Local Numbers

[Matt Morrison]

Right, I've been meaning to post this since forever, several different games have inspired me, most recently it was one of Gevin's NA games where he had the (if I remember correctly) same target in two consecutive rounds. So yeah - that's what I really enjoy, when you get several numbers rounds in a row that have really similar targets.

This 'game' has two parts then. The first is simply to find (or more likely, look out for in the future) games where consecutive numbers targets are very close together. Such as this one which features consecutive targets of 159, 183, 155, and 191.

Then on the other hand we need you maths geeks to help us find a formula that represents the factor of Number Locality (or whatever better name we can give it) for a particular set of rounds. So I can imagine this will take into account at least the following factors - the spread between the lowest and highest targets in the particular set of rounds; the number of rounds included in the set; and the maximum possible distinction between possible targets in that format.

So for the above example, we've got a spread of 32 over 4 rounds, in a format (regular) where the possible distinction is 899 (100 to 999).
Basically I'm just bandying this about, I think it's quite cool and hopefully it will catch the interest of some of you others. For example, what's cooler - two rounds which have the same consecutive number, or five rounds with a very small spread of 30 or something? I guess that has to factor into how we weight the formula. And can the factor remain accurate when applied to smaller and larger ranges (i.e. junior and hyper)?

I love you all.

[Howard Somerset]

I'm assuming that you're looking for something like the probablity that the spread is x or less when considering y consecutive rounds.

Or are you looking for the probablity that the somewhere in a game of z rounds there are y consecutive rounds in which the spread is x or less?

Or maybe you're looking for the expectation of the spread over y consecutive rounds, or possibly the expectation of the minimum spread when looking at each of the possible sets of y consecutive rounds within a game of z rounds.

Or is it something else altogether?

[Matt Morrison]

Or are you looking for the probablity that the somewhere in a game of z rounds there are y consecutive rounds in which the spread is x or less?

This. So that once we have the formula people can pick small sets of rounds and they can be given a Locality Factor or whatever, i.e. nothing to do with the game at large (only the picked rounds are relevant), and nothing to do with expectation (so just a means to judge precisely how local the chosen rounds were).

Something that can say for sure whether a set of three rounds where the target differs by 20 is 'more local' than a set of four rounds where the target differs by 40, or not.

[Gavin Chipper]

Sounds pretty good, although I'm fairly sure I can claim royalties because you're using my game as an example.

So from 101-999 there's 899 targets. Ignoring (for the minute - we'll have to come back to it) the fact that some targets are near one end of the range, to be within say 32 of the target, there are 65 options - 32 either side and the original target. The first target doesn't really matter but to get the other three within 32 it would be (65/899)^3. Which is 0.000378, or 1 in 2646.

It's fairly easy to work out manually each time how many numbers there are within a given range of the target, but as for a set formula, it might be a bit awkward.

Edit - I was thinking of within 32 of the first target but that's wrong.

[Matt Morrison]

This is a good start. And Michael Wallace is lurking.

[Michael Wallace]

This is a good start. And Michael Wallace is lurking.

[Gavin Chipper]

I deleted it from my last post but I think it is harder to all be within a certain range if they're near the edge of the 101-899 range. If the first number is 101, then all the others have to be 133 or less (going by the original example of 32), but if it's 500, then the second number at least can be +/- 32.

Edit - it's definitely the edges of the range that make this awkward (more awkward).

Edit 2 - I suppose you could argue that as it's more impressive to be within a certain spread when near the edge of the possible range, it deserves to be accounted for in the impressivness number. But that might be going a bit far.

[Gavin Chipper]

On a vaguely similar note, I was wondering if there might be a trap here that makes things seem better than they are. If you pick a random number from 0-100 (including non-integers) and decide that 0 is best and 100 worst, then if you pick 23, then obviously that's in the top 23%. And if you do it twice and get 23 and 14 you might be tempted to multiply and say you'd only do that well both times about 3% of the time. But you'd find that over time, you'd always get really impressive figures that way (because after 100 trials it is very unlikely if you repeated it that you'd do as well or better on each of the 100 trials). It's probably irrelevant to this though.

[Gavin Chipper]

OK, so if the lowest number is 101 then the probability of all four (or whatever) being within 32 (or whatever) of 101 is the probability of the lowest being 101 * (33/899)*3 (not the probability given the lowest is 101 but the probability of the lowest being 101 and the rest being within 32). The probability of the lowest one actually being 101 is 1-(898/899)^4. This works for all lowest targets up to 967, so 867 targets, so you multiply that by 867 and get something like 0.00019. Then for each target up you have to multiply by (32/899)^3, 31, 30... and and them all together. Is this any good?

[Gavin Chipper]

This 'game' has two parts then. The first is simply to find (or more likely, look out for in the future) games where consecutive numbers targets are very close together. Such as this one which features consecutive targets of 159, 183, 155, and 191. [...]
So for the above example, we've got a spread of 32 over 4 rounds

Isn't it 36 anyway? (not that it matters)