How to cast the I Ching: three coins or fifty yarrow stalks
Both methods do the same job: they produce six numbers, each 6, 7, 8 or 9, and those six numbers are your hexagram. They do not produce them with the same odds. This page walks through each method step by step, then works out exactly how often each number comes up and why.
The short version
Each throw gives one line, and you build the hexagram from the bottom up: the first number is line 1, the bottom line, and the sixth is the top. The four numbers mean:
- 7, young yang: a solid line that stays solid.
- 8, young yin: a broken line that stays broken.
- 9, old yang: a solid line that is moving, about to turn broken.
- 6, old yin: a broken line that is moving, about to turn solid.
If any line came out 6 or 9, flip every one of those lines and you get a second hexagram, the one your reading changes into. With coins, a moving line turns up once in every four lines on average. With yarrow stalks it is also once in four, but a moving solid line is three times as common as a moving broken one. The odds section shows where that comes from.
The three-coin method, step by step
- Take three coins. Any three of the same kind will do. Traditional Chinese coins had writing on one side only [WIKI-LOT].
- Decide which face counts as what before you throw. The common English description counts heads as 3 (yang) and tails as 2 (yin) [WIKI-DIV]. This site’s Oracle uses that convention.
- Throw all three together and add the faces. Three heads make 9, two heads make 8, one head makes 7, no heads make 6. No other total is possible.
- Write the total down as line 1, the bottom line. Throw five more times, writing each total above the last, so the sixth throw is the top line.
- Draw 7 and 9 as a solid line and 8 and 6 as a broken one. Mark the 6s and 9s: those are the moving lines.
- Look the six lines up (the lookup on the hexagram list takes the six numbers as you type them), then flip every 6 and 9 to find the hexagram it changes into.
Conventions vary, and some books swap the faces. Another description counts the coin’s written side as yin, worth 2, and the blank side as yang, worth 3 [WIKI-LOT]. If you call the written side heads, that is the opposite of the rule above. It makes no difference to the odds, because each face is equally likely, so pick one convention and keep to it for the whole cast.
Worked throw: six lines, one hexagram changing into another
| Line | Coins | Total | Drawn as |
|---|---|---|---|
| 1 (bottom) | H T T | 3 + 2 + 2 = 7 | solid |
| 2 | H H H | 3 + 3 + 3 = 9 | solid, moving |
| 3 | H H T | 3 + 3 + 2 = 8 | broken |
| 4 | H T H | 3 + 2 + 3 = 8 | broken |
| 5 | T T T | 2 + 2 + 2 = 6 | broken, moving |
| 6 (top) | T H T | 2 + 3 + 2 = 7 | solid |
moving lines 2 (○) and 5 (×)
Read from the bottom, the six totals 7, 9, 8, 8, 6, 7 draw two solid lines, three broken lines and a solid line on top: Mountain over Lake, hexagram 41. Lines 2 and 5 are moving. Flip just those two (the solid line 2 turns broken and the broken line 5 turns solid) and the figure becomes Wind over Thunder, hexagram 42. Decrease changing into Increase. With two moving lines, the classical rule is to read both of hexagram 41’s moving lines, with the upper one, line 5, as the main text. How to read changing lines covers every case from none to six.
The yarrow-stalk method, step by step
This is the older method, more than a thousand years older than coins [WIKI-DIV]. You need 50 stalks: dried yarrow stems, or any thin sticks of the same length. Each line takes three rounds of dividing and counting, called changes, so a whole hexagram takes eighteen. The steps below are the ones this site’s Oracle simulates when you choose Yarrow stalks.
- Set one stalk aside. It takes no further part. The other 49 are the bundle you work with.
- Divide. Split the bundle into two heaps, left and right, without counting.
- Take one. Take one stalk from the right-hand heap and hold it apart, traditionally between the fingers [WIKI-DIV].
- Count off by fours. Remove the left heap four stalks at a time until 1, 2, 3 or 4 remain. A heap that divides exactly leaves the last 4, never 0. Do the same with the right heap.
- Set the remainders aside. The held stalk plus the two remainders come to 5 or 9 on the first change. Put them aside and score the change: 5 counts 3, 9 counts 2.
- Repeat twice with what is left. Gather the stalks still in play (44 or 40) and do steps 2 to 5 again, then once more. On the second and third changes the set-aside bundle is 4 (counts 3) or 8 (counts 2).
- Add the three scores. 3 + 3 + 3 = 9, two 3s and a 2 = 8, one 3 and two 2s = 7, three 2s = 6. As a check, divide the stalks still in play by four: 36, 32, 28 or 24 stalks give the same 9, 8, 7 or 6.
- That is one line. Gather all 49 stalks and repeat for the other five lines, bottom to top.
Worked line: three changes that make an 8
| Change | Left / right | Set aside | Score |
|---|---|---|---|
| 1 | 22 / 27 of 49 | 1 + 2 + 2 = 5 | 3 |
| 2 | 19 / 25 of 44 | 1 + 3 + 4 = 8 | 2 |
| 3 | 17 / 19 of 36 | 1 + 1 + 2 = 4 | 3 |
The remainders are counted after the held stalk leaves the right heap: in change 1 the right heap is then 26 (remainder 2), in change 2 it is 24, which divides exactly and so leaves 4, and in change 3 it is 18 (remainder 2). Scores 3 + 2 + 3 = 8, a young yin. Check: 49 − 5 − 8 − 4 leaves 32 stalks in play, and 32 ÷ 4 = 8.
Where descriptions differ
The classical source, the Great Commentary (繫辭) attached to the Book of Changes, gives only an outline: 大衍之數五十,其用四十有九,分而為二以象兩,掛一以象三,揲之以四以象四時,歸奇於扐以象閏, “the number of the great expansion is fifty, of which forty-nine are used; divide them in two to represent the two; hang up one to represent the three; count them off by fours to represent the four seasons; return the remainder between the fingers to represent the intercalary month” [XICI]. The same passage says 十有八變而成卦, “eighteen changes make a hexagram”, which is three per line.
The detailed procedure above is a reconstruction from that outline. The version still used throughout East Asia is Zhu Xi’s, from the twelfth century, and the modern scholar Gao Heng made his own, which differs from Zhu Xi’s in places [WIKI-IC]. Two places where the outline leaves a choice:
- Which heap gives up the held stalk. The old text says only “hang up one”. The common English description, and this site’s Oracle, take it from the right-hand heap [WIKI-DIV]. As long as one stalk is held apart, the odds below come out the same either way.
- Scoring. Scoring each change 3 or 2 and adding is the same as dividing the stalks left in play by four. The old text supports the second form: it counts 乾之策,二百一十有六 (216 stalks for Qian, the all-yang hexagram) and 坤之策,百四十有四 (144 for Kun, the all-yin one) [XICI]. 216 is 6 × 36 and 144 is 6 × 24, the stalks left in play for six 9s and six 6s. That match is this page’s arithmetic, not a claim the text makes in so many words.
Why the odds differ
| Total | Line | Three coins | Yarrow stalks |
|---|---|---|---|
| 6 | old yin, moving | 1/8 12.5% | 1/16 6.25% |
| 7 | young yang | 3/8 37.5% | 5/16 31.25% |
| 8 | young yin | 3/8 37.5% | 7/16 43.75% |
| 9 | old yang, moving | 1/8 12.5% | 3/16 18.75% |
What this means: both methods give a moving line 1 time in 4 (coins 1/8 + 1/8, yarrow 1/16 + 3/16), but with yarrow a moving yang is three times as likely as a moving yin. Both also give solid and broken lines half the time each, so every one of the 64 hexagrams is equally likely as the cast hexagram under either method; what changes is which lines move.
The coin arithmetic
Three coins can land in 2 × 2 × 2 = 8 equally likely ways, if each coin is fair. Writing H for heads (3) and T for tails (2):
- TTT: total 6. 1 way in 8.
- HTT, THT, TTH: total 7. 3 ways in 8.
- HHT, HTH, THH: total 8. 3 ways in 8.
- HHH: total 9. 1 way in 8.
The yarrow arithmetic
Everything turns on what the left heap leaves when it is counted off by fours: 1, 2, 3 or 4. Once you know that, the right heap is fixed, because the total is fixed.
- First change (49 stalks, 48 after holding one). The two heaps together hold 48, a multiple of four, so the remainders go 1 + 3, 2 + 2, 3 + 1 or 4 + 4. With the held stalk, that sets aside 5, 5, 5 or 9. So the first change scores 3 with chance 3/4 and 2 with chance 1/4.
- Second and third changes (44 or 40 stalks, 43 or 39 after holding one, both 3 more than a multiple of four). The remainders go 1 + 2, 2 + 1, 3 + 4 or 4 + 3, and the bundles set aside are 4, 4, 8 or 8. So each later change scores 3 or 2 with chance 1/2 each.
The three changes are independent, so multiply along each order of scores and add the orders that give the same total:
- 6 = 2 + 2 + 2: 1/4 × 1/2 × 1/2 = 1/16.
- 7 = one 3 and two 2s: 3/4 × 1/2 × 1/2 (the 3 first) + 1/4 × 1/2 × 1/2 (second) + 1/4 × 1/2 × 1/2 (third) = 3/16 + 1/16 + 1/16 = 5/16.
- 8 = two 3s and a 2: 3/4 × 1/2 × 1/2 (2 in the third change) + 3/4 × 1/2 × 1/2 (2 in the second) + 1/4 × 1/2 × 1/2 (2 in the first) = 3/16 + 3/16 + 1/16 = 7/16.
- 9 = 3 + 3 + 3: 3/4 × 1/2 × 1/2 = 3/16.
The assumption behind these numbers: each time you divide, the left heap is equally likely to leave 1, 2, 3 or 4 when counted off by fours. A fair random split does that almost exactly. A real hand may not: someone who always splits the bundle near the middle could shift the odds a little. The Oracle’s simulation deals each stalk to the left or right heap on its own fair toss, which keeps the four remainders even to within a small fraction of a percent. Run over hundreds of thousands of lines, it lands on the table above. This page’s own test checks that on every build. The same 1/16, 5/16, 7/16, 3/16 split is the one given in the common English description [WIKI-DIV].
Cast it now
The Oracle does either method for you and draws both hexagrams. Under How to cast, Three coins is already selected. To use the stalks, pick Yarrow stalks before you press Cast the Oracle; the result then shows the three scores of every line.
Open the Oracle and choose your method under How to cast.
Threw your own coins or stalks? Type the six numbers into the lookup on the hexagram list. If some lines moved, how to read changing lines tells you which texts to read.
Sources
[WIKI-DIV] ‘I Ching divination’, English Wikipedia, sections ‘Yarrow stalks’ and ‘Three-coin method’: en.wikipedia.org/wiki/I_Ching_divination, revision 1377292647, read 2026-10-01.
[WIKI-LOT] ‘Binary lot’, English Wikipedia, which cites Richard Rutt’s Zhouyi (2002) for the written-side-yin convention: en.wikipedia.org/wiki/Binary_lot, revision 1365268014, read 2026-10-01.
[WIKI-IC] ‘I Ching’, English Wikipedia, section on divination: en.wikipedia.org/wiki/I_Ching, revision 1376299998, read 2026-10-01.
[XICI] Xici zhuan (繫辭上), chapter 9, the Chinese text: Chinese Wikisource transcription, revision 2611140, read 2026-10-01. The English renderings above are this site’s own, made for this page; they are not Legge’s.
The probability table is not copied from any source: it is worked out from the procedure as described on this page, and checked against the Oracle’s own simulation.