All Four Tarot Suits in One Spread: Exact Model Probabilities

Define the event before counting cards

“All four suits appear” means at least one Wands, Cups, Swords and Pentacles card occurs in the selected set. It does not mean that four cards have different names: four differently named Cups cards still represent only one suit. Nor does it mean that the cards have four different themes in an interpretation.

This calculation uses the named reference in the 78-card inventory checklist: four suits of fourteen, plus twenty-two Major cards outside those suits. Ace through Ten and the four courts belong within each suit. A King is not moved to the Major group because it has a prominent picture. Check your edition before using these counts.

Choose n distinct physical cards uniformly without replacement: selected cards stay out until this set is complete, and every n-card set has equal model probability. Ignore order and upright/reversed orientation for suit coverage. An app’s actual algorithm or a physical shuffle may not meet that assumption; this page has not audited either.

The Major-count guide counts one category against the rest. This problem requires four categories simultaneously. At least four selected cards are necessary; a three-card set cannot contain four different suits.

Count one card from each suit in a four-card set

With exactly four cards, every qualifying set has one from each suit and no Major. Choose a Wands card in fourteen ways, a Cups card in fourteen ways, then Swords and Pentacles likewise. Multiplying gives 14 × 14 × 14 × 14 = 38,416 qualifying sets.

The denominator is C(78,4), the number of unordered four-card sets, equal to 1,426,425. Here C(N,n) means “N choose n”: N!/[n!(N−n)!]. Dividing gives 38,416/1,426,425 = 2.693166%, rounded to six decimal places. No extra factor of 4! is needed: choosing one labelled suit card already identifies each unordered set once.

If you deliberately exclude all Majors before selecting, the qualifying four-card sets remain the same but there are only C(56,4) = 367,290 possible sets. The result becomes 10.459310%. That is the probability in the restricted experiment; it is not the probability for an ordinary full-pool draw.

You could also describe the restricted result as conditional on all four cards being Minor in the uniform full-pool model. State the condition explicitly. Do not substitute it for the full-pool probability simply because your observed cards happened to be Minor.

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Separate the two possible five-card compositions

Five cards can contain all four suits in two disjoint ways. First, select one from each suit and one of the twenty-two Majors. That contributes 14⁴ × 22 = 845,152 sets. Second, use only Minor cards: one suit contributes two cards and the other three contribute one each.

For the second case, choose which suit is doubled in four ways. Choose its unordered pair in C(14,2) = 91 ways, then one card from each other suit in 14³ = 2,744 ways. The count is 4 × 91 × 2,744 = 998,816. The doubled suit is unique, so this does not count a set twice.

Add 845,152 and 998,816 to obtain 1,843,968 qualifying full-pool sets. Divide by C(78,5) = 21,111,090: 8.734594%. A fifth card does not create a fifth Minor suit; it is either an unsuited Major or another member of one of the four suits.

For the deliberately restricted fifty-six-card pool, only the doubled-suit case is available. Its probability is 998,816/C(56,5) = 998,816/3,819,816 = 26.148275%. Both its numerator and denominator differ from the full-pool calculation.

On a phone, scroll the table sideways. Keyboard users can focus it and use the arrow keys.

Use an inclusion–exclusion count and a small checkable pool

For equal suit size s and total pool N, the number of n-card sets covering every suit is Σ(−1)ᵏ C(4,k) C(N−ks,n), for k from zero through four. Divide that count by C(N,n). Treat a combination as zero if it asks for more objects than are available.

Start with all sets. Subtract those missing each particular suit; sets missing two suits were subtracted twice, so add them back. Continue with the alternating triple and quadruple corrections. This is an event-counting correction, not an assumption that the four suit-presence events are independent.

A small original example makes the physical copies visible. Use nine distinct tokens labelled A1, A2, B1, B2, C1, C2, D1, D2 and M. The letters A through D stand for the four groups; M is outside them. No historical tarot image or actual draw is involved.

For four tokens, choose one of each letter: 2⁴ = 16 qualifying sets among C(9,4) = 126. For five tokens, include M with those sixteen sets, or double one letter group: 4 × C(2,2) × 2³ = 32. The total is 48 out of C(9,5) = 126. Listing every physical-token combination independently confirms both counts.

The reference six-card rows above were also checked by disjoint category-composition counts. The calculations verify the counting model. They do not estimate how you personally handle, cut or select a deck.

Keep inventory, selection and interpretation separate

If a card is missing or duplicated, write the actual suit counts and the other-card count. Unequal groups require an unequal-size inclusion–exclusion count: for each selected collection of absent suits, subtract their actual combined size from N. Merely changing N while retaining fourteen cards in every suit can describe an impossible pool.

If cards are returned between selections, unordered without-replacement sets are no longer the right sample space. If you choose cards deliberately, these uniform random probabilities do not describe the choice. For repeated identities under a separate replacement rule, use the existing repeated-card guide.

Keep the event and window fixed. “All four suits in these five cards” differs from finding all four somewhere across many spreads or after extra clarifiers. A low reference probability does not show that a message, relationship outcome or future event was established.

An optional record can note the actual pool, the rule, the observed suit labels and the model comparison. Put a creative association on another line. There is no required number of draws or reason to keep drawing until every suit appears.

Download the blank four-suit coverage record

The plain-text record is optional. It is a worksheet for your own notes, with no submission or scoring step. Finish when the event and model are clear.

Check four complete examples

These original cases use explicitly stated reference or token pools. Try the answer before opening the explanation.