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Probabilities for getting unchanging vs. changing hexagrams

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Freedda

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I've read in a few places the probabilities of getting different lines using different methods: yarrow stalk (or 16 objects), or three coins. For example, that the probability of getting an 8 is three out of sixteen, or of getting a 9 is 2 out of sixteen ....

But what about for the hexagrams themselves? Does anyone know the probability of getting an unchanging hexagram versus a changing one? One third of the time? Three out of sixteen? One in twenty? .....

Best, D.
 

Gmulii

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Probability is a very messy thing as depending where you view it from it will bring different results.
The good old story of the person that was asked "what is the chance if you get out of your home you will see a dinosaur walking around"(or something similar), and they answered 50%.

The idea was either they will see or won't(so 50/50). Very accurate by itself, yet most people would give it much lower probability based on where they want to look at it from(history/previous experience etc).

So if we take the end result and say that its as likely to get moving line as it is to get not moving line then it would be around 3%. That isn't the case, though but its nice first step.

50% chance to get not moving line(4 possible states, 2 of them not moving) once.

So getting 2 not moving lines would be 0.5 * 0.5 =0.25
And that would be 25%.
Next step 0.25 * 0.5 = 0.125
And that would be 12.5%
And with each line it goes lower by half and at the 6th will arrive at around 3%.
Its very different for the stalks, but people rarely use that.

That seems to low and if we look at the actual process that makes sense to be low, as in practice its not as likely to get 3 coins on the same side as it is to get 2 /1.

So getting 3 coins on the same side would be around 12%(0.5*0.5*0.5).
So if we take 12% for one side, 12% for the other we have something like
76% chance to be 2/1.

Using that we get to around 14% chance to get hexagram with not moving lines.


While that seems to make sense, at the end of the day I really like this point of view:

"Scientists have calculated that the chances of something so patently absurd actually existing are millions to one. But magicians have calculated that million-to-one chances crop up nine times out of ten." ― Terry Pratchett, Mort
 

Freedda

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Thanks Gmulii, you said
Using that we get to around 14% chance to get hexagram with no moving lines.
So, about an eight of the time, or a 2:16 probability of no moving lines.

I am not looking for any exactitude, and I understand 'probable.' It was just a point I was curious about based on something someone had said. It's 'close enough for government work' as they say.

Best, d.
 

Trojina

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For oneself is it perhaps something you can work out in the Resonance Journal. There's facilities to find the number of unchanging answers as compared to all line 1s I imagine. I don't know how but if anyone knows how to do that it's probably [MENTION=921]Liselle[/MENTION]
 

Liselle

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In advanced search, down in the "Changing Lines" section, there's a box to check off to search for only unchanging readings. But as far as I know, Resonance Journal doesn't provide counts of anything at the moment.

It'd probably be easiest to export your journal to Excel*, and sort and count using Excel features. Lots of interesting things could be done in Excel, I'm sure, not that I've done any of them yet. But I can vouch that the R.J. export feature works well, even for large-ish journals (mine's about 11,000 entries/rows). For instance I worried whether it would crash my laptop, and it didn't, and the resulting spreadsheet takes about 160 MB of memory, which also is fine. (This is by far the largest spreadsheet I've ever tried to work with, hence my trepidation, but it's perfectly fine.)

In R.J. itself, you could get a rough estimate by counting how many entries are a "screen-ful" in lists, and then scroll through one page at a time, counting pages (click in an empty part of the scroll bar, near the bottom but above the arrow), and then multiply. But Excel would be better, imo.





* when I say "Excel" - that's what it's called in R.J. in the Entries menu, but of course it works perfectly fine in any other spreadsheet program that can read .xls files. Personally I use LibreOffice Calc.
 
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moss elk

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I have to give you two different answers:

1- If making the error of assuming it is random,
it would be a 1.56% chance to get an unchanging hexagram. (4096 possible readings, 64 unchanging hexagrams)

2-100% chance to get the reading that you received. And this is the only important thing.
 

Freedda

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I have to give you two different answers:
Answer 2, most definately.

For answer 1, I'm thinking it's somewhere between 4.68% and 12.5%, but who's counting? ;)
 

jukkodave

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Hi everyone

Just came across this thread. Lots of interesting ideas and viewpoints.

Mathematically it is 1 in 64.
It doesnt matter if you are using stalks, coins or any other method, unless the method is skewed to favour certain lines, the odds are just the same.

Know this because it just maths.
Know this because I made cards with the options of all the possible line combinations and only one with no moving lines.
From another point of view every hexagram can potential move to any other hexagram, by any combination of moving lines, so 63 possible changes and only one that doesnt change.

So if anyone is getting more than 1 in 64 that is significant. Though why it is significant could be down to that the Yi is helping us out with more information that we need, or that we are "influencing" the reading, subconsiously and perhaps think that more information is somehow better than only a single hexagram.

The gnarly question of what the Yi is, and how and why the Yi is and how and why we use it just wont go away.

Dave
 

moss elk

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Mathematically it is 1 in 64.
I could make a career out of this...

There are 4096 possible readings,
64 chances to get an unchanging hexagram.

The formula for probability is:
64÷4096
which results in .015625
that reads as 1.56% chance.


But this is pointless though because, it isn't random,
among other reasons.
 

hilary

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Gmulii pointed out the problem with this thinking that leads to '1 in 64' as the answer. Yes, there are 64 possible outcomes (changing to any other hexagram, or not changing) but it doesn't follow that all outcomes are equally likely.

If that were the case, there would indeed be a 50% chance you'll see a dinosaur outside your window. Either you will or you won't, 2 possible outcomes. 1 out of 2 outcomes = half = 50% chance.

So Gmulii did the right sum but with the wrong numbers, because the odds of any given line changing are 1 in 4. This is true of yarrow or 3 coins, but easiest to describe with 3 coins. There are 8 ways they could fall:

HHH
TTT
HTT
THT
TTH
THH
HTH
HHT

Two of those eight, HHH and TTT, represent changing lines.

Odds of one line not changing: 75%.
Odds of two lines not changing: 75% of 75%.
Odds of 3 lines not changing: 75% of 75% of 75%.
And so on to power 6. I'm reliably informed that comes out as 729/4096, which is just under 18%.
 

jukkodave

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Hi Hilary

A rather different scenario with a dinosaur. To be applicable to only being 1 in 2 you would also have to donsider all sortsof other factors, how many dinosaurs on the planet, is the window facing the sea or land, am I eating my tea and not even looking out of the window.

It is mathematical question. You give TTH and THT as two different outcomes, but the "order" is not mathermatically relevant. TTH and THT are equivalent, as are HHT and HTH. In terms of coins is doesnt matter if you throw them one a time or 3 together, they land as all T's, all H's, 2H's and a T, or 2 T's and a H.

That gives the correct possibility of 50/ 50 for each line of having a moving line or not. For 6 lines that results in 1 in 64.

Using your description:
"Odds of one line not changing: 50%.
Odds of two lines not changing: 50% of 50%.
Odds of 3 lines not changing: 50% of 50% of 50%.
which comes out as 64/4096

As MossElf points out.

You may have been "reliably " informed, but the underlying principle, in this case a mathematical one, was incorrect and gives an incorrect result.

The answer is 1 in 64.

All the best

Dave
 

hilary

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There are 3 different coins. Each one could land either way up. If you throw HHT, the second coin was a head and the third was tails. This is a separate outcome from HTH, where the second coin was tails and the third was heads. Different things happened. 'Two heads and a tail' looks like one outcome, but there are actually three different ways the coins can fall to create this outcome. To calculate the probability, you need to count all three.

I'm sorry, I'm not a maths teacher and can't think of a better way to explain this. (I don't suppose we have a maths teacher in the house who would like to step in? ) But then again, you don't need me to. You can just sit down with 3 coins, cast a couple of dozen readings, and see what percentage come out unchanging.
 

Trojina

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jukkodave


As MossElf points out.

:rofl: OMG it's worth you being here just for the names. First 'Hairily' now 'MossElf'

An elf ? MossElf :rofl:

His name is Moss Elk. He is an Elk not an Elf.
 

Liselle

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In the new forum, I'd have been able to click the "lol" reaction button, instead of clicking generic "thanks," or posting a new post to say :rofl:. Am looking forward to the new forum.
 

hmesker

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:rofl: OMG it's worth you being here just for the names. First 'Hairily' now 'MossElf'
Not to mention the many times he calls me 'Harmon' :rofl:

Ah well. What's in a name. I've been called Herman, Harman, and several other variants.
 

jukkodave

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I hope that Moss Elk would not to caught up in a typographical error.

Aplogies to him for not being careful in my typing
 

jukkodave

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Sorry Hmesker,

I was reading through a post somewhere and that is the name someone used so I though it was correct.
What should I have been writing.
Why didnt you correct me if I was getting it wrong.

All the best

Dave
 

moss elk

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:rofl: OMG it's worth you being here just for the names. First 'Hairily' now 'MossElf'
An elf ? MossElf :rofl:

:rofl:
I like Moss Elf as much as when that angry forum user tried to insult me with Miss Elk, because I was chastising her boyfriend.
 

jukkodave

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Just so you know, it was an error, it was not meant to mean anything at all.
 

jukkodave

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Hi Moss Elk

I have no idea what that means. Could you clarify

Dave
 

jukkodave

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Moss Elk

That is just being facetious.

You obviously meant something by what you said and the emoji. Please do explain what that was.

Even your limitations might give me a clue as to what you meant. If you accepted them you wouldnt mind trying to explain what you intended, even if that made you look a fool.

Dave
 

Freedda

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So as the one whom started this thread - again, a reminder that it was a broad question, more just a curiousity, and I have nothing dependent or resting on getting a 'correct' answer.

But we now have a range here from about 1.56% to 18% of getting an unchanging hexagram.

I've read a few times that the probablity of getting a 6 or 9 (out of 6, 7, 8, 9) coin toss or using yarrow stalks is one in four, not 50% - and if true, it would favor a higher number. But not being much at math, and even less so with probabilities, I can't say for sure which 'probability' is probably the correct one.

Which leads me to my own on-the-ground, in-the-field observations: that for myself, and from what I've seen of others' sharing on this site, that we get an unchanging hexagram way more than 1.56% of the time, perhaps closer to 14-18% of the time.

And so I invite people to share their own experiences regarding this - not the stats or math, but what is true for you: how often do you know or feel you get an unchanging hexagram? Is it one out of 64 times? Or closer to 12 out of 64? Or something else entirely?

Best, D.
 

hilary

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Happy to report that maths works as expected in this part of the world ;) .

Yi works, too, and I find I get more unchanging hexagrams when I need a simple explanation, or an idea to take with me into complex situation.

This is where it goes beyond probability. You can calculate the probability of an unchanging reading, or of three unchanging readings in a row. What you can't calculate is the probability of those three readings being all around the same topic, or coming at a time when you most need simplicity, because those things can't be represented in a calculation.
 

hilary

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P.S. I know you said not to go into the probability again, but it's genuinely not complicated, you can work it out for yourself.

Here's an analogy that might help: if a couple have just two children, what are the odds they'll have a boy and a girl?

Well, they could have (1) two boys, or (2) two girls, or (3) one of each. Does this mean they have a one-in-three chance of having a boy and a girl? Of course not. They could have...

First child girl, second child boy.
First child boy, second child girl.
First and second both boys.
First and second both girls.

Hopefully that makes it clear. An older brother with younger sister is not the same thing as an older sister with a younger brother: those are two different things that could happen. That makes four possibilities, two of which give the boy+girl result, so they have a 50% chance of having one of each.

'If you toss two coins, what are the chances of getting a head and a tail?' is the same question, mathematically speaking. You could have...

First coin heads, second coin tails.
First coin tails, second coin heads.
First and second both heads.
First and second both tails.

That covers all the different things that could happen, so you can count them up and see that you get a head and a tail 2 out of 4 times, making a 50% probability.

Next question is 'What if you toss three coins, what's the probability of not getting them all the same?' - to answer that, see my previous post.

And for 6 lines in a row, you need to multiply up the probabilities, something you probably dimly remember from school ;) .
 

moss elk

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And just to add another example of how there is no probability involved in our readings:

About five years ago I had a plan.
I asked Yi if it was a good plan and received 17.5.
Six months passed and I procrastinated about starting work on that plan. I asked the exact same question and received 17.5 again. Another six months went by, and for the third time the same reading came for the same question

The 'probability' for that is very very very very very very very very very small. A mathmagician would be as surprised as I was.
 

Liselle

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Odds of one line not changing: 75%.
Odds of two lines not changing: 75% of 75%.
Odds of 3 lines not changing: 75% of 75% of 75%.
And so on to power 6. I'm reliably informed that comes out as 729/4096, which is just under 18%.
I just asked trusty spreadsheet to do this for my journal, and unchanging readings (edited: in my particular journal) are 18.01%, 1,823 out of 10,122 entries of type "Yijing Reading". (Yes, I cast too many readings; yes, we've had this discussion; yes; I've gotten better about it; yes, I still do it sometimes; yes, it's embarrassing.)

This is where it goes beyond probability. You can calculate the probability of an unchanging reading, or of three unchanging readings in a row. What you can't calculate is the probability of those three readings being all around the same topic, or coming at a time when you most need simplicity, because those things can't be represented in a calculation.
What's also fascinating is how Yi manages to agree with theoretical probability in this instance and talk to me meaningfully, that is, give me unchanging readings when I need them.

(Also, Hilary, I don't know if I've ever heard you say you're not good at math, but you are henceforth not allowed to say it. I've had a bit of university-level statistics and gave up on David's question. :bows:)
 
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Liselle

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What's also fascinating is how Yi manages to agree with theoretical probability in this instance and talk to me meaningfully, that is, give me unchanging readings when I need them.
And just to add another example of how there is no probability involved in our readings:

About five years ago I had a plan.
I asked Yi if it was a good plan and received 17.5.
Six months passed and I procrastinated about starting work on that plan. I asked the exact same question and received 17.5 again. Another six months went by, and for the third time the same reading came for the same question

The 'probability' for that is very very very very very very very very very small. A mathmagician would be as surprised as I was.
That exactly. And yet I bet if you calculated the percentage of times you've received 17.5 > 51 out of all your readings, it would hew to the expected mathematical probability. :eek:
 
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