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Although players can get a significant edge from blackjack ace sequencing, also called ace tracking or ace location, you need to know how to do it right. David McDowell's Blackjack Ace Prediction gets the ace sequencing math and methods wrong. Players cannot get an edge using the blackjack ace location or prediction methods provided in this book.
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(From Blackjack Forum XXIV #2, Spring 2005)
© 2005 Blackjack Forum

I purchased my copy of David McDowell’s Blackjack Ace Prediction at the Gambler’s Book Shop in Las Vegas last month (January 2005). No errata sheet was provided and there is no instruction anywhere in the book to go to Dalton’s Web site for the errata sheet.

I have recently been alerted to the errata sheet for McDowell’s Blackjack Ace Prediction on Dalton’s Web site. Unfortunately, as ETFan has pointed out on other Web sites, the error in the page 114 formula that I addressed in my initial article is still there in the errata sheet. In fact, the situation with the errata sheet "correction" is even worse.

This math is not rocket science. It is simple, straightforward, and logical. It is based on the fundamental principles of blackjack math. David McDowell does not understand the mathematical logic of blackjack. So, I will try to explain this math. Dalton seems to think that McDowell has already addressed my concern. He has not. Please just try to follow...

According to the Errata sheet, McDowell says he made an error in his formula on page 114 (which is the formula I found fault with in my initial review of this book). The errata sheet "corrects" the formula by stating that instead of the dealer and player getting 6 additional aces each, they will be sharing a total of 6 extra aces, with only three extra aces each per 100 aces bet. The errata sheet "corrects" the player’s win rate with this assumption from 4.2% (as per the book) to 1.3% (per the errata sheet formula).

Again, McDowell is making a HUGE error. The same error with slightly different numbers.

Please follow the logic.

The errata sheet now assumes that the player and dealer will each get only 3 extra aces per 100 aces bet. Dalton’s errata sheet assumes, as per BJAP, that the ace is worth 51% when it lands on the player’s hand, and -34% when it lands on the dealer’s hand. The errata sheet now uses McDowell’s assumption that there are 7 "random" aces going to both the player and dealer per 100 aces bet, and 3 "predicted" aces going to each, for a total of 10 aces each per 100 aces bet. The errata sheet presents the math for this as follows:

E(X) = (+0.51 x 0.10) + (-0.34 x 0.10) + (-0.005 x 0.80)

= +0.051 - 0.034 - 0.004

= +.0130

Or, a 1.3% advantage when betting on an ace.

I do not know how to explain the math on this any more clearly than I did in my initial BJF article, but I will try. The MAJOR error McDowell is making is that he is assuming that the 80 hands where neither the player nor dealer get a first-card Ace are played with a house edge of only 0.50%.

This is a very important concept. Please pay attention to this: If neither the player nor the dealer are dealt a first-card Ace on any of these 80 hands, then the house edge is no longer 0.50% on these hands for the basic strategy player. That -0.50% basic strategy expectation assumes that the player and dealer will each get the random number of aces as a first card that would be expected to be dealt from a full six-deck shoe. As soon as you remove all of the random first-card aces from this set of 80 hands per 100 (which McDowell has done), then we must reconsider what the cost of these 80 hands with no first-card aces is.

Let me give an example that should clarify this error. The average number of "random" aces dealt as a first card to both the player and dealer is 1/13, or 7.7 first-card aces each per 100 hands. Exactly 1 out of every 13 cards in a six-deck shoe is an ace, so if we’re simply playing random hands, with no predictions, we would expect 7.7 aces per 100 hands since 7.7 is 1/13 of 100. The dealer would also get 7.7 aces per 100 hands.

Is this clear?

Here is how you find the error in McDowell’s formula:

McDowell estimates that the average number of aces per 100 hands is only 7, not 7.7. This is a minor error, but the math on it is so simple, I do not know why he did not just say 7.7 instead of 7. We are attempting to estimate an advantage here, so why not use the most precise number?

McDowell then says (as per the Errata sheet now) that the player and the dealer will each receive an extra 3 aces, for a total of 10 aces each. In his formula, he provides this number as a percentage (0.10) of the 100 aces bet on. He then figures out the expectation if the player and dealer each get 10 aces (0.10) per 100, and no aces are dealt on the other 80 (0.80) of the 100 aces bet on.

His mistake is in assuming that these 80 hands are played against the standard house edge of -0.50%. Note in the formula the last item: (-0.005 x 0.80), which is the notation for his mistaken calculation that 80% of the hands will be played at -0.005 (or -0.50%).

Here’s how you know the formula is WRONG.

What if we use this formula to calculate the player’s advantage when the player and dealer are each dealt exactly 7.7 aces per 100 aces bet (the exact number of aces that would occur at random with no prediction)? Here is what the formula gives us:

E(X) = (+0.51 x 0.077) + (-0.34 x 0.077) + (-0.005 x 0.846)

= +0.03927 - 0.02618 - 0.00423

= +.00886

If this were correct, it would mean that a basic strategy player, getting just the random number of aces (7.7 per 100), and with a dealer also getting 7.7 random aces, would be winning at almost a +0.9% win rate. But we’re assuming that in this game the house has a 0.50% advantage over the player. So, why isn’t our result on this completely random basic strategy game -0.50%?

The reason is because of McDowell’s incorrect assumption about the 84.6% of the hands (100% -7.7% -7.7%) that are played when neither the player nor the dealer is dealt a first-card Ace. He is making the mistake of assuming that on these other 84.6 hands per 100, the player’s expectation is the same as in a 6-deck game where those random first-card aces are included.

The house edge on hands which do not contain an ace is not the same as the edge on hands where a random ace occurs. It’s only when you combine the house edge on the ace hands with the house edge on the non-ace hands that you get an overall player expectation of -0.50%.

If we separate out our basic strategy edge on the 7.7 hands per 100 when the player is dealt a random ace (51% player expectation) and the 7.7 hands per 100 when the dealer is dealt a random ace (-34% player expectation), and we want to know the overall expectation for the basic strategy player on all 100 hands played, then we have to figure out what the player expectation is when neither the player nor the dealer is dealt an ace. This is a fairly simple calculation if we use Griffin’s numbers from Theory of Blackjack. If neither the player nor dealer is dealt an ace as a first card, the house edge is approximately 2.13%. Let’s try this number in McDowell’s expectation formula:

E(X) = (+0.51 x 0.077) + (-0.34 x 0.077) + (-0.0213 x 0.846)

= +0.03927 - 0.02618 - 0.01802

= -0.00493

Which shows the basic strategy player’s expectation to be about -0.49%, just about exactly what we expect.

You do not need a computer simulation to know this. This is long-established, fundamental blackjack math.

So, if the dealer and player are each getting 10 aces per 100 aces bet, as per McDowell’s errata sheet, then the correct math is:

E(X) = (+0.51 x 0.010) + (-0.34 x 0.010) + (-0.0213 x 0.80)

= +0.051 - 0.034 - 0.01704

= -0.00004

which is a -0.004% expectation for the player.

In other words, if the dealer and player each get 10 aces per 100 bet, instead of 7.7 aces per 100 bet, then the house no longer has a 0.50% advantage over the player, but only a 0.004% advantage.

Unfortunately, 10 aces per 100 is not quite enough aces to get the player an advantage over the house, if the dealer is also getting 10 aces per 100. The house still has a slight edge. Does this sound impossible to you? Consider what 10 aces per 100 means to the player who would normally get only 7.7 per 100.

This is an extra 2.3 aces (not 3, as McDowell claims) per 100 aces bet on. That means that for every 43 times that you bet on an ace, you will get one more ace than normal expectation. So, if you are betting on 4 aces per shoe (as per McDowell), you will get one extra ace every 11 shoes.

Unfortunately, the player advantage from getting this extra ace will be cut by the dealer also getting one extra ace every 11 shoes.

I do not know how to make the math any more clear than that.

If McDowell’s "correction" is right, and the dealer and player expect to get 10 aces each per 100 bet by the player, then there is no advantage whatsoever to the player using McDowell’s system.

McDowell then goes on in Dalton’s errata sheet to show what happens "if the dealer can be prevented from getting the ace." This section is entirely without explanation. This is not acceptable. It is extremely important to know how the dealer is prevented from getting the ace, and an expectation cannot be calculated without specifying this information. Since the first formula shows the dealer and player splitting the 6 extra aces, then in order to prevent the dealer from getting his share of extra aces, either the player must spread to more hands in order to get the aces that would have gone to the dealer, or the aces must go to a hand played by someone else, possibly a civilian. In the first example, where the player spreads to two hands to get the ace on a big bet, we could give the player 13 aces, but we would have to show that the player is now placing 200 bets (not 100) in order to capture the dealer’s aces. In the second example, where a civilian hand is catching the dealer’s share of extra aces, then the formula should show the player is getting his 10 aces, and the dealer is getting only his 7.7 random aces.

(I want to make it clear that I am NOT saying that by spreading to two hands you can prevent the dealer getting extra aces. I am using two hands in this example only to illustrate the principle, not to provide a practical guideline.)

Unfortunately, neither of these methods are shown in the errata sheet formula. The player is shown getting all 6 extra aces, but only playing a single hand (100 hands total). The dealer is shown getting only 7 (not 7.7) aces. And the other 80 hands where no first-card aces are dealt are still being shown with a house edge of only 0.50%, not 2.13%.

I don’t even know how to correct this formula unless McDowell can describe the mechanism by which "the dealer can be prevented from getting the ace."

If you want to put the numbers into the formula yourself, go ahead. I’ll spare you the math. If the player is spreading to two hands, placing 200 bets total to get all the extra aces, he actually does have a win rate: +0.12%.

On the other hand, if there is a convenient civilian at the table, ready to catch the dealer’s aces at no charge to the player, so the player is placing only 100 bets but still getting just a total of 10 aces to the dealer’s 7.7, the player also has a positive advantage: +0.07%.

So, in this example, spreading to two hands is the better strategy, assuming that you prefer to make +0.12% instead of +0.07%. But neither strategy is of any interest whatsoever to the player who is trying to make money by ace tracking.

Here is the problem: David McDowell didn't go through the hard work necessary to test the theories he presented in Blackjack Ace Prediction. As a result, he never learned how to track or sequence aces. He never came to understand the techniques, the problems, or fundamental blackjack math, a problem we find with many non-players who profess to be "blackjack experts." ♠

Spring 2005 Blackjack Forum

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