Fascinating, Garyo! And if you remove the doubles, you get a pool of 45, of which 10 now are repeats. (I'm taking them out, because the straights vs. box odds for doubles is different than singles.) What percentage is 10 of 45?... Appx. 22%, very close to 1/6th... which is what we'd expect to shake out when it comes to boxes vs. straights (6 boxes, of which 1 would be a straight match). This fits my theory, then, that the numbers are (a) chosen first as to positioning, above/below/beside, and then (b) shuffled randomly to arrive at one of the 6 box final positions as played—of which appx. 1/6th would be straights, and the rest various boxes.
But as to "First, Second, Third choice" or "Position A, B, C"... these are non-abstract concepts, whereas the game works with the utter abstractness of numbers.
Still... if one was to assume a limited pool to begin with, for whatever reason—a pre-selected pool of numbers from which to pick which number's remaining digit—if one speaks temporally, one would want the limited pool picked first; that's because, it would have the least impact on the greater pool; it would diminish the end result least. If the limited pool were digits 0, 1, 2, 3, 4, you'd want that limited pool picked first, because it'd enable greater range of play (i.e., if second and third "positions" went first and second, you might get 1 and 4; this limits the scope of a limited pool, because now its choices are but three, 1, 2, 3; but if the limited pool is picked first, say 1, it limits the range of second and third picks no more than if the first pool were limited, or not limited).
Why do I say a pool is limited, and it's probably the first one? Because I just believe it, for one thing. But also, it would account for a very observable phenomenon in the game, that of digits being long absent.
If we have digits 0-9 (a-j); and then we have, as the pool for First choice digit (A), a limitation to say five digits (¹); then subsequent pick Second (B), has a range of:
B = A - a¹.
Favoring one way, means not favoring another. Choice Second (B) now has a full range of choices, including those of the original pool (minus A); but choice Third (C) becomes dependent upon the second choice, too, so its pool is diminished by two - A, and B.
Which means, if we assume a limited pool from which the First digit is chosen... then the Second Choice has the largest range of optional digits, and it's here that most of the long-outs would be sourced. (Because the limitations of A, mean their chance of dominating any individual play is greater than the impact of digits outside that pool—hence, you get so many repeats and often-playeds, and the long-outs, too. Under this system, you'd get less long-outs as a whole, than often playeds or repeats [indeed, you usually only get an average of about two long-out digits at any one time.])
... This is all, if I'm doing my math right.