Major Arcana Probability in a Three-Card Spread: Exact Odds

The four possible Major Arcana counts

The four rows cover every possible count: zero, one, two or three Major cards. Their exact probabilities add to one. Percentages are rounded to six decimal places for checking; that many decimal places does not make an unknown selection procedure more certain.

Three other ways to ask the same question belong in this table: all Minor is the zero-Major row, at least one Major is 1 − P(K = 0) = 63.562753%, and at least two Majors is P(K = 2) + P(K = 3) = 19.028340%. “All Major” is the final row, not a separate kind of probability.

Check the deck and the draw rules first

This calculation uses a specific structure: 78 different physical cards, 22 classified as Major and 56 as Minor. Court cards belong to the Minor group in this structure. The site’s Major and Minor guide explains the categories; the deck checklist lets you check names, duplicates and edition-specific extras before using the reference count. [2]

• Three cards, without replacement: put each selected card aside until all three have been chosen.
• Uniform selection: every possible set of three cards has the same chance.
• Category count only: K counts Major cards, not their names, positions or upright/reversed orientation.

If you choose the cards deliberately, restrict the pool, leave extras in it or do not know how an app selects, record that rule. The reference percentages do not become measured probabilities for that process. Counting an altered deck gives you new inputs; it does not by itself establish uniform selection.

OpenStax describes a hypergeometric experiment as sampling from two groups without replacement, where later selections depend on what has already been removed. That is the model used here. [1] It is different from asking whether the same card identity appears again across reset sessions.

Where the formula and the 76,076 denominator come from

A combination is a selection where order is ignored. C(N, n), read “N choose n,” counts how many different groups of n objects can be selected from N different objects. There are

C(78, 3) = (78 × 77 × 76) / (3 × 2 × 1) = 76,076

three-card sets. Division by six removes the six possible orders of each same set. Order can matter to a reading position, but it does not change how many Major cards the set contains.

To obtain exactly k Major cards, choose k from the 22 Majors and the other 3 − k from the 56 Minors. Multiply those independent counts of choices, then divide by all possible sets:

P(K = k) = C(22, k) × C(56, 3 − k) / C(78, 3).

For example, exactly two Major cards has 231 possible Major pairs and 56 possible accompanying Minors: 231 × 56 = 12,936. A particular named three-card set is only one of those combinations. Do not confuse the broad category event with one specific set you name in advance.

The displayed formula is derived by counting the sets. For an altered pool with N cards, M in the category of interest and n selected without replacement, use C(M, k)C(N − M, n − k)/C(N, n). The possible range is max(0, n − (N − M)) through min(n, M). Use actual category counts and a justified selection model rather than silently retaining 22 and 78.

A six-card example you can verify by listing every pair

Use six named tokens, not an actual tarot draw: two Major-labelled tokens M1, M2 and four Minor-labelled tokens m1, m2, m3, m4. Select two without replacement, ignoring order. There are C(6, 2) = 15 possible pairs:

Six pairs contain no Major, eight contain one, and one contains two. The probabilities are therefore 6/15 = 40%, 8/15 = 53⅓% and 1/15 = 6⅔%. The same formula gives C(2, 0)C(4, 2)/15, C(2, 1)C(4, 1)/15 and C(2, 2)C(4, 0)/15.

This smaller example does not estimate the 78-card percentages. It shows why counting the group choices works. You can inspect all fifteen possibilities rather than trusting an unexplained calculator output.

Without replacement, the next category chance changes

Before the first draw, the chance of a Major is 22/78. If the first is Major, only 21 Majors remain among 77 cards, so the next chance is 21/77. If the first is Minor, all 22 Majors remain among 77, so it is 22/77. Treating each step as an independent 22/78 chance would describe a different model.

You can also count with order retained. Exactly one Major has three possible category arrangements: Major–Minor–Minor, Minor–Major–Minor and Minor–Minor–Major. Each has 22 × 56 × 55 ordered card selections. Together they give 203,280 out of 78 × 77 × 76 = 456,456 ordered selections—the same 33,880/76,076 probability as the unordered calculation.

The question “Does position 2 contain a Major?” is different from “How many Majors are in all three positions?” Assigning a position a meaning does not change its category. Do not count a dramatic image twice or exclude a court card because it feels important.

What changes if you return the card before every selection?

Define a separate model in which each card is returned before the next selection, the selection is uniform and the three selections are independent. A Major then has probability q = 22/78 every time, and

P(K = k) = C(3, k)q^k(1 − q)^{3 − k}.

The differences are modest here, but the formulas answer different procedures. Returning a card makes that card available again; it does not certify independent or uniform physical mixing. The repeated-card probability guide focuses on matching identities across a defined replacement process. Its 1/78 next-match example is not the probability of a Major category, which contains 22 identities.

A rare composition does not establish a message or a future event

Three Majors is less likely than one Major in this model. That statement concerns the composition of a sample. It does not make a three-Major reading more accurate, diagnose a turning point or measure how serious a real situation is. Those claims would require evidence beyond the card count.

Nor is 2.02% a promise to see an all-Major spread once every fifty readings. A model probability is not a schedule. If you reset to the full deck between otherwise independent spreads, earlier misses do not make the next all-Major spread due. If you keep drawing from the depleted deck instead, the inputs and conditional chances have changed.

Noticing an unusual set after several spreads is also different from defining one event and one three-card window before selection. Extra clarifiers, discarded spreads and other patterns you inspected are additional opportunities. Do not attach the probability for one specified spread to “anything unusual happened somewhere.”

Keep one clear record, then choose an ordinary next step

If you already want to record a spread, keep the mathematical question small. There is no required number of draws and no need to obtain a rare category count.

1. Write the deck or tool and what you know about its selection rules.
2. State whether cards stay out until the three-card set is complete.
3. Record the three card names and their categories; keep orientation separate.
4. Count K. Label a comparison “reference model” if the actual uniformity is unknown.
5. If using the cards for reflection or fiction, write your association on another line.

A completed example might say: “Reference set: 78 different cards, 22 Major; three selected without replacement; actual mixing uniformity unknown. The named cards include one Major and two Minors. The table gives the reference model for K = 1; it does not explain why these particular cards appeared.” This is an illustrative record, not a real experiment.

Stop after the task is clear. A useful next step can be checking a card’s category, writing one fictional paragraph or making no further draw. Use actual information and appropriate support for consequential decisions; the category count cannot settle them.