ICG 50-Sticks

An Approach to the 'I-CHING'

by Frank C. Fung ( 1st published in December, 2004. )

Section 6 : { Combinatorial Mathematics } in the 'I-CHING'

Topic I	Introduction & Summary for the Section
Topic II	Recalling the { Ceremony with the 50 Sticks }
Topic III	The { 1st Round }
Topic IV	The { 2nd Round }
Topic V	The { 3rd Round }
Topic VI	Recalling the Summary Table
Topic VII	Calculating the Probabilities ( 1 of 4 )
Topic VIII	Calculating the Probabilities ( 2 of 4 )
Topic IX	Calculating the Probabilities ( 3 of 4 )
Topic X	Calculating the Probabilities ( 4 of 4 )
Topic XI	A Statement of the Probabilities
Topic XII	Re-cap & further Questions

Topic I - Introduction & Summary for the Section :

In this section, we shall re-look at the { Ceremony with the 50 Sticks } from a { Combinatorial Mathematics } stand-point.

By the end of this Section, we will have come up with the values for :

the probability of a { 'Old Yang Line' } , ( a { moving [ 9 ] } ) ,
the probability of a { 'Young Yin Line' } , ( a { stationary [ 8 ] } ) ,
the probability of a { 'Young Yang Line' } , ( a { stationary [ 7 ] } ) ,
the probability of a { 'Old Yin Line' } , ( a { moving [ 6 ] } ) .

This then demonstrates the depth of understanding of { Combinatoiral Mathematics } required for fuller appreciation the 'I-CHING'.

Topic II - Recalling the { Ceremony with the 50 Sticks } :

Let us recall, from Section 4, the { Ceremony with the 50 Sticks }.

And the very 1st thing that we did here was to put aside { 1 stick } and work with the remaining { 49 sticks }.

Let us also recall that we have identified three (3) spaces, for easier-interpretation purposes :

[ Space-A ] : the space between the { Little Finger } and the { 4th Finger } on the Left Hand,
[ Space-B ] : the space between the { 4th Finger } and the { Middle Finger } on the Left Hand,
[ Space-C ] : the space between the { Middle Finger } and the { Index Finger } on the Left Hand.

Topic III - The { 1st Round } :

Let us recall that we started out with { 49 Sticks } in the { 1st Round } , for the combinatorial counting for [ 1st Hole ] in the Box .

The possible outcomes for the { 49 sticks } here are as follows :

Number of Sticks				TOTAL in 1st Hole = { Space-A } + { Space-B } + { Space-C }
in { Space-A }	in { Space-B }	in { Space-C }	remaining from the 'take 4-sticks at-a-time' procedure	TOTAL in 1st Hole = { Space-A } + { Space-B } + { Space-C }
1	1	3	44	5
1	2	2	44	5
1	3	1	44	5
1	4	4	40	9

Several things to note here :

By definition, there is always { 1 stick } in [ Space-A ] ,
the { Number of Sticks } remaining, from the { 'take 4 sticks at-a-time' } procedure is always a { multiple of '4' } ,

because you can only add { 4 sticks at-a-time } to this pile.
The { Number of Sticks in the 1st Hole } is therefore either [ 5 ] or [ 9 ] as per the last column on-the-right.
The number of Sticks remaining, for the next round, is either [ 44 ] or [ 40 ] .

Topic IV - The { 2nd Round } :

Let us recall that we can start out with either { 44 Sticks } or { 40 Sticks } in the { 2nd Round } , for the combinatorial counting for [ 2nd Hole ] in the Box .

The possible outcomes for the { 44 sticks } here are as follows :

Number of Sticks				TOTAL in 2nd Hole = { Space-A } + { Space-B } + { Space-C }
in { Space-A }	in { Space-B }	in { Space-C }	remaining from the 'take 4-sticks at-a-time' procedure	TOTAL in 2nd Hole = { Space-A } + { Space-B } + { Space-C }
1	1	2	40	4
1	2	1	40	4
1	3	4	36	8
1	4	3	36	8

The possible outcomes for the { 40 sticks } here are as follows :

Number of Sticks				TOTAL in 2nd Hole = { Space-A } + { Space-B } + { Space-C }
in { Space-A }	in { Space-B }	in { Space-C }	remaining from the 'take 4-sticks at-a-time' procedure	TOTAL in 2nd Hole = { Space-A } + { Space-B } + { Space-C }
1	1	2	36	4
1	2	1	36	4
1	3	4	32	8
1	4	3	32	8

Several things to note here :

By definition, there is always { 1 stick } in [ Space-A ] ,
the { Number of Sticks } remaining, from the { 'take 4 sticks at-a-time' } procedure is always a { multiple of '4' } ,

because you can only add { 4 sticks at-a-time } to this pile.
The { Number of Sticks in the 2nd Hole } is therefore either [ 4 ] or [ 8 ] as per the last column on-the-right.
The number of Sticks remaining, for the next round, is either [ 40 ] , [ 36 ] or [ 32 ] .

Topic V - The { 3rd Round } :

Let us recall that we can start out with either { 40 Sticks } , { 36 Sticks } , or { 32 Sticks } in the { 3rd Round } , for the combinatorial counting for the [ 3rd Hole ] in the Box .

The possible outcomes for the { 40 sticks } here are as follows :

Number of Sticks in				TOTAL in 3rd Hole = { Space-A } + { Space-B } + { Space-C }
in { Space-A }	in { Space-B }	in { Space-C }	remaining from the 'take 4-sticks at-a-time' procedure	TOTAL in 3rd Hole = { Space-A } + { Space-B } + { Space-C }
1	1	2	36	4
1	2	1	36	4
1	3	4	32	8
1	4	3	32	8

The possible outcomes for the { 36 sticks } here are as follows :

Number of Sticks				TOTAL in 3rd Hole = { Space-A } + { Space-B } + { Space-C }
in { Space-A }	in { Space-B }	in { Space-C }	remaining from the 'take 4-sticks at-a-time' procedure	TOTAL in 3rd Hole = { Space-A } + { Space-B } + { Space-C }
1	1	2	32	4
1	2	1	32	4
1	3	4	28	8
1	4	3	28	8

The possible outcomes for the { 32 sticks } here are as follows :

Number of Sticks				TOTAL in 3rd Hole = { Space-A } + { Space-B } + { Space-C }
in { Space-A }	in { Space-B }	in { Space-C }	remaining from the 'take 4-sticks at-a-time' procedure	TOTAL in 3rd Hole = { Space-A } + { Space-B } + { Space-C }
1	1	2	28	4
1	2	1	28	4
1	3	4	24	8
1	4	3	24	8

Several things to note here :

By definition, there is always { 1 stick } in [ Space-A ] ,
the { Number of Sticks } remaining, from the { 'take 4 sticks at-a-time' } procedure is always a { multiple of '4' }.

because you can only add { 4 sticks at-a-time } to this pile.
The { Number of Sticks in the 1st Hole } is therefore either [ 4 ] or [ 8 ] as per the last column on-the-right.
The number of Sticks remaining, at the end, is either [ 36 ] , [ 32 ] , [ 28 ] , or [ 24 ] .

Topic VI - Recalling the Summary Table :

Let us now re-look at this table previously presented in Section 4 :

Number of Sticks in			[ QI ] or [ OU ] designation			Resultant Line	Number of Sticks remaining
1st hole	2nd hole	3rd hole	1st hole	2nd hole	3rd hole	Resultant Line	after 1st Split	after 2nd Split	after 3rd Split
5	4	4	QI	QI	QI	OLD YANG	44	40	36
5	4	8	QI	QI	OU	YOUNG YIN	44	40	32
5	8	4	QI	OU	QI	YOUNG YIN	44	36	32
5	8	8	QI	OU	OU	YOUNG YANG	44	36	28
9	4	4	OU	QI	QI	YOUNG YIN	40	36	32
9	4	8	OU	QI	OU	YOUNG YANG	40	36	28
9	8	4	OU	OU	QI	YOUNG YANG	40	32	28
9	8	8	OU	OU	OU	OLD YIN	40	32	24

which is consistant with our findings above.

Topic VII - Calculating the Probabilities : ( 1 of 4 )

We note, first of all, that :

There is always one (1) , and only one, stick in [ Space-A ] ,
The minimum { number of sticks } in [ Space-B ] is [ 1 ] ,
The minimum { number of sticks } in [ Space-C ] is [ 1 ] .

For the { 49 Sticks } Split :

{ 1 Stick } must go into [ Space-A ] ,
{ 1 Stick } must go into [ Space-B ] ,
{ 1 Stick } must go into [ Space-C ] ,
and the remaining { 46 Sticks } may then go into either [ Space-B ] or [ Space-C ] .

For the { 44 Sticks } Split :

{ 1 Stick } must go into [ Space-A ] ,
{ 1 Stick } must go into [ Space-B ] ,
{ 1 Stick } must go into [ Space-C ] ,
and the remaining { 41 Sticks } may then go into either [ Space-B ] or [ Space-C ] .

For the { 40 Sticks } Split :

{ 1 Stick } must go into [ Space-A ] ,
{ 1 Stick } must go into [ Space-B ] ,
{ 1 Stick } must go into [ Space-C ] ,
and the remaining { 37 Sticks } may then go into either [ Space-B ] or [ Space-C ] .

For the { 36 Sticks } Split :

{ 1 Stick } must go into [ Space-A ] ,
{ 1 Stick } must go into [ Space-B ] ,
{ 1 Stick } must go into [ Space-C ] ,
and the remaining { 33 Sticks } may then go into either [ Space-B ] or [ Space-C ] .

For the { 32 Stick } Split :

{ 1 Stick } must go into [ Space-A ] ,
{ 1 Stick } must go into [ Space-B ] ,
{ 1 Stick } must go into [ Space-C ] ,
and the remaining { 29 Sticks } may then go into either [ Space-B ] or [ Space-C ] .

Let us now assume, the remaining { 46 sticks } / { 41 Sticks } / { 37 Sticks } / { 33 Sticks } / { 29 sticks } each have an equal probability of going into either [ Space-B ] or [ Space-C ] .

i.e. , for each of the remaining sticks :

probabilty of the stick going to [ Space-B ] = 0.50 ,
probabilty of the stick going to [ Space-C ] = 0.50 .

( The reader may wish to relax this assumption at a later stage. )

branch to Combinatorics Tables

Based on this assumption, we then provide herewith a branching to Combinatorics Tables for the numbers { 28 } thru { 49 } :

Braching to Combinatorics Tables for a given Value of [ N ]
The Value of [ N ]			49	48	47	46	45	44
	43	42	41	40	39	38	37	36
	35	34	33	32	31	30	29	28
	------	------	------	------	------	------	------	------

Topic VIII - Calculating the Probabilities : ( 2 of 4 )

Let us now focus on :

the number of sticks going into [ Space-B ] , additional to the one (1) single stick that is already a 'must' for [ Space-B ] .

And we shall designated this quantity as [ ANSB ] .

( [ ANSB ] then stands for { Additional Number of Sticks for [ Space-B ] )

We then present this Summary Table for the entire range from { N = [ 28 ] } to { N = [ 49 ] } , although the only numbers relevent here are the numbers [ 46 ] , [ 41 ] , [ 37 ] , [ 33 ] , & [ 29 ] , as marked below.

Value of 'N'	Probability of [ ANSB ] being equal to { 0 mod 4 }	Probability of [ ANSB ] being equal to { 1 mod 4 }	Probability of [ ANSB ] being equal to { 2 mod 4 }	Probability of [ ANSB ] being equal to { 3 mod 4 }	Value of DELTA
49	0.25 + DELTA	0.25 + DELTA	0.25 - DELTA	0.25 - DELTA	0.25 * { 1 / ( 2^24 ) }
48	0.25 + DELTA	0.25	0.25 - DELTA	0.25	0.25 * { 1 / ( 2^23 ) }
47	0.25 + DELTA	0.25 - DELTA	0.25 - DELTA	0.25 + DELTA	0.25 * { 1 / ( 2^23 ) }
46	0.25	0.25 - DELTA	0.25	0.25 + DELTA	0.25 * { 1 / ( 2^22 ) }
45	0.25 - DELTA	0.25 - DELTA	0.25 + DELTA	0.25 + DELTA	0.25 * { 1 / ( 2^22 ) }
44	0.25 - DELTA	0.25	0.25 + DELTA	0.25	0.25 * { 1 / ( 2^21 ) }
43	0.25 - DELTA	0.25 + DELTA	0.25 + DELTA	0.25 - DELTA	0.25 * { 1 / ( 2^21 ) }
42	0.25	0.25 + DELTA	0.25	0.25 - DELTA	0.25 * { 1 / ( 2^20 ) }
41	0.25 + DELTA	0.25 + DELTA	0.25 - DELTA	0.25 - DELTA	0.25 * { 1 / ( 2^20 ) }
40	0.25 + DELTA	0.25	0.25 - DELTA	0.25	0.25 * { 1 / ( 2^19 ) }
39	0.25 + DELTA	0.25 - DELTA	0.25 - DELTA	0.25 + DELTA	0.25 * { 1 / ( 2^19 ) }
38	0.25	0.25 - DELTA	0.25	0.25 + DELTA	0.25 * { 1 / ( 2^18 ) }
37	0.25 - DELTA	0.25 - DELTA	0.25 + DELTA	0.25 + DELTA	0.25 * { 1 / ( 2^18 ) }
36	0.25 - DELTA	0.25	0.25 + DELTA	0.25	0.25 * { 1 / ( 2^17 ) }
35	0.25 - DELTA	0.25 + DELTA	0.25 + DELTA	0.25 - DELTA	0.25 * { 1 / ( 2^17 ) }
34	0.25	0.25 + DELTA	0.25	0.25 - DELTA	0.25 * { 1 / ( 2^16 ) }
33	0.25 + DELTA	0.25 + DELTA	0.25 - DELTA	0.25 - DELTA	0.25 * { 1 / ( 2^16 ) }
32	0.25 + DELTA	0.25	0.25 - DELTA	0.25	0.25 * { 1 / ( 2^15 ) }
31	0.25 + DELTA	0.25 - DELTA	0.25 - DELTA	0.25 + DELTA	0.25 * { 1 / ( 2^15 ) }
30	0.25	0.25 - DELTA	0.25	0.25 + DELTA	0.25 * { 1 / ( 2^14 ) }
29	0.25 - DELTA	0.25 - DELTA	0.25 + DELTA	0.25 + DELTA	0.25 * { 1 / ( 2^14 ) }
28	0.25 - DELTA	0.25	0.25 + DELTA	0.25	0.25 * { 1 / ( 2^13 ) }

We have taken the liberty to mark :

those probabilities exactly equal to [ 0.25 ] , or [ one-quarter ] , in { green-color } ,
Those probabilities greater than [ 0.25 ] in { red-color } ,
those probabilities less than [ 0.25 ] in { blue-color } .

The most important thing to note here is that :

the probabilities of :
- [ ANSB ] = 0 modulo 4 ,
- [ ANSB ] = 1 modulo 4 ,
- [ ANSB ] = 2 modulo 4 ,
- [ ANSB ] = 3 modulo 4 ;
  
  are never { all equal to exactly 0.25 } .

The reader may now notice a { cycle of '8' } pattern in the above table, namely :

that the pattern for the numbers [ 33 ] , [ 41 ] , & [ 49 ] are the similar ,
that the pattern for the numbers [ 32 ] , [ 40 ] , & [ 48 ] are the similar ,
And so-on-and-so-forth .

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Topic IX - Calculating the Probabilities : ( 3 of 4 )

Let us now summarize the probabilities as follows :

For the { 49 Sticks Split } for the [ 1st Hole ] :

the probability of a [ 5 ] is [ { 0.25 } * { 3 - ( 1 / 2^22 ) } ] ,
the probability of a [ 9 ] is [ { 0.25 } * { 1 + ( 1 / 2^22 ) } ] .

Value of [ ANSB ]	Probability	Number of Sticks			TOTAL number of Sticks	[ QI ] or [ OU ]
Value of [ ANSB ]	Probability	in Space A	in Space B	in Space C	TOTAL number of Sticks	[ QI ] or [ OU ]
{ 0 modulo 4 }	0.25	1	1	3	5	[ QI ]
{ 1 modulo 4 }	0.25 * { 1 - ( 1 / 2^22 ) }	1	2	2	5	[ QI ]
{ 2 modulo 4 }	0.25	1	3	1	5	[ QI ]
{ 3 modulo 4 }	0.25 * { 1 + ( 1 / 2^22 ) }	1	4	4	9	[ OU ]

For the { 44 Sticks Split } for the [ 2nd Hole ] :

the probability of a [ 4 ] is [ { 0.50 } * { 1 - ( 1 / 2^20 ) } ] ,
the probability of a [ 8 ] is [ { 0.50 } * { 1 + ( 1 / 2^20 ) } ] .

Value of [ ANSB ]	Probability	Number of Sticks			TOTAL number of Sticks	[ QI ] or [ OU ]
Value of [ ANSB ]	Probability	in Space A	in Space B	in Space C	TOTAL number of Sticks	[ QI ] or [ OU ]
{ 0 modulo 4 }	0.25 * { 1 - ( 1 / 2^20 ) }	1	1	2	4	[ QI ]
{ 1 modulo 4 }	0.25 * { 1 - ( 1 / 2^20 ) }	1	2	1	4	[ QI ]
{ 2 modulo 4 }	0.25 * { 1 + ( 1 / 2^20 ) }	1	3	4	8	[ OU ]
{ 3 modulo 4 }	0.25 * { 1 + ( 1 / 2^20 ) }	1	4	3	8	[ OU ]

For the { 40 Sticks Split } for the [ 2nd Hole ] and/or [ 3rd Hole ] :

the probability of a [ 4 ] is [ { 0.50 } * { 1 - ( 1 / 2^18 ) } ] ,
the probability of a [ 8 ] is [ { 0.50 } * { 1 + ( 1 / 2^18 ) } ] .

Value of [ ANSB ]	Probability	Number of Sticks			TOTAL number of Sticks	[ QI ] or [ OU ]
Value of [ ANSB ]	Probability	in Space A	in Space B	in Space C	TOTAL number of Sticks	[ QI ] or [ OU ]
{ 0 modulo 4 }	0.25 * { 1 - ( 1 / 2^18 ) }	1	1	2	4	[ QI ]
{ 1 modulo 4 }	0.25 * { 1 - ( 1 / 2^18 ) }	1	2	1	4	[ QI ]
{ 2 modulo 4 }	0.25 * { 1 + ( 1 / 2^18 ) }	1	3	4	8	[ OU ]
{ 3 modulo 4 }	0.25 * { 1 + ( 1 / 2^18 ) }	1	4	3	8	[ OU ]

For the { 36 Sticks Split } for the [ 3rd Hole ] :

the probability of a [ 4 ] is [ { 0.50 } * { 1 - ( 1 / 2^16 ) } ] ,
the probability of a [ 8 ] is [ { 0.50 } * { 1 + ( 1 / 2^16 ) } ] .

Value of [ ANSB ]	Probability	Number of Sticks			TOTAL number of Sticks	[ QI ] or [ OU ]
Value of [ ANSB ]	Probability	in Space A	in Space B	in Space C	TOTAL number of Sticks	[ QI ] or [ OU ]
{ 0 modulo 4 }	0.25 * { 1 - ( 1 / 2^16 ) }	1	1	2	4	[ QI ]
{ 1 modulo 4 }	0.25 * { 1 - ( 1 / 2^16 ) }	1	2	1	4	[ QI ]
{ 2 modulo 4 }	0.25 * { 1 + ( 1 / 2^16 ) }	1	3	4	8	[ OU ]
{ 3 modulo 4 }	0.25 * { 1 + ( 1 / 2^16 ) }	1	4	3	8	[ OU ]

For the { 32 Sticks Split } for the [ 3rd Hole ] :

the probability of a [ 4 ] is [ { 0.50 } * { 1 - ( 1 / 2^14 ) } ] ,
the probability of a [ 8 ] is [ { 0.50 } * { 1 + ( 1 / 2^14 ) } ] .

Value of [ ANSB ]	Probability	Number of Sticks			TOTAL number of Sticks	[ QI ] or [ OU ]
Value of [ ANSB ]	Probability	in Space A	in Space B	in Space C	TOTAL number of Sticks	[ QI ] or [ OU ]
{ 0 modulo 4 }	0.25 * { 1 - ( 1 / 2^14 ) }	1	1	2	4	[ QI ]
{ 1 modulo 4 }	0.25 * { 1 - ( 1 / 2^14 ) }	1	2	1	4	[ QI ]
{ 2 modulo 4 }	0.25 * { 1 + ( 1 / 2^14 ) }	1	3	4	8	[ OU ]
{ 3 modulo 4 }	0.25 * { 1 + ( 1 / 2^14 ) }	1	4	3	8	[ OU ]

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Topic X - Calculating the Probabilities : ( 4 of 4 )

For our next table, derived from the 5 tables immediately above, we then set up the following values for easier-notation purposes :

ZETA	=	1 / ( 2^22 )
ALPHA	=	1 / ( 2^20 )
BETA	=	1 / ( 2^18 )
GAMMA	=	1 / ( 2^16 )
PSI	=	1 / ( 2^14 )

We then have the table below for the probabilities at the { 1st / 2nd / 3rd Splits } :

I.D.	Line Type	# of stick at Split			Probability at 1st split	Probability at 2nd split	Probability at 3rd split
I.D.	Line Type	1 s t	2 n d	3 r d	Probability at 1st split	Probability at 2nd split	Probability at 3rd split
5-4-4	Old YANG	49	44	40	0.25 * ( 3 - ZETA )	0.50 * ( 1 + ALPHA )	0.50 * ( 1 - BETA )
5-4-8	Young YIN	49	44	40	0.25 * ( 3 - ZETA )	0.50 * ( 1 + ALPHA )	0.50 * ( 1 + BETA )
5-8-4	Young YIN	49	44	36	0.25 * ( 3 - ZETA )	0.50 * ( 1 - ALPHA )	0.50 * ( 1 + GAMMA )
5-8-8	Young YANG	49	44	36	0.25 * ( 3 - ZETA )	0.50 * ( 1 - ALPHA )	0.50 * ( 1 - GAMMA )
9-4-4	Young YIN	49	40	36	0.25 * ( 1 + ZETA )	0.50 * ( 1 - BETA )	0.50 * ( 1 + GAMMA )
9-4-8	Young YANG	49	40	36	0.25 * ( 1 + ZETA )	0.50 * ( 1 - BETA )	0.50 * ( 1 - GAMMA )
9-8-4	Young YANG	49	40	32	0.25 * ( 1 + ZETA )	0.50 * ( 1 + BETA )	0.50 * ( 1 - PSI )
9-8-8	Old YIN	49	40	32	0.25 * ( 1 + ZETA )	0.50 * ( 1 + BETA )	0.50 * ( 1 + PSI )

We shall now proceed to calculate the probabilities, i.e. :

{ Probability } = { Probability at 1st Split } * { Probability at 2nd Split } * { Probability at 3rd Split }

But for the Calculations Table below, let us first take-out the common factor :

the value { 0.25 * 0.50 * 0.50 } = 0.0625 or [ 1 / 16 ]

for the time-being.

		Negative Powers of Two
		0	14	16	18	20	22	32	34	36	38	40	42	44	54	56	58	60	62	64	66
5-4-4	Old YANG	+3			-3	+3	-1				-3	+1	-1					+1
5-4-8	Young YIN	+3			+3	+3	-1				+3	-1	-1					-1
5-8-4	Young YIN	+3		+3		-3	-1			-3	-1		+1				+1
5-8-8	Young YANG	+3		-3		-3	-1			+3	+1		+1				-1
9-4-4	Young YIN	+1		+1	-1		+1		-1		+1	-1				-1
9-4-8	Young YANG	+1		-1	-1		+1		+1		-1	-1				+1
9-8-4	Young YANG	+1	-1		+1		+1	-1		-1		+1			-1
9-8-8	Old YIN	+1	+1		+1		+1	+1		+1		+1			+1
Sum of Column		16	0	0	0	0	0	0	0	0	0	0	0		0	0	0	0

We note here that :

the 1st { data Column on-the-left } , marked [ 2^0 ] , adds-up to [ 16 ] ,
all other { data Columns }, adds-up to [ 0 ] ,

And this is consistant with the fact that we have taken-out the factor of [ 1 / 16 ] for this table, i.e. :

The sum of the probabilities for all entries { adds-up-to-[ 1 ] } , always.

We then arrive, after further calculations, at this Summary Table below :

	Probability is equal to :
	+ { 1 / 16 } *	+ { 1 / 16 } * { 1 / 2^22 } *	+ { 1 / 16 } * { 1 / 2^42 } *	+ { 1 / 16 } * { 1 / 2^60 } *
Old YANG	+3	-37	-45	+1
Young YIN	+7	+287	-408	-31
Young YANG	+5	-523	-639	-34
Old YIN	+1	+273	+1092	+64

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Topic XI - The Statement of the Probabilities :

Let us now restate the above probabilities in the more familiar format :

Probability of a { 'OLD YANG Line' }

{ 3 / 16 }

* { 1 / 2^26 }

* { 1 / 2^46 }

* { 1 / 2^64 }

Probability of a { 'YOUNG YIN Line' }

{ 7 / 16 }

287

* { 1 / 2^26 }

408

* { 1 / 2^46 }

* { 1 / 2^64 }

Probability of a { 'YOUNG YANG Line' }

{ 5 / 16 }

523

* { 1 / 2^26 }

639

* { 1 / 2^46 }

* { 1 / 2^64 }

Probability of a { 'OLD YIN Line' }

{ 1 / 16 }

273

* { 1 / 2^26 }

1092

* { 1 / 2^46 }

* { 1 / 2^64 }

Incidently, { 2^64 } , in [ base 10 ] , is [ 18,446,744,073,709,551,616 ] , a { twenty-digits-number } .

We can then set up the four (4) values, [ N1 ] , [ N2 ] , [ N3 ] , & [ N4 ] respectively :

N1	=	3	* 2^60	-	37	* 2^38	-	45	* 2^18	+	1
N2	=	7	* 2^60	+	287	* 2^38	-	408	* 2^18	-	31
N3	=	5	* 2^60	-	523	* 2^38	-	639	* 2^18	-	34
N4	=	1	* 2^60	+	273	* 2^38	+	1092	* 2^18	+	64

so that :

Probability of a { OLD YANG Line }	=	N1	/ 2^64
Probability of a { YOUNG YIN Line }	=	N2	/ 2^64
Probability of a { YOUNG YANG Line }	=	N3	/ 2^64
Probability of a { OLD YIN Line }	=	N4	/ 2^64

We note, of course, that :

[ N1 ] + [ N2 ] + [ N3 ] + [ N4 ] = 2^64

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Topic XII - Re-cap & further Questions :

From a rather simple procedure, the { Ceremony with the 50 Sticks } , we went thru a rather long procedure in { Combinatorial Mathematics } to arrive at four (4) probabilities.

Several questions remain :

What significance, if any, can be attached to the numbers [ N1 ] , [ N2 ] , [ N3 ] , & [ N4 ] ?
Does the tiny fraction attached to each of the four (4) probabilites link us onto [ C.H.A.O.S. ] ?
- There is an old Chinese saying :
  
  { Differ by a tiny fraction , miss it by a thoudand miles }
Suggested key-searchwords on { C.H.A.O.S. } : [ chaos ] , [ fractel ] , [ lorenz ] .

An Approach to the 'I-CHING'

by Frank C. Fung ( 1st published in December, 2004. )

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Section 6 : { Combinatorial Mathematics } in the 'I-CHING'

Topic I - Introduction & Summary for the Section :

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Topic II - Recalling the { Ceremony with the 50 Sticks } :

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Topic III - The { 1st Round } :

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Topic IV - The { 2nd Round } :

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Topic V - The { 3rd Round } :

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Topic VI - Recalling the Summary Table :

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Topic VII - Calculating the Probabilities : ( 1 of 4 )

branch to Combinatorics Tables

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Topic VIII - Calculating the Probabilities : ( 2 of 4 )

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Topic IX - Calculating the Probabilities : ( 3 of 4 )

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Topic X - Calculating the Probabilities : ( 4 of 4 )

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Topic XI - The Statement of the Probabilities :

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Topic XII - Re-cap & further Questions :

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go to the next Section : Section 7 - The { Parity of '4' } in the 'I-CHING' & { Non-parity }

go to the last Section : Section 5 - The { ICG Hexagram Placement Scheme }

return to the ICG HomePage

original dated 2004-12-18