REPEATED-CARD RECORD — OBSERVATION BEFORE INTERPRETATION Deck or tool used; actual selection rules known / unknown: Distinct identities available N, if known: Orientation recorded separately / included in identity definition: Cards returned before each draw / one block without replacement / other: Uniform selection and independence assumed / justified / unknown: Event chosen before this record, if any: Window: start, end and number of draws: Every draw in order, including clarifiers and discarded results: What repeated, using the original event definition: Was the pattern selected only after viewing the results? Procedure changes or missing entries: My optional image association; separate from a cause claim: Real-world facts, unknowns and one modest action or pause: What this record cannot establish: MODEL ASSUMPTIONS Uniform independent draws, replacement, N distinct identities; orientation ignored unless explicitly included. N=78 in the examples is an assumption, not a tested selection mechanism. Next matches the first card already seen | 1/78 | 1.282051% A particular named card on both planned draws | 1/6,084 | 0.016437% Three planned draws all the same, any card | 1/6,084 | 0.016437% Three planned draws all a particular named card | 1/474,552 | 0.000211% At least one named card in seven draws | 1−(77/78)^7 | 8.636473% Any repeat among seven draws | 1−(78×77×76×75×74×73×72)/78^7 | 24.197236% QUESTIONS — TRY BEFORE READING ANSWERS 1. The first card has already been drawn. Under the N=78 uniform independent replacement model, what is the chance the next identity matches it? My answer and reason: 2. Before either draw, you name one card and ask whether both planned draws will be that card. Is 1/78 the right probability? My answer and reason: 3. You draw three unique physical cards without returning them from a deck with one copy of each identity. Can the same physical identity occur twice in that block? My answer and reason: 4. Your seven-draw record is A, B, C, A, D, E, F. Does it show three consecutive matching identities? My answer and reason: 5. An app produces a repeat. Can this calculator certify that the app chooses uniformly and independently? My answer and reason: 6. A card repeats and the probability for a chosen event looks small. Does that prove you failed to act on its lesson or that a predicted event must occur? My answer and reason: ANSWER NOTES 1. 1/78, about 1.28%. The first identity is already known. One of 78 equally likely next identities matches it. Do not require the first card to be a name specified before drawing. 2. No. It is 1/78² = 1/6,084 under the stated model. Both draws must satisfy the named-card condition. This is a smaller event than two cards matching each other with any identity allowed. 3. No, not under that without-replacement procedure. Once selected, that unique identity is unavailable for the rest of the block. Repeats across sessions after returning cards are a different event and need their own procedure. 4. No. It shows an identity repeated somewhere in the window. A occurs twice on nonadjacent draws. An any-repeat event, an adjacent match and an all-matching block should not be substituted for one another. 5. No. Those are assumptions, not an audit result. The calculator uses N and m; it does not inspect the app, test its selection mechanism or infer it from a short history. Label an unknown procedure as unknown. 6. No. Keep the outcome separate from an optional interpretation. A model probability does not identify a cause or validate a prediction. You may choose a modest reflection action or pause, without claiming the draw established a missed lesson or future fact. No record or probability establishes a warning, missed lesson, future event or another person’s thoughts. VERIFIED SOURCE PAGES https://www.itl.nist.gov/div898/handbook/eda/section3/eda366i.htm https://openstax.org/books/introductory-statistics-2e/pages/3-2-independent-and-mutually-exclusive-events ADDITIONAL DUPLICATE-COPY MODEL — ORIGINAL EXAMPLE 79 physical copies: two named A, 77 single-copy names. Select each copy uniformly; orientation ignored. WITH REPLACEMENT AND INDEPENDENCE: 6,241 ordered copy pairs; 81 match by name; four are A–A. Any-name match: 81/6,241; both A: 4/6,241. After observed A: 2/79; after observed single-copy name: 1/79. WITHOUT REPLACEMENT: 6,162 ordered pairs; two match by name, giving 1/3,081. After observed A: 1/78; after observed single-copy name: 0. Copy identity cannot repeat. This model is not encoded by the existing uniform-distinct-identity N-only calculator and does not describe an audited actual shuffle. CASE 7 In the hypothetical 79-copy pool, name A has two copies and each of 77 other names has one. With independent uniform-copy selections and replacement, is the any-name match probability 1/78? ANSWER: No. It is 81/6,241, about 1.297869%. Name A contributes four matching ordered copy pairs and the 77 singles contribute one each, among 6,241 total pairs. Names are not equally likely. CASE 8 Without replacement from that same 79-copy pool, you first select a single-copy name. Can the next name match it? ANSWER: No. That name has no remaining copy. After A the conditional next-name match would instead be 1/78. The unconditional match before either selection is 2/(79×78), not either conditional value.