K-man's pogo.com Informational Pages

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 --

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