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Turbo 21 (Laffman's Letter)

Laffman (another player on pogo.com) was kind enough to send me this letter pointing out some advanced math concepts about Turbo 21 (and a couple errors on my main page!). With his permission, I'm including the entire text of his letter here:

To: lemke@plainsboro.com
From: laffman-91@usa.net
Subject: More thoughts on Turbo 21
Date: Sun, 03 Dec 2000 15:36:40 -0800

Hi, K-man!

Since you were so helpful at Turbo 21, and your tips page was instrumental 
to my first attempts at an actual strategy, I thought you might be 
interested in some of my own thoughts on it.  I'm still not what you'd 
call proficient, but I did get hit the 500 level (17 21's) for the first 
time, so I'm getting better.

At any rate, when I went to bed last night, Turbo 21 kept playing in my 
head, as is often the case when I'm newly addicted to some game.  Using 
your analysis as a starting point, I started to consider some of the 
mathematics of the game.  I'm not a formal mathemetician, so I can't put 
things as a proof, but I think I've come up with some "rules" to add to 
what you already figured out and wrote down for the rest of us.  This is 
my first go-round at getting it down in writing, so please excuse the 
roughness.  I may jump around a bit.  I'm also not entirely consistent of 
my use of numerals versus spelled-out numbers, but I do try.

First, let me invent some definitions for the purpose of this 
explanation.  I'm going to call all two-card combinations whose sum is 11, 
"natural pairs".  With respect to each other, I'll call them 
"partners".  For instance, 7 is 4's partner, and vice versa, and together 
they form a natural pair.  Also, I'll call all 21's "blackjacks", 
regardless of the combination of cards which are used to actually reach 
that value.  A blackjack which can be broken down into a 10 and an 11 are 
"natural blackjacks".  A perfect score is 18 blackjacks, and all that 
follows applies to properties of perfect scores.

By considering the game in terms of 10's and 11's, I came up with 
variations on your principles, which say roughly the same thing, but I 
find a little easier to remember.  Your explanation of the "remaining 8" 
and possible combinations was very useful, but hard to remember.  In the 
haze of near sleep, it occurred to me that most of the time, all but one 
of the possible 18 blackjacks are natural blackjacks.  The remaining 
blackjack consists of a 9, and some combination of cards whose sum is 
12.  Look at your own chart, and you'll see this is always the case.

As you pointed out, there are 16 cards worth 10.  Each of those must be 
paired with 11's - either Aces or natural pairs - to achieve a perfect 
score.  Assuming we accomplish that, that leaves two blackjacks to come up 
with.  One will be a 9 and some 12-combination, as previously noted, and 
the other, most of the time, can be broken down into another natural 
blackjack.  For instance, 6-4-5-6 can be considered as (6+4) + (5+6).  Put 
another way, you'll have some 10-sum combination, plus a natural 
pair.  Now for the interesting stuff...

Looking at the 9 + 12 blackjack that we'll have to achieve, there are four 
ways to get that 12 sum in just two cards:

A: 6 + 6
B: 7 + 5
C: 8 + 4
D: 9 + 3

(Technically, we could also say 10 + 2, or Ace + Ace work, but those 
couldn't be used in a perfect game.)  What's interesting about these 
pairings is that if you look at the sum of their respective partners, it 
will always be 10.  Looking at just the partners of A-D, then, we get:

A: 5 + 5
B: 4 + 6
C: 3 + 7
D: 2 + 8

Remembering that there are 17 natural blackjacks in a perfect game, but 
only 16 cards worth 10, here's where the additional 10 must come from, to 
be used in constructing that 17th natural blackjack!  If we use a two-card 
12, then we can immediately deduce what the corresponding two-card 10 has 
to be for that perfect game, and vice versa.  This means that if we're 
fast enough with calculations and card-counting (which comes with 
practice), then it's not as important to have the cards fall just right 
and do all the fancy combinations with a "remaining 8" at the end of the 
game.  It still takes luck, to be sure, but not quite as much.

There's also some 3-card combinations worth considering, but the same 
principles end up applying.  Let's see the ways to make a 12 to go with our 9:

F: 2 + 2 + 8
G: 2 + 3 + 7
H: 2 + 4 + 6
I:  2 + 5 + 5

These four combinations, if you take a close look, end up being nothing 
more than natural blackjacks in disguise.  Put the 9 in the mix, and it 
will pair up with it's partner 2, and the remaining two cards form one of 
the 10's as seen in A-D.  So, let's look at the ways for making 3-card 
12's that don't include a 2:

J: 3 + 3 + 6
K: 3 + 4 + 5
L: 4 + 4 + 4

Each of these three combinations prove exceptions to our usual rule that 
17 of the 18 blackjacks in a perfect game have to be naturals.  Looking at 
the partners of these 3-card 12's, they end up making another blackjack, 
though not a natural one.  Here's how the parnters of J-L end up looking:

J: 8 + 8 + 5 = 21
K: 8 + 7 + 6 = 21
L: 7 + 7 + 7 = 21

 From this, we can conclude that if a 3-card combination is used to 
 provide the necessary 12 to go with the 9, then a perfect game will 
 consist of 16 natural blackjacks, the 9-12 blackjack, and one of the 
 3-card blackjacks listed in J-L.  Just like the 2-card 10's were 
 logically linked to corresponding 2-card 12's, these 3-card 12's are 
 logically linked to corresponding "unnatural" blackjacks.  If one occurs, 
 so must its corresponding other, in order for a perfect game to be achieved.

One other thing I noticed is that your rule about every Ace having to go 
with a 10-value single card isn't entirely accurate.  If you lay out the 
deck and arrange them into 18 blackjacks, you will find that you can swap 
the Ace with any 2-card 10 that was used, and you will still have 18 
blackjacks.  The key is that you don't have enough cards to swap out more 
than one Ace like this.  Also, it assumes that 17 of your blackjacks are 
naturals, because if you end up with a 3-card 12 and unnatural blackjack 
like J-L, then there is no 2-card 10 with which to swap the ace.  Note 
that if you do swap that Ace, then you end up with three 2-card blackjacks 
(A-10), and the remaining fifteen blackjacks are all 3-card 
combinations.  The rule about having at least one 4-card blackjack, then, 
also doesn't hold true if the ace is swapped.

All this leads me to the following revised rules to describe a perfect game:

1.  One 2 must be left over, and every other card must be used.
2.  All Aces must be used as hard 11's.
3.  At least three Aces must be paired with single, 10-value cards.
4.  If all 4 Aces are paired with single, 10-value cards, then exactly one 
of your blackjacks must contain 4 cards, and all those cards must be in 
the 2-9 range (i.e., no Aces or 10's).
5.  If one Ace is used in a 3-card blackjack, then exactly one of your 
blackjacks must be 8-8-5, 8-7-6, or 7-7-7.  It also means there can be no 
4-card blackjacks.

That's the basics.  Although it doesn't seem any easier to remember than 
your chart at first glance, I find it makes more sense to me at least, 
because instead of trying to remember specific card combinations, I have 
some easier-to-remember card relationships to consider, and with practice, 
expect to be able to do the calculation part faster and faster to generate 
the actual numbers I'll need to look for.

I'm not dedicated enough to go collecting data and figuring probabilities 
on all this, but I imagine that the above rules would lead to certain 
conclusions about what strategies are most effective.  As with your rules, 
they're easiest to follow when the cards fall nicely and leave you with a 
convenient "remaining 8" to sort out at the end, but there are probably 
also sound reasons for playing a certain sequence of cards a certain way 
if they come early.  Any time you have seven or eight cards falling in the 
2-9 range, there is probably a "proper" way to play them, whether that 
happens at the beginning, middle, or end of the game.  The trick, of 
course, is that you never know when (or if) that sequence is about to 
happen, so by the time it falls in the beginning or middle, you may have 
already misplayed them.  When they come at the end, you know more or less 
what's left and can play accordingly.

Here's a few more ruminations that I couldn't quite decide how to fit in, 
but might interest you...

There are four natural pairs: 6 & 5, 7 & 4, 8 & 3, and 9 & 2.  Each occurs 
four times, for a total of 16 natural pairs.  Among these, we know that we 
can't use one of the 9 & 2 pairs and still achieve a perfect game, leaving 
15 natural pairs.  Counting each of the Aces as an 11, that gives us 19 
possible 11's, which is actually one over what we need!

When we split two natural pairs into 2-card combinations to get our 10 and 
12, that leaves exactly 17 natural pairs.  Paired with the sixteen 10's 
and the 2-card 10 we just made, that gives us our 17 natural blackjacks, 
and the 12 goes with the leftover 9 to make the eighteenth.

And here's the hardest to describe: when we split three natural pairs on 
the way to a 3-card 12 to go with the 9, we also end up with a 3-card 
unnatural blackjack (unless one of the pairs is 9-2, in which case we end 
up with a natural blackjack in disguise.)  The splitting of those three 
natural pairs sets up two blackjacks, leaving the sixteen intact ones to 
be used in natural blackjacks.

Well, I hope that interested you.  Catch ya later...

-- Laffman --