read about how your Ties Win Blackjack is on field trial in Laughlin, Nevada. What kind of permit is required and how much did it cost?

"Anonymous" .

A new game trial period permit was required. This is opposed to a "variation" permit, which is less expensive. For a new game the cost is \$3000, I had to fill out lots of forms, including an employment and residence history going back 20 years. The waiting time was six months, which was shorter than what I was expecting.

I read your article entitled Marketing New Casino Games, and I am a bit discouraged because I just invented a new game and I am actually thinking of marketing it. In your article you mentioned that new table games are rented out to casinos for about \$300 to \$500 per table per month. I thought that there is big money in this business if you're fortunate enough to invent a really good game. I was told that Derek Webb, the guy who invented 3-card poker made millions from the game. Is this not true?

"Anonymous" .

Truly top games like Three Card Poker can get up to \$1,500 to \$2,000 per month, from what I hear. I don't know exactly how much Webb made but whatever it was he had to spend a lot of it on lawyers fees defending the game. There is an article about Webb and Three Card Poker in the August 2004 issue of Playboy.

I was recently in Las Vegas, and the casinos’ methods for playing two pair in Pai Gow Poker vary somewhat significantly from your methodology. I was wondering whether your rules are devised to optimize the player’s chances because you know how the house will play, or whether your methodology is simply a better strategy than the house method. If the answer is the former, then if a player banks, the player should play the house way instead of your way? If it is the latter, then your method should always be used. However, if it is the latter, why don’t the casinos use your methodology?

"Anonymous" .

My two pair rule is optimized to play against the house way. However I think it is probably just any reasonable strategy. For example I would use it when banking against other players. The reason the casinos use a more complicated and less powerful rule is probably out of tradition. Whoever invented the game probably came up with that strategy rather arbitrarily and it since become a hard habit to break. Two other rules I find ridiculous are counting A2345 (known as "the wheel") as the second highest straight and bothering to state an exception in the house way that if the dealer has five aces with a pair of kings to play the pair of kings in the low hand. The probability of getting this hand is 1 in 25,690,513. In my estimation this hand may have come up about 100 times in the history of the game, but has probably never affected the outcome of a hand compared to the alternative of playing a full house in the high hand. Yet every single dealer to have dealt the game had to be bothered with learning the exception.

Hi - Thank you for your web site. I'm wondering if you can tell me what are the odds if you are dealt Q-Q that any of the remaining 8 people at the table would be dealt A-A, A-K, K-K, or A-Q? Thank you!

"Anonymous" .

For any given person the probability of having AA is combin(4,2)/combin(50,2) = 6/1,225 = 0.0049 because there are 6 ways to pick 2 aces out of 4, and 1225 ways to pick any 2 cards out of the 50 left in the deck. The probability is the same for a pair of kings. For A-K the probability is 4*4/1,225=0.0131, because there are 4 ways to get an ace and 4 ways to get a king. For A-Q the probability is 4*2/1225=0.0065, because there are 4 aces but only 2 queens left in the deck. So the probability any given player will have one of these hands is (6+6+16+8)/1225 = 0.0294. Now the next step is clearly not perfect because if one player doesn't have one of these hands the odds the next player does is a little bit higher. Forgetting this for the sake of simplicity the probability no player has one of these hands is (1-0.0294)8 = 78.77%. So the probability at least one player has one of these hands is 21.23%.

Imagine a island that is inhabited by 10 people, and the politics is such that each day an islander is chosen at random to be chief for exactly one day; after the day has elapsed another islander is chosen at random (so the same islander who was just chief has a 1/10 chance of being chief again). The question to be solved: on average, how many days would have to elapse before each islander would have been chief at least once?

"Anonymous" .

It will only take 1 day so that 1 person has served as chief. For the second day the probability of a new chief is 0.9. The expected number of days it will take to get a new chief, if the probability each day is 0.9 is 1/0.9 = 1.11. This is true for any probability: the expected number of trials until a success is 1/p. So after 2 people have served the probability of a new chief on the next day is 0.8. So the waiting period for a 3rd chief is 1/0.8 = 1.25 days. The answer is the sum of the waiting periods, which is 1/1 + 1/.9 + 1/.8 + ... + 1/.1 = 29.28968 days.

If I buy two quick-pick lottery tickets what is the probability I get the same number on both cards. Assume a 6/49 lottery.

"Anonymous" .

The probability of winning correctly picking 6 numbers out of 49 is 1 in combin(49,6) = 1 in 13,983,816. This is also the probability of your two tickets matching.