myrmepropagandistFeb 20

@Bumblefish

Which one is random?
(data sets are 100 numbers 1 to 6)

listA=[2,3,5,1,2,2,4,2,4,5,2,3,3,4,5,6,4,2,6,2,2,1,3,4,5,5,6,3,3,6,1,4,2,1,4,5,2,2,3,3,3,5,6,3,2,4,5,5,1,1,1,6,1,4,3,5,5,3,1,1,1,6,1,4,6,6,3,6,6,2,4,4,4,5,1,5,6,2,6,1,1,2,4,2,2,3,4,4,5,6,1,3,3,3,5,4,6,5,1,6]

listB=[4,2,5,6,3,5,3,1,3,4,2,3,4,3,4,5,5,1,3,3,2,1,1,6,1,3,2,2,2,6,1,5,6,3,6,3,2,3,2,4,6,1,1,6,3,2,4,1,6,1,3,1,5,6,2,3,3,5,1,6,4,5,2,5,1,1,5,3,6,2,3,3,6,5,2,3,3,1,6,3,2,3,2,1,6,6,4,4,6,2,4,5,4,5,3,4,6,5,3,2]

Show threadAbyssal RookFeb 21

@futurebird Before I look at where the answer shows up, my guess would be that List A is random.

The odds of both dice being the same number when you roll 2 dice is 1/6 (36 possibilities, 6 desired results). For 3, that becomes 1/36. (6*6*6 possibilities, 6 desired).

What we have here is 98 consecutive possible places for a 3-of-a-kind to start. The odds that you would only draw the 1/36 chance ONCE (The 3 2's near the beginning of B) is something like....8%?

Show threadInga stands with 🇺🇦 🇵🇸Feb 21
@AbyssalRook @futurebird I see two mistakes in your reasoning.
One is technical: events "numbers with position N, N+1 and N+2 are the same" for different values of N are _not_ independent of each other. (For example, if we know that this statement is true for N=10, then there likelihood of it being true for N=11 is 1/6, not 1/36.)
Another symbolizes a deeper problem with a lot of modern research that relies heavily on p-values: consider how many statements of this kind, containing the same amount of information, could you make? Unless you commit to a specific statement beforehand, before seeing the data: "this statement would only be true in 8% of cases for truly random data" does not really mean anything if it's just one out of 20 equally "interesting" statements one could make about the data (e.g. "how many triplets of incrementing numbers (modulo six) are there", "how many decrementing triplets are there", etc), each only 8% likely. Because of course it is expected that for most random sequences, a few of these individually not very likely statements will be true.
Show threadAbyssal Rook
@IngaLovinde I'm not following the first problem in the logic. The situation you're describing might be important if we're looking at more and more instances of it happening, but looking at it happening at least once (~94%) doesn't change at all, and it happening ONLY once might jiggle the ~8% estimate I had, but not significantly move it.
Feb 21 at 12:37pmWeb
Show threadAbyssal RookFeb 21

@IngaLovinde As for the latter, that is entirely true from a research perspective, but I picked the 3-of-a-kind pattern because I assumed the non-random list was entirely human constructed, and that particular pattern is one that sticks out to us the most. Someone making a list by hand is more likely to see "6-6-6" as less random than "6-1-2" or "3-4-5".

I did not clock 'Which is random?' as one being a dice roll and the other being a shuffled deck of prescribed cards.

Show threadInga stands with 🇺🇦 🇵🇸Feb 21
@AbyssalRook but the same goes for numbers repeating twice, or four times, or ascending (2, 3, 4), or descending (4, 3, 2), or repeated pairs (5, 1, 5, 1), etc.
One can come up with many patterns or tests like that, each with similarly low probability; but that one of them matches the data doesn't mean anything, because it is expected that some of them will match. _Especially_ if you only come up with a specific pattern _after_ seeing the data.
Show threadInga stands with 🇺🇦 🇵🇸Feb 21

@AbyssalRook okay let's calculate it:
Let a_n be the probability that the sequence of length n does not contain triplets of identical numbers, and does not end with two same numbers; b_n, the same, but ends with two same numbers.
Then a_1 = 1, a_2 = 5/6, b_2 = 1/6; a_(n+1) = a_n * 5/6 + b_n * 5/6; b_(n+1) = a_n * 1/6.
Or, expanding b_n, we get a_(n+2) = a_(n+1) * 5/6 + a_n * 5/36.
Plugging these numbers into Wolfram alpha (`LinearRecurrence[{5/6, 5/36}, {1, 5/6}, 100]`), we obtain a_100 ~= 0.0762866, a_99 ~= 0.0781878, and therefore the probability that the sequence of 100 random numbers does not contain triplets of the same number is a_100 + a_99/6 ~= 0.0893 = 8.93%.

By contrast, the probability that out of 98 random (and independent) triplets none will consist of three same numbers is (35/36)^98 ~= 6.32%.

That's a pretty large difference, and not just a jiggle.

(I understand that this is not the number you were looking at, but it's the easiest way to illustrate that there is a significant difference between answering questions about triplets of repeating number among 98 independent random triplets and among 98 sub-triplets of the sequence with 100 independent random numbers.)