Consecutive Tarot Card Repeats: A Run Within a Longer Record

Define one run length inside one fixed window

“Three cards in a row were the same” can describe a short streak somewhere inside a longer record. It is different from all cards in the entire record matching, and different from one identity appearing twice with other cards between them. Choose the exact event before attaching a probability.

Let N be the number of distinct equally likely identities, m the total number of picks in the stated window, and r the qualifying consecutive-run length. A run of at least r identical consecutive picks qualifies, including a longer run. Count the window once if it qualifies, even if several starting positions show a streak.

The reference table uses N=78, independent picks and replacement before every pick. Orientation is ignored when defining an identity. This is a hypothetical selection model, not a measured result for a deck, this site’s tool or a person’s record. Returning a card makes it available; uniformity and independence are additional assumptions.

The existing repeated-card guide already calculates next matches, whole-window all-matches and any identity repeat. Its calculator explicitly excludes a shorter streak somewhere inside a longer record. This page supplies that separate calculation.

Keep nonadjacent repeats, whole-window matches and runs separate

In the fictional record A–B–B–C–D–E–F, an identity repeats and a length-two run occurs. No length-three run occurs. In A–B–C–A–D–E–F there is a nonadjacent repeat but no equal adjacent pair. Those records cannot be classified with the same event label.

For one specified block of three independent uniform picks, all three identities match with probability 1/N²: the first can be anything, while the next two must match it. For all seven picks to match each other, the expression is 1/N⁶. Neither expression answers whether some length-three run appears anywhere among seven picks.

Under the stated N=78 model, at least one length-three run somewhere in seven picks has probability 0.081332%. At least one adjacent match in the same seven-pick window has probability 7.449934%. The broader event has more qualifying outcomes. It is not evidence that a particular person should seek a streak.

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

The last column is an upper bound formed by adding the individual candidate-window probabilities, capped at one. It is not the exact result. Six decimal places help compare the calculations; they do not establish that an unknown selection process deserves that precision.

See why overlapping windows cannot simply be added

Use a small original two-identity model: each independent pick is A or B with equal probability. Observe four picks and ask whether three consecutive picks match. There are 2⁴ = 16 equally likely ordered records.

A run can start at position one or two. The first window matches in AAAA, AAAB, BBBB and BBBA. The second matches in AAAA, BAAA, BBBB and ABBB. Adding four plus four counts AAAA and BBBB twice.

The union therefore has 4+4−2 = 6 records out of sixteen, or 3/8 = 37.5%. Merely adding two probabilities of 1/4 would incorrectly give 50%. A four-pick run contains two matching three-pick windows but qualifies as one record.

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

This example shows dependence for the overlapping three-pick windows. It does not imply that every kind of overlap must have the same correction. The reference result below uses a prefix count that handles any qualifying run without having to label all its overlapping windows.

Count prefixes that have not yet reached the run

For a length-three event, keep two safe states. State one counts records whose final run has length one; state two counts records whose final run has length two. Neither state has already contained a length-three run anywhere. Records that hit the event leave these safe states.

After one pick there are N safe records in state one and zero in state two. From either safe state, N−1 different identities break the final run and return to state one. The one same identity extends state one to state two. A same-identity extension from state two hits length three, so it is excluded from the safe count.

If the current safe counts are a₁ and a₂, the next counts are new a₁ = (N−1)(a₁+a₂) and new a₂ = a₁. Both new values use the previous counts; do not update a₁ and then use that new value for a₂.

For N=2 the states after one, two, three and four picks are (2,0), (2,2), (4,2) and (6,4). Ten four-pick records avoid the run, so sixteen minus ten gives the six qualifying records listed above.

For general r, keep safe states for final run lengths one through r−1. Different identities reset to state one; a same identity extends a safe run by one, unless it reaches r. After m picks, subtract the sum of safe states from Nᵐ and divide by Nᵐ.

The displayed results were calculated with exact integer counts and checked independently by listing all ordered records in several small models. The calculation is not a simulation or a sample of actual tarot draws.

Use the simpler adjacent-match formula and keep the limits visible

For r=2, a record avoids the event only when every new identity differs from the preceding one. The first has N choices, and each later pick has N−1 choices. Thus the no-adjacent-match probability is (1−1/N)ᵐ⁻¹, and its complement is 1−(1−1/N)ᵐ⁻¹. This agrees with the safe-prefix count.

For seven picks and N=78, use 1−(77/78)⁶ = 7.449934%. This is the chance of at least one adjacent match anywhere in the fixed window. The chance that just the next pick matches a known preceding identity remains 1/78. A longer window supplies more opportunities, not a different next-pick probability.

If unique physical cards stay out without replacement, that physical identity cannot form a repeat run in that block. Duplicate printed names, unequal selection weights or changing pools need their own actual inputs. Do not change N alone and pretend a weighted-name model became uniform.

Choose the window and event before looking at the record if you want this comparison. If you already saw a qualifying run, the event is known to have occurred; this table is not the chance that the recorded fact is true. An incomplete or retrospectively selected history cannot be assigned the reference percentage without further information.

A small model probability does not identify a cause, validate a prediction or create an obligation to continue drawing. An optional creative association can remain separate from the observation. Stop when the counting question is clear.

Download the blank consecutive-run record

Read the original safe-prefix counting notes

Both are plain-text files for a bounded study task, with no submission or scoring step. The notes state the recurrence and complete sixteen-outcome counterexample.

Check four complete run questions

These original examples test the event definition and the counting procedure. They do not grade intuition or recommend a number of draws.