Repeated label puzzle: count each outcome once

Repeated label puzzle is a small paper question for a quiet evening: if four fictional draws each receive one of six possible symbols, how likely is at least one repeat? Under the explicit model used here, the answer is thirteen eighteenths. Counting six possible matching pairs and adding their chances would instead give one, which answers the wrong question.

Imagine four adults, Nia, Omar, Tess and Rae, considering the puzzle together. They are invented people, and their guesses are part of this authored example. No actual gathering, random draw or learning result is reported. The paper problem can be inspected without buying equipment or assigning real people any personal information.

Agree on the fictional model first

Use six symbols: A, B, C, D, E and F. Each of four draws independently receives one symbol, with each symbol having probability one sixth on every draw. Earlier assignments do not remove a symbol from later possibilities. These are stipulated mathematical conditions, rather than findings about any physical randomizer.

Keep the four positions in order. A, B, A, C is one possible outcome; A, A, B, C is another. They both contain a repeat, but they are different complete records because the second and third positions differ. The event of interest includes either outcome without treating them as the same ordered record.

The question asks whether any symbol appears more than once. It does not ask whether A specifically repeats, whether the first two positions match, or whether exactly one pair matches. Those narrower events would require their own counts. Write the full question above the four empty positions before calculating.

Make a guess without calling it evidence

In the invented scene, Nia expects four draws to leave plenty of room among six symbols. Omar wonders whether every pair of positions supplies another chance to match. Tess writes A, B, A, C on a spare line, and Rae asks how many times that one outcome should count.

Their remarks identify different parts of the problem. Four being smaller than six does allow an all-distinct outcome, but it does not establish its probability. The example with two As establishes a possible repeat, without establishing how often the defined model assigns one. Keep those observations separate from a completed calculation.

The friends leave their initial guesses beside the worksheet. There is no need to make the guesses agree before proceeding. The count will concern the supplied model, not whose first impression sounded most confident or which invented speaker has the tidiest handwriting.

Count the complete ordered outcomes

The first position has six possibilities. For each of those, the second still has six, and the same remains true for the third and fourth. Multiplying six by itself four times gives 1,296 ordered outcomes. The stated independence and equal probabilities give each outcome weight one over 1,296.

Those equal weights follow from the assumptions. They are not produced by drawing four neat boxes or by selecting familiar letters. A different probability assignment could use the same boxes while giving their complete records different weights. Counting records alone would then require a different justification.

For this model, the full record A, B, A, C has the same weight as A, A, A, A. The latter looks more repetitive but is still one particular sequence. Its appearance cannot make it count as several complete outcomes when the sample space was defined as ordered four-symbol records.

Count no repeats before counting any repeat

It is easier here to count the outcomes in which all four symbols differ. The first position can use any of six symbols. To remain all-distinct, the second then has five acceptable choices. The third has four, and the fourth has three. Multiplying six, five, four and three gives 360.

The numbers five, four and three count acceptable choices for this event. They do not change the underlying model, in which every position still has six possibilities. The earlier symbols remain available; choosing one of them would simply move the completed record outside the all-distinct event.

Thus the no-repeat probability is 360 divided by 1,296, or five eighteenths. You can also write the successive factors as six sixths, five sixths, four sixths and three sixths. Every denominator remains six because the stipulated assignment rule has not removed any symbol.

Every complete record either has all four symbols distinct or contains at least one repeat. These two possibilities do not overlap and cover the full set. Subtracting 360 from 1,296 leaves 936 repeat-containing outcomes. Their probability is 936 over 1,296, which simplifies to thirteen eighteenths.

Count a repeat-containing outcome once

Return to A, B, A, C. The first and third positions match. This record belongs to the any-repeat event, so it contributes one outcome to the 936. The question does not award another outcome for underlining the same two As in another color or noticing the match a second time.

Now inspect A, A, A, A. Every pair of positions matches. Nevertheless it is still one ordered record and contributes one outcome to the same any-repeat event. The event records whether a repeat exists, rather than how many matching pairs the record contains.

That distinction is the reason to keep the complete sequence beside a short match label. A note saying only “repeat” deliberately discards information about which positions matched. Such a label can classify an outcome without preserving the number of matching pairs inside it.

See where adding six pair chances goes wrong

Four positions contain six different position pairs: first with second, third and fourth; second with third and fourth; and third with fourth. Each selected pair has matching probability one sixth under the given model. Once one member's symbol is fixed, the other independently has one matching choice out of six.

Adding those six probabilities gives six sixths, or one. But the pair-match events overlap. A, A, A, A lies in all six, so that addition counts it six times. The any-repeat event counts that record once. The sum therefore cannot stand in for the probability of at least one match.

Other outcomes make the difference visible too. A, A, B, B has two matching position pairs. A, A, A, B has three. A, B, A, C has one. All three outcomes satisfy “any repeat,” while contributing different amounts to a tally that adds one for every matching pair.

Across all 1,296 equally weighted records, the total number of matching pairs is also 1,296. Dividing gives a mean of one matching pair per modeled outcome. That mean is compatible with a repeat-event probability of thirteen eighteenths because it counts a different quantity. It does not say every record has one pair.

Keep the full count as a cross-check

The complete enumeration partitions outcomes by their number of matching pairs. There are 360 with zero, 720 with one, ninety with two, 120 with three and six with six. These categories add to 1,296. Adding the nonzero categories gives 936, agreeing with the complement calculation.

Weighting those categories by their pair counts gives 720 plus 180 plus 360 plus 36, or 1,296 matching pairs altogether. This checks the separate mean just described. Neither calculation changes the central event: any record with one or more matching pairs enters the repeat event only once.

In the fictional discussion, Rae circles both results and writes a different heading over each. Nia revises her initial guess after checking the 360 all-distinct outcomes. They keep their earlier guesses on the worksheet, with the completed count beside them.

Change the number of draws carefully

With two draws under the same independent six-symbol rule, the repeat probability is one sixth. With three draws, the no-repeat count is six times five times four out of six cubed, or five ninths. The three-draw repeat probability is therefore four ninths, below one half.

Four draws give thirteen eighteenths, above one half. This comparison describes these specified models; it does not establish a frequency for real birthdays, names, equipment outputs or social groups. The six symbols have equal probabilities here because the exercise assigns them those probabilities.

If the rule instead removed a chosen symbol before the next draw, four draws could all be distinct with certainty under that new no-replacement model. Its repeat probability would be zero. Label the rule change rather than keeping the earlier thirteen-eighteenths answer after replacing the assumption that produced it.

Seven draws from only six available symbols must include a repeat, regardless of their assigned probabilities, because seven distinct labels cannot fit within that six-label set. This is a separate boundary fact. It does not make a repeat certain in the original four-draw problem.

Keep the native organizer accounts limited

The Hyjoy desk organizer account names two-tier shelving. Its sole image shows a complete dark desk-clamp organizer with trays and posts. Books, plant and cup of stationery illustrate use; they are not verified supplied stocks. No clamp fit, load, guardrail dimension, material or performance is tested here.

The Raindrops compartment organizer account retains three-piece wording. Its sole image shows one complete empty black compartment unit. That view does not independently verify three delivered units, dimensions, cleaning ease, material or gift suitability. The catalog maker is not authenticated through a product trial.

Neither selected account supplies the fictional symbols or probabilities. The organizer references provide optional tabletop context, with no claim that either generated random labels or was used by the invented friends. No purchase is needed for the puzzle.

Leave neighboring group questions separate

The two dice totals guide maps equally weighted ordered pairs onto unevenly weighted sums. Here the central comparison is an any-repeat event versus overlapping pair-match counts across four positions. A familiar six-value model does not make the event definitions interchangeable.

The hairbrush gift decision guide gives three fictional people fixed preference rankings and a cyclic majority. Those are stipulated choices, not independent random labels. Its pairwise voting comparisons supply no collision probability for this separate leisure puzzle.

Keep an actual order outside the game

If you separately consider an organizer order on republeq.com and want to ask about USDT, begin with the exact offer and intended quantity. The fictional symbols, draw counts and probabilities in this worksheet are not prices, cryptocurrency amounts or payment instructions. Correct arithmetic establishes no checkout method or actual order status. Read the payment options and conditions presented for the intended transaction, and ask the responsible support contact to clarify essential missing information before deciding whether to proceed. If a cryptocurrency option is offered, establish the specified asset and network, requested amount, any stated validity period and the reference associating payment with the order. Do not infer a transfer destination from a product image, worksheet heading or fictional label. Resolve material product questions separately, including actual supplied contents and any intended use you need the item to support. Ask about applicable delivery, cancellation, return and refund conditions, including the denomination of any approved refund and relevant deductions. Keep each answer with the offer it concerns. Information about one account does not automatically describe another quantity or purchase, and a changed order may require reviewing the resulting terms again. This article promises no present USDT availability, fee level, conversion result, confirmation time, refund amount or delivery outcome. You can inspect every fictional label record without purchasing anything. Preserve any eventual order acknowledgement and transaction record separately from the worksheet. Those records concern an actual transaction under its documented conditions; the calculations compare explicitly defined events and counts. A preferred payment asset or a plausible probability cannot fill a gap in actual order information.

Keep the event beside the answer

Repeated label puzzle outcomes include 936 records with a repeat out of 1,296 equally weighted records, giving thirteen eighteenths. Some records contain several matching pairs, but each enters that event once. The friends leave both headings on the page: chance of any repeat and mean number of matching pairs.