# Thread: Challenge to the calculator illustrates the advantages of Reinin Dichotomies

1. ## Challenge to the calculator illustrates the advantages of Reinin Dichotomies

Lets suppose we grouped people into 8 different types (for example, INTx, ESFx, etc.), and had three dichotomies used for typing people (for example, I/E, T/F, N/S). If we could guess which side of each dichotomy a person was on with 90% accuracy, we could guess his type with 72.9% accuracy. That is, we might be 90% sure a person is a feeler, 90% sure that he's an extrovert, and 90% sure that he's a sensor, so we'd be 0.9^3 = 72.9% sure that all three of our guesses are correct and that he's an ESFx.

Based on these 3 basic dichotomies can add 4 more control (extra) in a similar way as they are introduced Reynin.
So, here is a combination of dichotomies types:

Where 1,2,3 - basic dichotomy
4,5,6,7 - control, where:
4 = 1 * 2
5 = 1 * 3
6 = 2 * 3
7 = 1 * 2 * 3
Horizontal lines: number of dichotomy
Vertical lines: number type.
Assume that the probability of a precise definition of control is also dichotomies 0.9.
Suppose that the true type - a type of number 1.

But we can also add more dichotomies to increase our chances of a correct typing; say we add four more dichotomies. These are superfluous in a sense, since they don't give us any extra information (for example, one dichotomy may have IxTxs and ExFxs on one side and ExTxs and IxFxs on the other, but we can figure out on which side of this dichotomy an individual belongs solely based on the earlier I/E and T/F dichotomies). Suppose that we can guess on which side of these new dichotomies a person lies with 90% accuracy. Now we will guess all 7 dichotomies correctly 0.9^7 = 47.8% of the time.

But even if we only guess 6 dichotomies correctly, we'll still get the type right, because of the 8 types, only one type will agree with 6 of our dichotomies, while the others will agree with 4. This depends on how the dichotomies are chosen; basically, they must be chosen so that every two types agree on exactly 3 dichotomies (see the chart with pluses and minuses for how this is possible). That way, if we get 6 out of 7 dichotomies right for a person of type X, only 3 of those dichotomies will be in common with a person of type Y; then even if we mistype on the 7th dichotomy, the person will have at most 4 dichotomies in common with type Y. So how often will this occur? For example, what are the odds of guessing the first dichotomy wrong and the rest right? That's simply 0.1 * (0.9)^6 = 5.3%. But you can also guess the second dichotomy wrong and the rest right with a probability of 5.3%; in total, then, there is a 7 * 0.053 = 37.1% of guessing exactly one dichotomy wrong, but getting the type right. So there is approximately a 37.1% + 47.8% = 85% chance of getting at most one dichotomy wrong, and therefore getting the type right.
For convenience of comparison of the results counted, just as it makes the algorithm Calc2008.txt

With this method of formation of any type of dichotomies is different from any other 4 dichotomies ie: the minimum distance = 4, which can be seen from the example of a type.
If all of the dichotomy identified correctly, then the code will be received by type 1 and type will be determined with certainty.
If a single mistake is made then get the following picture: the difference between the code of the first type will be equal to 1, between any other difference would be 3 and a type that will eventually set to as the most likely - the first, so that the bug will be fixed too.
In all such combinations = 7
Probability of occurrence of combinations of dichotomies without error = 0.478
Probability of occurrence of combinations of dichotomies with a single error = 0.053
Total probability of correct typing = 0.478 + 7 * 0.053 = 0.85
What is already more than the first: 0.729
Next: When receiving a combination of dichotomies with a double error type is classified as a "type not defined" because there are always two types, the minimum distance to which is the same.
The number of such combinations 7C2 = 21
The probability of this outcome = 7C2 * 0.9 * 0.1 ^ 5 ^ 2 = 0.124
The very same probability of error in this case is = 0.026
It turned out that the probability of correct identification of TIM is 97% and 3% error.
Progress is very significant.

2. 2 Traveler:
sling mud at others - it's what you can do? - Troll!

3. lol

4. Originally Posted by Yaaroslav
2 Traveler:
sling mud at others - it's what you can do? - Troll!
It seems like you mean well in your posts, but maybe you should make posts that are less mathematical heavy and more descriptive instead, that way you can get your message across better to the general public.

5. Originally Posted by Traveler
Originally Posted by Yaaroslav
2 Traveler:
sling mud at others - it's what you can do? - Troll!
It seems like you mean well in your posts, but maybe you should make posts that are less mathematical heavy and more descriptive instead, that way you can get your message across better to the general public.
If you consider that my articles can be simpler - do them simpler!
For example: I can't find the function which can present my table with 8 types looking like cells, so were the columns and rows. If you know how to do them - do them. I will be grateful for anyone who will contribute in the development socionic science

6. Originally Posted by Yaaroslav
Originally Posted by Traveler

It seems like you mean well in your posts, but maybe you should make posts that are less mathematical heavy and more descriptive instead, that way you can get your message across better to the general public.
If you consider that my articles can be simpler - do them simpler!
For example: I can't find the function which can present my table with 8 types looking like cells, so were the columns and rows. If you know how to do them - do them. I will be grateful for anyone who will contribute in the development socionic science
I don't think the problem is that it's too complex. The math makes it more interesting, as long as the writing is clear. But the problem is that there's an implicit assumption that the Reinin dichotomies can be assessed reliably. What I've read is that Reinin merely demonstrated the combinations, and that the descriptions came from Augusta based on observations of a couple of dozen students. That's just not a scientific enough way to determine what the behaviors are for the dichotomies. To make matters worse, many people assess dichotomies based on the words used to name them rather than on any more substantial description.

You yourself pointed out that the inter-rater reliability of expert Socionists is low. How much lower would it be for dichotomies about which there's even less agreement or expertise?

So you can do all those calculations, but it's "garbage in, garbage out." That's why I generally don't bother to read the mathematical stuff about Reinin; if the data are reliable to start with, why process it?

7. Originally Posted by Yaaroslav
If you consider that my articles can be simpler - do them simpler!
For example: I can't find the function which can present my table with 8 types looking like cells, so were the columns and rows. If you know how to do them - do them. I will be grateful for anyone who will contribute in the development socionic science
ok ok I'll see what I can do:

The table tags contain the table, the "|" bars separate the cells horizontally, and vertical separation is just done by hopping down a line and all

8. Originally Posted by woofwoofl
Originally Posted by Yaaroslav
If you consider that my articles can be simpler - do them simpler!
For example: I can't find the function which can present my table with 8 types looking like cells, so were the columns and rows. If you know how to do them - do them. I will be grateful for anyone who will contribute in the development socionic science
ok ok I'll see what I can do:

The table tags contain the table, the "|" bars separate the cells horizontally, and vertical separation is just done by hopping down a line and all
Nice job woof! You should translate his theory into layman terms, an LII and an SEE working together? Who would of knew?

9. Originally Posted by Traveler
Originally Posted by Yaaroslav
2 Traveler:
sling mud at others - it's what you can do? - Troll!
It seems like you mean well in your posts, but maybe you should make posts that are less mathematical heavy and more descriptive instead, that way you can get your message across better to the general public.
He's basically saying the following: suppose we grouped people into 8 different types (for example, INTx, ESFx, etc.), and had three dichotomies used for typing people (for example, I/E, T/F, N/S). If we could guess which side of each dichotomy a person was on with 90% accuracy, we could guess his type with 72.9% accuracy. That is, we might be 90% sure a person is a feeler, 90% sure that he's an extrovert, and 90% sure that he's a sensor, so we'd be 0.9^3 = 72.9% sure that all three of our guesses are correct and that he's an ESFx.

But we can also add more dichotomies to increase our chances of a correct typing; say we add four more dichotomies. These are superfluous in a sense, since they don't give us any extra information (for example, one dichotomy may have IxTxs and ExFxs on one side and ExTxs and IxFxs on the other, but we can figure out on which side of this dichotomy an individual belongs solely based on the earlier I/E and T/F dichotomies). Suppose that we can guess on which side of these new dichotomies a person lies with 90% accuracy. Now we will guess all 7 dichotomies correctly 0.9^7 = 47.8% of the time.

But even if we only guess 6 dichotomies correctly, we'll still get the type right, because of the 8 types, only one type will agree with 6 of our dichotomies, while the others will agree with 4. This depends on how the dichotomies are chosen; basically, they must be chosen so that every two types agree on exactly 3 dichotomies (see the chart with pluses and minuses for how this is possible). That way, if we get 6 out of 7 dichotomies right for a person of type X, only 3 of those dichotomies will be in common with a person of type Y; then even if we mistype on the 7th dichotomy, the person will have at most 4 dichotomies in common with type Y. So how often will this occur? For example, what are the odds of guessing the first dichotomy wrong and the rest right? That's simply 0.1 * (0.9)^6 = 5.3%. But you can also guess the second dichotomy wrong and the rest right with a probability of 5.3%; in total, then, there is a 7 * 0.053 = 37.1% of guessing exactly one dichotomy wrong, but getting the type right. So there is approximately a 37.1% + 47.8% = 85% chance of getting at most one dichotomy wrong, and therefore getting the type right.

Now what if we guess 5 dichotomies correctly? Well, then there's three types which are equally viable (for example, suppose we type someone as --+++++ using the table in the original post; then types 1, 7, and 8 are equally plausible, since they agree with our typing in 5 out of 7 dichotomies). The same is true in general; so how often does this occur? Well, there's 7C2 = 21 ways to get two dichotomies wrong. For example, say you get dichotomies x and y wrong; there are 7 choices for what x can be and 6 choices for what y can be (since x and y cannot be the same), but you must divide by 2 to avoid double-counting (e.g., getting dichotomies 1 and 6 wrong is the same as getting dichotomies 6 and 1 wrong). What are the odds of getting, say, dichotomies 1 and 2 wrong and getting the rest right? That's simply 0.1^2 * 0.9^5 = 0.59%. So the total chance of making exactly two errors is 21 * 0.0059 = 12.4%. From here, I think the original poster made a mistake. At this point, we have three types that fit our data equally well, so we can't do any better than randomly guessing amongst them. Hence, the chance of making two mistakes but guessing the correct type is 0.124/3 = 4.1%. Adding this all up, we have approximately a 85%+4.1% = 89% chance of guessing the correct type with 7 dichotomies, as opposed to a 72.9% chance with only 3 dichotomies.

10. Originally Posted by Traveler
Nice job woof! You should translate his theory into layman terms, an LII and an SEE working together? Who would of knew?
Thanks I think me and LIIs confuse each other more than out-and-out conflict

The biggest thing I take issue with here is the idea of "N" and "S" as hugely important groupings in and of themselves; a large amount of mistypings that I see come when "N" and "S" are looked at with more weight than the Judicious/Decisive Reinin dichotomy. I always make it a point to use the three letter typings, so those things are pushed to the background; to distance things from MBTI, to place more importance on quadras, on functional axes, on Reinin dichotomies, etc....

And yes, as usual, I think the problem isn't with having too much data, but with not having enough, and the time for me to post a math-related thing I've been working on for a while is fast approaching...

11. 2 woofwoofl Thanks!
I will correct my message!

Originally Posted by Begoner
Originally Posted by Traveler

It seems like you mean well in your posts, but maybe you should make posts that are less mathematical heavy and more descriptive instead, that way you can get your message across better to the general public.
He's basically saying the following: suppose we grouped people into 8 different types (for example, INTx, ESFx, etc.), and had three dichotomies used for typing people (for example, I/E, T/F, N/S). If we could guess which side of each dichotomy a person was on with 90% accuracy, we could guess his type with 72.9% accuracy. That is, we might be 90% sure a person is a feeler, 90% sure that he's an extrovert, and 90% sure that he's a sensor, so we'd be 0.9^3 = 72.9% sure that all three of our guesses are correct and that he's an ESFx.

But we can also add more dichotomies to increase our chances of a correct typing; say we add four more dichotomies. These are superfluous in a sense, since they don't give us any extra information (for example, one dichotomy may have IxTxs and ExFxs on one side and ExTxs and IxFxs on the other, but we can figure out on which side of this dichotomy an individual belongs solely based on the earlier I/E and T/F dichotomies). Suppose that we can guess on which side of these new dichotomies a person lies with 90% accuracy. Now we will guess all 7 dichotomies correctly 0.9^7 = 47.8% of the time.

But even if we only guess 6 dichotomies correctly, we'll still get the type right, because of the 8 types, only one type will agree with 6 of our dichotomies, while the others will agree with 4. This depends on how the dichotomies are chosen; basically, they must be chosen so that every two types agree on exactly 3 dichotomies (see the chart with pluses and minuses for how this is possible). That way, if we get 6 out of 7 dichotomies right for a person of type X, only 3 of those dichotomies will be in common with a person of type Y; then even if we mistype on the 7th dichotomy, the person will have at most 4 dichotomies in common with type Y. So how often will this occur? For example, what are the odds of guessing the first dichotomy wrong and the rest right? That's simply 0.1 * (0.9)^6 = 5.3%. But you can also guess the second dichotomy wrong and the rest right with a probability of 5.3%; in total, then, there is a 7 * 0.053 = 37.1% of guessing exactly one dichotomy wrong, but getting the type right. So there is approximately a 37.1% + 47.8% = 85% chance of getting at most one dichotomy wrong, and therefore getting the type right.

Now what if we guess 5 dichotomies correctly? Well, then there's three types which are equally viable (for example, suppose we type someone as --+++++ using the table in the original post; then types 1, 7, and 8 are equally plausible, since they agree with our typing in 5 out of 7 dichotomies). The same is true in general; so how often does this occur? Well, there's 7C2 = 21 ways to get two dichotomies wrong. For example, say you get dichotomies x and y wrong; there are 7 choices for what x can be and 6 choices for what y can be (since x and y cannot be the same), but you must divide by 2 to avoid double-counting (e.g., getting dichotomies 1 and 6 wrong is the same as getting dichotomies 6 and 1 wrong). What are the odds of getting, say, dichotomies 1 and 2 wrong and getting the rest right? That's simply 0.1^2 * 0.9^5 = 0.59%. So the total chance of making exactly two errors is 21 * 0.0059 = 12.4%. From here, I think the original poster made a mistake. At this point, we have three types that fit our data equally well, so we can't do any better than randomly guessing amongst them. Hence, the chance of making two mistakes but guessing the correct type is 0.124/3 = 4.1%. Adding this all up, we have approximately a 85%+4.1% = 89% chance of guessing the correct type with 7 dichotomies, as opposed to a 72.9% chance with only 3 dichotomies.
supposed that in case of result "type not defined" - Socionics just repeat the experiment that is These behaviors are excluded from the calculation

To all:
If the members of this forum will vote that Begoner's interpretation is more understandable - I will change the top of my theme

12. Originally Posted by Yaaroslav

To all:
If the members of this forum will vote that Begoner's interpretation is more understandable - I will change the top of my theme

13. Originally Posted by Analyst Trevor
Originally Posted by Yaaroslav

To all:
If the members of this forum will vote that Begoner's interpretation is more understandable - I will change the top of my theme
Done

15. Originally Posted by Galen

Galen gets it.

16. Originally Posted by Yaaroslav
Lets suppose we grouped people into 8 different types (for example, INTx, ESFx, etc.), and had three dichotomies used for typing people (for example, I/E, T/F, N/S). If we could guess which side of each dichotomy a person was on with 90% accuracy, we could guess his type with 72.9% accuracy. That is, we might be 90% sure a person is a feeler, 90% sure that he's an extrovert, and 90% sure that he's a sensor, so we'd be 0.9^3 = 72.9% sure that all three of our guesses are correct and that he's an ESFx.

Based on these 3 basic dichotomies can add 4 more control (extra) in a similar way as they are introduced Reynin.
So, here is a combination of dichotomies types:

Where 1,2,3 - basic dichotomy
4,5,6,7 - control, where:
4 = 1 * 2
5 = 1 * 3
6 = 2 * 3
7 = 1 * 2 * 3
Horizontal lines: number of dichotomy
Vertical lines: number type.
Assume that the probability of a precise definition of control is also dichotomies 0.9.
Suppose that the true type - a type of number 1.

But we can also add more dichotomies to increase our chances of a correct typing; say we add four more dichotomies. These are superfluous in a sense, since they don't give us any extra information (for example, one dichotomy may have IxTxs and ExFxs on one side and ExTxs and IxFxs on the other, but we can figure out on which side of this dichotomy an individual belongs solely based on the earlier I/E and T/F dichotomies). Suppose that we can guess on which side of these new dichotomies a person lies with 90% accuracy. Now we will guess all 7 dichotomies correctly 0.9^7 = 47.8% of the time.

But even if we only guess 6 dichotomies correctly, we'll still get the type right, because of the 8 types, only one type will agree with 6 of our dichotomies, while the others will agree with 4. This depends on how the dichotomies are chosen; basically, they must be chosen so that every two types agree on exactly 3 dichotomies (see the chart with pluses and minuses for how this is possible). That way, if we get 6 out of 7 dichotomies right for a person of type X, only 3 of those dichotomies will be in common with a person of type Y; then even if we mistype on the 7th dichotomy, the person will have at most 4 dichotomies in common with type Y. So how often will this occur? For example, what are the odds of guessing the first dichotomy wrong and the rest right? That's simply 0.1 * (0.9)^6 = 5.3%. But you can also guess the second dichotomy wrong and the rest right with a probability of 5.3%; in total, then, there is a 7 * 0.053 = 37.1% of guessing exactly one dichotomy wrong, but getting the type right. So there is approximately a 37.1% + 47.8% = 85% chance of getting at most one dichotomy wrong, and therefore getting the type right.
For convenience of comparison of the results counted, just as it makes the algorithm Calc2008.txt

With this method of formation of any type of dichotomies is different from any other 4 dichotomies ie: the minimum distance = 4, which can be seen from the example of a type.
If all of the dichotomy identified correctly, then the code will be received by type 1 and type will be determined with certainty.
If a single mistake is made then get the following picture: the difference between the code of the first type will be equal to 1, between any other difference would be 3 and a type that will eventually set to as the most likely - the first, so that the bug will be fixed too.
In all such combinations = 7
Probability of occurrence of combinations of dichotomies without error = 0.478
Probability of occurrence of combinations of dichotomies with a single error = 0.053
Total probability of correct typing = 0.478 + 7 * 0.053 = 0.85
What is already more than the first: 0.729
Next: When receiving a combination of dichotomies with a double error type is classified as a "type not defined" because there are always two types, the minimum distance to which is the same.
The number of such combinations 7C2 = 21
The probability of this outcome = 7C2 * 0.9 * 0.1 ^ 5 ^ 2 = 0.124
The very same probability of error in this case is = 0.026
It turned out that the probability of correct identification of TIM is 97% and 3% error.
Progress is very significant.
I read the first paragraph and you asked me to assume a bunch of stuff. No.

Then saw something to do with maths.

17. The short version is: if you are able to reliably guess with high accuracy 3 Jungian and several (carefully selected) Reinin dichotomies of a person you're trying to type, your chance of typing them correctly as compared to using just the 3 Jungian dichotomies is higher.

Making similar assumptions to the OP, if you can correctly identify 90% of the time the club, quadra and temperament of the individual in question,

Probability of occurrence of combinations without error = 0.9^3 = 0.729
Probability of occurrence of combinations of small groups with a single error = 0.9^2 x 0.1 x 2 = 0.162
Total probability of correct typing = 0.891

Which, if Begoner's calculations are correct, is the same as yours.

Now, you might say that club, quadra etc. are confounded with your original dichotomies of N/S, T/F etc. and therefore should have lower accuracies, but Reinin dichotomies aren't independent of Jungian dichotomies either and you've given them 90%.

And given that realistically, the chances of typing club/quadra/temperament correctly are much higher than typing Reinin dichotomies correctly, and you need to be able to type all 15 dichotomies accurately to get your 89% accuracy (as opposed to only 3 small groups in my example), I think small groups is a much better typing method.

18. Yeah I know, I had created a test based on this simple property and a process of further elimination. I can't find the thread now though, damn it.

19. Originally Posted by octopuslove
Making similar assumptions to the OP, if you can correctly identify 90% of the time the club, quadra and temperament of the individual in question,

Probability of occurrence of combinations without error = 0.9^3 = 0.729
Probability of occurrence of combinations of small groups with a single error = 0.9^2 x 0.1 x 2 = 0.162
Total probability of correct typing = 0.891
I don't think your numbers are correct. Say you type someone as an Alpha IP researcher, but have made a mistake in one of the categories. Of course, you don't know where you made a mistake, but you evidently made a mistake somewhere. You could be wrong about the Alpha part, in which case you have an INTp; you could be wrong about the IP part, in which case you have an INTj or ENTp; or you could be wrong about the researcher part, in which case you have an ISFp. So there are four possible types you have to consider.

Also, the probability of making a single error is 3 * 0.9^2 * 0.1 = 24.3%. However, having identified only two out of the four groups correctly, you will have four equally plausible candidates. Thus, probability of a correct typing having made one error would be about 6%, so the total probability of a correct typing would only be about 79%.

Edit: Now that I think about it, there needn't always be four possibilities. For example, if you type someone as an Alpha IP humanitarian, there are only two possibilities: ISFp and INFp. In fact, there are always either two or four possibilities (according to whether the quadra/club combination exists or not), and they occur equally frequently. Thus, we need divide 24.3% by 3 rather than 4, and we obtain a total probability of a correct typing of 81%. Note that this occurs because we are using groups of four, which are equivalent to two dichotomies; that is, typing according to 3 groups is equivalent to typing according to 6 dichotomies.

20. Originally Posted by Begoner
Originally Posted by octopuslove
Making similar assumptions to the OP, if you can correctly identify 90% of the time the club, quadra and temperament of the individual in question,

Probability of occurrence of combinations without error = 0.9^3 = 0.729
Probability of occurrence of combinations of small groups with a single error = 0.9^2 x 0.1 x 2 = 0.162
Total probability of correct typing = 0.891
I don't think your numbers are correct. Say you type someone as an Alpha IP researcher, but have made a mistake in one of the categories. Of course, you don't know where you made a mistake, but you evidently made a mistake somewhere. You could be wrong about the Alpha part, in which case you have an INTp; you could be wrong about the IP part, in which case you have an INTj or ENTp; or you could be wrong about the researcher part, in which case you have an ISFp. So there are four possible types you have to consider.
These kind of versions are considered: "Type not identified" and would be filtered, or the experiment of typing would be repeated.
So these cases - would be filtered

Originally Posted by Begoner
Also, the probability of making a single error is 3 * 0.9^2 * 0.1 = 24.3%.

However, having identified only two out of the four groups correctly, you will have four equally plausible candidates.
Using only two groups (Tetratomies) with a probability correct typing 0,9. I will achieve the total probability of typing 0,9^2 = 0,81
Back to example with a 3 Tetratomies:
(1) Typing without error 0,9^3 = 0,729
(2) Typing with a single error is 3 * 0.9^2 * 0.1 = 0,243 = "Type not identified"
(3)Two or more errors: 1 - 0,729 - 0,243 = 0,028
We do not calculate cases when "Type not identified", so then:
probability of true typing:
(Value1):=(Value1)/(1 - (Value2)) = 0,729/(1 - 0,243) = 0,963
The probability of error:
(Value3):=(Value3)/(1 - (Value2)) = 0,243/(1 - 0,243) = 0,036

The progress is significant, But the typist just became to be less productive

Originally Posted by Begoner
Thus, probability of a correct typing having made one error would be about 6%, so the total probability of a correct typing would only be about 79%.
Where have you get these numbers?

Originally Posted by Begoner
Edit: Now that I think about it, there needn't always be four possibilities. For example, if you type someone as an Alpha IP humanitarian, there are only two possibilities: ISFp and INFp. In fact, there are always either two or four possibilities (according to whether the quadra/club combination exists or not), and they occur equally frequently. Thus, we need divide 24.3% by 3 rather than 4, and we obtain a total probability of a correct typing of 81%. Note that this occurs because we are using groups of four, which are equivalent to two dichotomies; that is, typing according to 3 groups is equivalent to typing according to 6 dichotomies.
This logic is not correct

21. Originally Posted by Yaaroslav
Where have you get these numbers? This logic is not correct

I obtained them by assuming that one could not repeat the test an indefinite amount of times and obtain a different result each time. Thus, one would have to make an educated guess on the basis of a single test. The logic is quite correct, I assure you.

22. Originally Posted by Begoner
Originally Posted by Yaaroslav
Where have you get these numbers? This logic is not correct

I obtained them by assuming that one could not repeat the test an indefinite amount of times and obtain a different result each time. Thus, one would have to make an educated guess on the basis of a single test. The logic is quite correct, I assure you.
Give me calculations with formulas: Where you had get these numbers?

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