**By ETFan**

(From *Blackjack Forum* XXIV #2, Spring 2005)

© 2005 Blackjack Forum
First of all, it's great to see *Blackjack Forum* back in business, and wonderful to see the greatest blackjack writer of all time "back in the saddle again"!

Next, I need to say I don't have David McDowell's book. I was too cheap to buy it without more review, and believe it or not, Arnold, a few months after it was out I was warned off it by a math wiz/tracker. But I can't comment on the book, or how fairly you've stated his arguments. [Ed. Note: ETFan purchased *Blackjack Ace Prediction* shortly after this post and wrote the article titled "Convexing Calculations," a link for which can be found at the left of this article.]

But here's another way to do it ...

The assumption underlying Arnold's and David's calculations is the linear assumption. For example, it's assumed adding 0.2 to the probability of getting an ace will have twice as much effect as adding 0.1. This is a good approximation of reality when N is large, which may have originated with Thorp, and runs all through Griffin (eg. the tables in Chapter 6 of ToB).

If the linear assumption is good, then there *must* be a solution for P, U and H in the following equation:

P(PA)*P + P(UA)*U + P(HA)*H + C = ev(P)/bet

where P(PA) = Probability Player gets a given Ace, P(UA) = Probability the Ace lands on the Upcard, and P(HA) = Probability the Ace lands on the Holecard.

We can solve this linear equation in three unknowns by putting three more constraints on it, from data in books or sims, and solving the resulting 3 linear equations in 3 unknowns. Here's one approach (I'll assume 4 decks, because I have the figures handy, and the number of decks wasn't mentioned. The numbers are from Braun):

If an ace is removed from the shoe, we have:

P(PA)=P(UA)=P(HA)=0, therefore C = normal exp off the top + EOR(A) = -.532% -.150% = -.682%

If the player gets the ace in one of the two

first cards, we have: 1*P + C = 51.1%, or P = .511 - .00682 = 0.518

If the ace lands on the upcard, we have:

1*U + C = -34.4%, or U = .344 - .00682 = -0.337

Lastly, for normal distribution off the top, we have:

(2/208)*P + (1/208)*U + (1/208)*H +C = normal exp off the top

or: (2/208)*51.8% + (1/208)*(-33.7%) + (1/208)*H = -EOR(A)

or: 2*51.8% - 33.7% + H = -208*(-.150%)

[Note the accuracy of EOR(A) is critical]

or: H = -.387

Yeilding our linear equation:

.518*P(PA) - .337*P(UA) - .387*P(HA) - .007 = ev(P)/bet

To satisfy our curiousity, we can calculate the player's expectation when the dealer gets an ace in the hole:

- .387*(1) - .007 = -39.4%

Caveats:

1) The linear assumption is, of course, debatable.

2) The EOR(A) and off the top figures are from table 2-2 in Thorp's <I>The Mathematics of Gambling</I>. He credits the table to Braun. The EOR(A) figure is Braun's -.598 divided by four, since the "small quantity" of cards deleted (p.13) is <I>apparantly</I> four aces. Again, you need to be very accurate with your EOR(A) if you rely on this system to get your H coefficient.

Now, this may look all needlessly complicated, but the point is, once you get this equation, you can quickly solve hypos like McDowell/Snyder's.

The only problem with using the equation directly for McDowell's hypo is we aren't given P(PA) etc. directly. Instead of the probability of getting the one tracked ace in a given position, we are given the inflated probability that *any* ace will be the player's first card (0.13). It's a little tricky to convert, so rather than bore you with more calculations, I'll just give them to you in a table. I'm assuming here that the first card dealt to the dealer is the upcard, (though I have seen it the other way once or twice).

Probabilities for four decks:

Pl's C1 Pl's C2 Up Card Hole Card Another place

tracked ace 0.062031 0.004252 0.062031 0.004252 0.004252 x 204

another ace 0.067969 0.072156 0.067969 0.072156 0.072156 x 204

another card 0.870000 0.923592 0.870000 0.923592 0.923592 x 204 You can verify these must be correct by adding up the rows and columns, by noting that (tracked ace) + (another ace) = 0.13 as required, by noting the distribution for Player's C1 is the same as the Up Card, as required, noting that the distribution for Player's C2 is the same as the Hole Card or any other place, and checking that (another card)/(another ace) = 192/15 in all columns. *Has* to be right.;-)

So our equation becomes:

.518x(.062031+.004252) - .337x(.062031) - .387x(.004252) - .007 = 0.004785 =~ 0.48%

Since my assumptions are slightly different from Arnold's (51.1% vs. 51%, -.532% vs. -.5%), and since any error in EOR(A) gets magnified by a factor of 208, I don't expect accuracy to the hundredths of a percent. So I think this figure agrees with Arnold's 0.45%. However, I *have* verified tenths of a percent with simulations, so 4.2% is definitely off.

Now, I want to say the assumptions under the McDowell/Snyder hypo are very weak! One would certainly hope you could do better than a 13% chance of catching *any* ace, with an equal chance to the dealer! One fellow I know claims there is an 80% chance a tracked ace will fall within the first four cards following a key, with one common shuffle.

I also agree with MathBoy's comment about how easy it is to see the roses and miss the thorns. At least a dozen times over the years I've become very excited about some new AP technique, only to write a sim and be disabused. The one or two times an idea has panned out, well ... I'm keeping that to myself for now.

ETFan ♠

For accurate information on advanced professional gambling strategies, written by pros who actually use these strategies in casinos, see *Advanced Tactics in Casino Advantage Play*
by Abram Alexander, as well as The Blackjack Shuffle Tracker's Cookbook
by Arnold Snyder.

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