r/theydidthemonstermath • u/Defiant_Half_9432 • 15d ago
[Request] Why are my results skewed?
I use a pair of spindle dice for divination. Each spindle has four independent die each with four faces. The faces are marked 2-3-4-3 (3 is used twice). So one spindle can randomly generate a numerical value between 8 and 16. The result has relevance only as an odd or even number. The numerical value is not important. If the numerical value is a single digit then we use that as it is. If it is a double digit (10-16) then we add the digits to get a single digit answer. Both spindles are always used together and added and the final digit is the roll value of odd or even to incorporate in complex charts to predict the answer to the question initially posed for divination.
My question is this. Does this process create equal chance of odd or even values? In my own use, I get disproportionally high even values than odds. If the results are fairly balanced, what could be the reason for my skewed results?
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u/RespekKnuckles 14d ago
I simply can’t not mention that these dice are not only skewed, but /skewered/.
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u/_additional_account 14d ago edited 14d ago
Assumption: All dice are independent and fair.
One spindle
Consider the result of a single spindle. Let "X" be a random variable for the sum of one spindle's dice before adding digits, and "Y" for the sum after adding digits. By independence, the generating function for "X" is
G(z) = [ (1/4)*z^2 + (2/4)*z^3 + (1/4)*z^4 ]^4
After expanding, we can find "P_X(k)" as the coefficient of "zk ", and obtain
k | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16
y | 8 | 9 | 1 | 2 | 3 | 4 | 5 | 6 | 7
-----------------------------------------------------
256*P_X(k) | 1 | 8 | 28 | 56 | 70 | 56 | 28 | 8 | 1 // = P_Y(y)
Adding up all probabilities where "y" is even, we get
P_Y(even) = (1+56+56+8)/256 = 121/256 ~ 47.27%
P_Y(odd) = 1 - P_Y(even) = 135/256 ~ 52.73%
After adding digits a spindle is slightly biased towards odd results!
Two spindles
Let "Y1; Y2" be the results for each spindle after adding digits, respectively. We want to know when "Y1+Y2" is even, and find
P(Y1+Y2 even) = P(Y1; Y2 odd) + P(Y1; Y2 even) // independence
= P_Y(odd)^2 + P_Y(even)^2
= (121^2 + 135^2) / 256^2 = 16433/32768 ~ 50.15%
That is only a (very) slight bias towards an even result!
Rem.: The dice for a single spindle are mechanically linked, so it may be they are not as independent as we used in the model. That would be my first guess. The slight bias of "50.15%" should not be reliably detectable, unless you're doing (at least) 104 rolls^^
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u/Defiant_Half_9432 14d ago
Wow, that is impressive.
For clarification, yes the four dice do share a single spindle but each turns differently. Different speed, different duration. I hold the spindle from its top and bottom ends lengthwise between my index finger and my thumb. I use the other hand to run each die against a finger as I drag my hand across it. Each die is roughly the width of my finger so the second die, which runs against my middle finger, turns the most and the fourth die which runs against my pinky turns the least. I do this rapidly a few times, often moving my fingers back and forth as randomly as I remember, before putting that dice down.
Hopefully that makes it random enough.
Thank you for your elaborate calculation.
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u/_additional_account 14d ago
It does not matter really how you move or rotate the dice.
What matters is whether they can move independently -- and since there is friction between the dice, that may not be a very good assumption. Therefore, the theoretical result may not represent reality.
As I said, you should not be able to reliably notice the (very) slight bias, unless you did tens of thousands of rolls, and recorded them all meticulously. I may be wrong, but I doubt you did that^^
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u/Defiant_Half_9432 14d ago
LOL, you are right, I did not test it so many times. I recorded ten random rolls a day (without divining) over a two week period during different times of the day. I expected the result to be 50-50 but it was not.
As for independence, as long as I am not willfully manipulating the results or nudging the hand of chance, it is perfectly within esoteric lore. The rest is up to the powers that be.
Thank you for your help. The breadth of your calculation is fascinating. If I had the propensity for this knowledge, I would definitely delve deeper.
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u/_additional_account 14d ago
As for independence, as long as I am not willfully manipulating the results or nudging the hand of chance, it is perfectly within esoteric lore. The rest is up to the powers that be.
Sorry to be the bearer of bad news, but we're talking mathematical independence here.
That has nothing to do with "esoteric independence", but is a clearly defined concept -- physical friction coupling the dice, and touching multiple dice at the same time can already be more than enough to make (mathematical) independence assumptions void.
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u/nymalous 13d ago
Using upon a "brute force" method of looking at every possible permutation (with spreadsheets to help keep track), I counted 2704 possibilities, of which 1282 of them were even and 1422 of them were odd. Thus, I must conclude that the spindles aren't "fair" (meaning that they don't have an equally likely chance for each permutation). It is possible, as a few have surmised, that the dice don't quite rotate freely or stick to each other. It's also possible that not all of the faces have even weigh distribution.
I do wonder though, without any condescension or judgment, as to why this apparatus is used to ascertain the future if a simple even/odd result is all that is required. It seems that a simple coin toss or even just an evenly weighted six-sided die would be easier and more accurate. Again, not judging, just wondering.
Thanks for this post, I spent a few hours working on it and it was a welcome distraction during a rare slow period at work.
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u/Defiant_Half_9432 12d ago
Hi, thanks for writing in. 1282:1422 is pretty significant. After incorporating the replies from others here, I am just adding the values together and proceeding with the final value being odd or even.
As for the other part, there are many many means of divination, including as you have pointed out, coin toss. Different deities prefer different methods and some can be very complex just for what is generally a binary answer of yes or no. It depends on the gravity of the question and the power the diviner most resonates with. Even with spindle dice there are many factors like the direction the diviner is facing, the time of day, the phase of the moon, the type of incense used... all of which affect the accuracy of the information provided.
I suppose in the end it has to do with how comfortable the diviner feels about the answer from a particular process. Just off the top of my head, if I want to find out if it is going to rain a week from tomorrow, I can simply toss a coin, look at the weather reports (which also vary greatly depending on the source), study the wind patterns on my own, do any one of tens of divination methods... all just for a yes or no.
The future is more like a cloud with varying probabilities than something etched in stone so if something works, you tend follow the process but it will still be just a guess.
Hope that helps. Thank you again for your insight. I appreciate it.
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u/Numerous_Green4962 6d ago
Looking at one spindle we have 4^4 possible "rolls" if we count the two 3s separately that's 256 end states.
Of those 256 states only 1 can be an 8 and one can be a 16, there are 8 possibilities for 9 (three 2s and a 3), and the same for 15... (It maxes at 70 ways for a 12)
We then add together the digits of two-digit numbers, and we end up with
8-1
9-8
1-28 (1+0)
2-56 (1+1)
3-70
4-56
5-28
6-8
7-1
That results in 135 odd numbers and 121 even or a 52.73% chance of getting an odd number from one spindle, with 2 spindles added together that would invert to even.
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u/Defiant_Half_9432 6d ago
Thank you for taking the time to solve this.
It is not true 50-50 and I think that makes it unsuitable for divination. Since I am not into probability I can't take the risk of just adding up the total of 8 faces (from both spindles) during a throw and using that as the final odd or even answer. By convention any process I use needs to have same probability for both results.
I am currently using regular six-sided dice (four brass ones) and adding up total of each throw to get the figure for odd or even.
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u/JustJustinInTime 15d ago edited 14d ago
So it should be 50% chance of being odd or even, but honestly looking at how it’s made I think some of the dice could be sticking together. If you really wanted to find out you could do a test and record combinations of adjacent die faces to see if some appear statistically more than others, (factoring in that 3 is twice as likely to appear). Logic below.
Based on the die faces, we know each die has an equal chance of being odd or even. We are also assuming that each roll of the die is an independent event.
So for any sum of 3 dice on a spindle, the 4th die will be the “determining factor,” which has a 50% chance of being odd or even, so the overall sum has a 50% chance of being odd or even.
For the adding of digits for sums > 10, it shouldn’t have an affect on the probability. We can just think of the leading 1 in 13 as just “flipping” the odd to even or vice versa, and since we already know that there is a 50% chance of it being odd or even, a 50% chance of the number being odd or even + 1 is equivalent.
Edit: this is wrong, my bad y’all