Practical guide
Shared Cards Between Two Tarot Spreads: Reset, Overlap and Order
Two independently reset three-card spreads from a uniform 78-identity pool share at least one identity with probability 11.240076%. The same three names in any order and the same three assigned positions are smaller, different events. State the reset rule before comparing them.
Write the reset rule before comparing two spreads
A card can reappear in a later spread even when each individual spread contains no duplicate physical card. The key question is what happened between them. Did you return every card to the full pool before the second selection, or did the first spread remain out?
This guide uses a precise reference: seventy-eight distinct identities, three cards chosen uniformly without replacement for each spread, the entire pool restored between spreads, and the two selections independent. Within a spread the cards are not independent picks; after selecting one, it is unavailable for that spread. Between the two restored spreads, independence is a separate model assumption.
Fix the first three cards as A, B and C. They can stand for any three distinct identities. Ask how many of those appear anywhere in the second three-card set. Count each shared identity once, ignore orientation for this question, and leave assigned position for a separate comparison.
The repeated-card guide explains independent single-card picks with replacement. Treating six cards across these two spreads as six such picks would allow repeats within a spread and describe a different experiment. Neither calculation certifies a physical shuffle or the actual selection algorithm of an app.
Count zero, one, two or three shared identities
After fixing A, B and C, the restored pool contains three identities from that first set and seventy-five outside it. For exactly K shared identities, choose K from the three and the other 3−K from the seventy-five. Divide by all C(78,3) = 76,076 possible second sets.
The expression is P(K=k) = C(3,k)C(75,3−k)/C(78,3). C(N,n) means an unordered choice of n distinct objects from N. Each qualifying second set has one unique shared subset, so the multiplication counts it once. The first spread does not also need to be a particular named set chosen before either draw; it is fixed for this comparison.
For example, exactly one shared identity can be A, B or C, giving three choices. Choose two outsiders in C(75,2) = 2,775 ways: 3 × 2,775 = 8,325. Exactly two shared identities gives C(3,2) × 75 = 225. All three shared gives only the set A, B, C.
The zero-overlap count is C(75,3) = 67,525. Its complement gives at least one shared identity: (76,076−67,525)/76,076 = 8,551/76,076 = 11.240076%. “At least one” includes the one-, two- and three-overlap rows; it is not the chance of exactly one.
On a phone, scroll the table sideways. Keyboard users can focus it and use the arrow keys.
| Shared identities K | Qualifying second sets | Exact probability | Rounded percentage |
|---|---|---|---|
| 0 | 67,525 | 67525/76076 | 88.759924% |
| 1 | 8,325 | 8325/76076 | 10.943004% |
| 2 | 225 | 225/76076 | 0.295757% |
| 3 | 1 | 1/76076 | 0.001314% |
Separate shared names from matching assigned positions
First spread
Position 1: A. Position 2: B. Position 3: C.
Second spread
Position 1: B. Position 2: D. Position 3: E.
Comparison
B is shared, so K=1. No identity occupies the same numbered position.
Two spreads can share all three identities while assigning them to different positions. If the first ordered spread is A–B–C, the second set {A,B,C} has six possible orders: ABC, ACB, BAC, BCA, CAB and CBA. Under uniform ordered selection, only one matches all three original positions.
The chance of the same three identities in any order is 1/C(78,3) = 1/76,076 = 0.001314%. The chance of A, then B, then C in the corresponding positions is 1/(78×77×76) = 1/456,456 = 0.000219%. The second denominator counts ordered distinct selections.
Conditional on already knowing that the same three identities returned, the chance of the matching order is 1/6 under that ordered model. That conditional statement is not the original chance of all three position matches before the second selection.
Use the position-and-interpretation exercise if your task is comparing how one image answers different assigned questions. A probability about card placement does not decide which interpretation is correct or establish a future event.
Verify the overlap logic with six named tokens
Use six distinct tokens A, B, C, D, E and F. Select two without replacement, restore all six, then select another two uniformly. Fix the first set as {A,B}. There are C(6,2) = 15 possible second sets, small enough to list completely.
The six zero-overlap sets choose both tokens from C,D,E,F. The eight one-overlap sets pair A or B with one of those four outsiders. The only two-overlap set is {A,B}. No other second set exists; the counts sum to fifteen.
On a phone, scroll the table sideways. Keyboard users can focus it and use the arrow keys.
| Overlap K | Second sets | Count / 15 |
|---|---|---|
| 0 | CD, CE, CF, DE, DF, EF | 6 / 15 |
| 1 | AC, AD, AE, AF, BC, BD, BE, BF | 8 / 15 |
| 2 | AB | 1 / 15 |
If you list every possible first set as well, there are 15 × 15 = 225 ordered pairs of sets. Ninety pairs have zero overlap, 120 have one and fifteen have two. Dividing those counts by 225 gives the same distribution. The first and second sets are an ordered pair of experiments even though position within each set is ignored here.
This is an original counting example, not a sample of real tarot sessions. It confirms the formula because every possible token outcome can be inspected. It does not estimate how well any person mixes cards.
A mean count is different from the chance of any overlap
In the six-token example, the chance of any overlap is (8+1)/15 = 60%. The expected overlap count is (0×6 + 1×8 + 2×1)/15 = 2/3 of a token per comparison. An expected count of 2/3 does not mean there is a 66⅔% chance of any overlap; the two-overlap outcomes contribute twice to the mean.
More generally, compare a fixed first set of size a with a uniformly selected second set of size b from a restored pool of N distinct identities. Each of the a first-set identities has probability b/N of appearing in the second. Adding their contributions gives mean overlap ab/N. The full count distribution is C(a,k)C(N−a,b−k)/C(N,b).
For two reference three-card spreads the mean is 9/78, about 0.115385 shared identities per comparison. That is an average over model comparisons, not a fractional card you will observe and not the 11.240076% chance of any overlap. An actual pair has an integer K: zero, one, two or three.
If the first set stays excluded, it cannot overlap with the next draw from the remaining unique identities. If a printed name has two physical copies, or your edition has extras, distinguish physical copies from name labels and use actual contents. The inventory checklist supports that separate task.
Keep the reset rule, both complete spreads and the event definition in one optional record. Do not repeatedly redraw until the sets overlap or stop after the first memorable pair while treating an incomplete history as one predeclared experiment.
Download the blank two-spread comparison record
Record an observation separately from an optional creative association. Neither an overlap count nor an unusual ordering identifies a cause or settles a consequential real-world decision.
Check four different comparison questions
These original cases use named tokens or the stated 78-identity reference. Try an answer before opening the explanation.
Case 1
The first ordered spread is A–B–C and the second is B–D–E. What is the shared-identity count, and how many corresponding positions match?
Show answer and explanation for case 1
K=1; zero corresponding positions match.
B appears in both sets but moved from position 2 to position 1. Count the set intersection separately from equality at matching position numbers. One shared identity does not guarantee one position match.
Case 2
Both spreads use the same three identities, but the second order is C–A–B. Does the matching-position probability apply to this observed outcome?
Show answer and explanation for case 2
No. The identities match as a set, but the three positions do not.
Under the uniform reset model, any-order set recurrence has probability 1/76,076; all three corresponding positions matching has probability 1/456,456. The observed C–A–B belongs to the first event and not the second. These reference probabilities do not establish a meaning for the outcome.
Case 3
Three unique physical cards remain on the table, and the second spread is selected from the other 75 without replacing them. Can any of those original physical identities occur in it?
Show answer and explanation for case 3
No, under that stated procedure.
The three originals are excluded from the available pool, so physical-identity overlap is zero. The 11.240076% table assumed full restoration. A duplicate printed name in a different pool is a different identity-counting problem, not an exception that restores an excluded physical card.
Case 4
In the six-token/two-card model, why is mean overlap 2/3 not the 60% chance of any overlap?
Show answer and explanation for case 4
The mean counts a two-overlap outcome twice; the event probability counts that outcome once.
The mean is (8 + 2×1)/15 = 2/3. The probability of at least one is (8+1)/15 = 9/15. Both follow from the same three-row distribution but answer different questions. Label a mean as a count, not a percentage probability.
Sources and further reading
- Tarot Tools: repeated-card rules and event definitionsExisting complete guide separates replacement picks, unique physical identities, duplicate names and whole-window events. This article derives the distinct reset-spread intersection model.
- Python documentation: unordered combination countCombination definition and factorial expression; complete function definition checked.
- Python documentation: ordered distinct selectionsPermutation definition and N!/(N−n)! expression; the position-match comparison uses ordered selections.