## Wednesday, September 10, 2008

### Brain Teaser: Seemingly Random Number List Selection

Hey there,

We're just about done with this series of somewhat disjointed posts on number pools and guaranteed matches. Hopefully, each post in the series has been somewhat entertaining and far enough removed from the starting post that we kept up some sense of originality. It wasn't too hard (unless we screwed up ;) since there were so many unique concepts to cover within the realm of reaching the Objective of the original post. Please refer back to our post on number pools for definitions of all the major terms and a listing of the Objective (although that definition is repeated in a few other posts along the way).

We're just about at the end of the road, so I thought I'd make today's post a bit of a Brain Teaser. The answer to this is actually pretty interesting and (from my struggling with it) you'll either get it right away or it'll take you a good amount of time (kind of like the M I U Puzzle). I, for one, spent way too much time reading up on number sequences and probability. I'm not an engineering major, so I actually had to reference other materials to decipher the Greek alphabet notation. It was interesting for me (especially the insomniac in me), but not something I'd recommend doing unless you're really into it. One thing I noticed, although this is slightly off topic; when people see you sitting in your easy chair chugging diet Pepsi, popping Diphenhydramine and typing in your terminal window with your lap covered with shreds of paper with interpretive numeric scrawl written all over them, they tend to not ask too many questions. It's a great way to guarantee you'll have the evening to yourself ;)

Anyway; for this post's meat: The brain teaser we have today is pretty straightforward, by way of explanation. Here's the setup:

1. Like we did in our posts on deriving list permutations, sorting lists and sorting lists within lists (not necessarily in that order), we've generated a list of 56 5 digit lists using an 8 digit pool of numbers (1, 2, 3, 4, 5, 6, 7, 8). This list represents all possible permutations of the 8 digit list within 5 digit lists while "absolutely removing duplicates" (i.e. "1, 2, 3, 4, 5" and " 5, 3, 2, 4, 1" would be considered equal and would both be removed). That list of numbers follows here (Note that each line is numbered):

`     1 1 2 3 4 5     2 1 2 3 4 6      3 1 2 3 4 7     4 1 2 3 4 8     5 1 2 3 5 6     6 1 2 3 5 7     7 1 2 3 5 8      8 1 2 3 6 7     9 1 2 3 6 8    10 1 2 3 7 8    11 1 2 4 5 6    12 1 2 4 5 7    13 1 2 4 5 8    14 1 2 4 6 7    15 1 2 4 6 8    16 1 2 4 7 8    17 1 2 5 6 7    18 1 2 5 6 8    19 1 2 5 7 8    20 1 2 6 7 8    21 1 3 4 5 6    22 1 3 4 5 7    23 1 3 4 5 8    24 1 3 4 6 7    25 1 3 4 6 8    26 1 3 4 7 8    27 1 3 5 6 7    28 1 3 5 6 8    29 1 3 5 7 8    30 1 3 6 7 8    31 1 4 5 6 7    32 1 4 5 6 8    33 1 4 5 7 8    34 1 4 6 7 8    35 1 5 6 7 8    36 2 3 4 5 6    37 2 3 4 5 7    38 2 3 4 5 8    39 2 3 4 6 7    40 2 3 4 6 8    41 2 3 4 7 8    42 2 3 5 6 7    43 2 3 5 6 8    44 2 3 5 7 8    45 2 3 6 7 8    46 2 4 5 6 7    47 2 4 5 6 8    48 2 4 5 7 8    49 2 4 6 7 8    50 2 5 6 7 8    51 3 4 5 6 7    52 3 4 5 6 8    53 3 4 5 7 8    54 3 4 6 7 8    55 3 5 6 7 8    56 4 5 6 7 8`

2. Using the technique we looked at in our post on determining maximum sets of lists within pools using binomial coefficients, we've deduced that, since we want to guarantee (within the lists in this list of lists) unique 3 digit combinations, we can do so most economically by selecting 8 of the 5 digit lists from the 56. This is not the exact formula we followed in the noted post (as it showed that the binomial coefficient of 8 over 3 is 56). In fairly loose terms, if we want to maximize the odds that we'll get any given 3 digit list (and not use all 56 lists, which works, but is expensive), we can do so with a minimum of 8 5 digit lists. Those 8 lists follow here:

`     1 2 3 4 6     1 2 3 5 8     1 2 4 7 8     1 3 6 7 8     1 4 5 6 7     2 3 4 5 7     2 5 6 7 8     3 4 5 6 8`

3. Using the numbers from point 1 and 2, above, the following chart shows you at what line (in the list of 5 digit lists) each of the 8 lists we determined would give us the best odds of getting a maximum return on a 3 digit list match up.

`     2 1 2 3 4 6  1 2 3 4 6     7 1 2 3 5 8  1 2 3 5 8    16 1 2 4 7 8  1 2 4 7 8    30 1 3 6 7 8  1 3 6 7 8    31 1 4 5 6 7  1 4 5 6 7    37 2 3 4 5 7  2 3 4 5 7    50 2 5 6 7 8  2 5 6 7 8    52 3 4 5 6 8  3 4 5 6 8`

4. Using the numbers from point 1 and 2, above, the following chart shows you at what line (in the list of 5 digit lists) each of the 20 lists we can also determine would give us the best odds of getting a maximum return on a 4 digit list match (We never covered this, but I just added it to give you some frame of reference if the above 3 points aren't enough - BTW, I generated this list because they weren't enough for me, so no offense intended :)

`     1 1 2 3 4 5   1 2 3 4 5     8 1 2 3 6 7   1 2 3 6 7     9 1 2 3 6 8   1 2 3 6 8    14 1 2 4 6 7   1 2 4 6 7    15 1 2 4 6 8   1 2 4 6 8    18 1 2 5 6 8   1 2 5 6 8    19 1 2 5 7 8   1 2 5 7 8    21 1 3 4 5 6   1 3 4 5 6    22 1 3 4 5 7   1 3 4 5 7    23 1 3 4 5 8   1 3 4 5 8    26 1 3 4 7 8   1 3 4 7 8    35 1 5 6 7 8   1 5 6 7 8    38 2 3 4 5 8   2 3 4 5 8    39 2 3 4 6 7   2 3 4 6 7    42 2 3 5 6 7   2 3 5 6 7    45 2 3 6 7 8   2 3 6 7 8    46 2 4 5 6 7   2 4 5 6 7    49 2 4 6 7 8   2 4 6 7 8    52 3 4 5 6 8   3 4 5 6 8    53 3 4 5 7 8   3 4 5 7 8`

Now, here's the fun part. Considering that:

a) no matter how many times we do this using the exact parameters, the results (the lists and the placement of the 8 most economical choices) are totally consistent.
b) If we change anything (like make our list of "1, 2, 3, 4, 5, 6, 7, 8" into "1, 4, 3, 2, 5, 6, 7, 8") about our 8 digit list (assuming that we don't resort everything to numeric order before generating our 8 best 5 digit lists), the outcome will be affected in that the 8 lists will not only contain different numerals, but also end up matching at different numbered locations on master list of 56 lists.

Considering that, how am I determining where to place my 8 5 digit lists in relation to my 56 list list. Or, in other terms, how am I picking the 8 5 digit lists I feel will give me the best odds of matching a 3 digit combination of any 3 unique digits from the 8 digit list used to seed these 5 digit selections?

Think about it, but not too hard. This is only fun if you're having fun with it. If it starts to upset you, watch something violent, yet satisfying (like a good revenge flick), on television and get some sleep ;)

Cheers,

, Mike

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