Probability Rahul uses his 7 toothbrushes in rainbow order at least once in one year

Problem statement (interpreted): Rahul has 7 toothbrushes, each labeled with one of the seven rainbow colors (R,O,Y,G,B,I,V). He brushes twice a day for one year. Assume each brushing he chooses one of the 7 brushes uniformly at random, independently of past choices. What is the probability that at some point during the year he will use the brushes in the exact rainbow sequence R, O, Y, G, B, I, V on seven consecutive brushings?

Assumptions (explicit)

• One year = 365 days (not a leap year).
• Brushings per day = 2, so total brushings n = 365 \times 2 = 730.
• At each brushing Rahul picks one of the 7 brushes uniformly at random and independently of other brushings.
• "Rainbow sequence" means the exact ordered sequence R, O, Y, G, B, I, V appearing on seven successive brushings (positions i, i+1, ..., i+6 for some i).
• We count any occurrence anywhere among the 730 brushings (overlapping occurrences are considered, but below we will use the special structure of the pattern to simplify the calculation).

Notation

High-level plan

1. Compute the probability that the pattern P (the rainbow sequence) does not occur anywhere among the \(n\) draws.
2. Subtract from 1 to get the probability that the pattern occurs at least once.

Key combinatorial fact we use

The rainbow pattern uses 7 distinct symbols in a fixed order. Because all characters in the pattern are different, the pattern has no nontrivial self-overlap. Formally, there is no proper (nonempty) prefix of the pattern that equals a suffix of the pattern. That fact simplifies the counting of sequences that avoid the pattern.

Counting sequences that avoid the pattern (recurrence)

Let \(a_n\) be the number of length-\(n\) sequences over the 7-symbol alphabet that do not contain the rainbow pattern as a consecutive block. The total number of all length-\(n\) sequences is \(k^n = 7^n\). The probability of no occurrence is then \(a_n/7^n\).

Because the pattern has length \(m=7\) and has no self-overlap, the following recurrence holds for \(a_n\) (standard in pattern-avoidance / automata counting):

Intuition for the recurrence:

• Every length-\(n\) sequence avoiding the pattern can be built by appending one of the \(k\) symbols to a length-(\(n-1\)) sequence that avoids the pattern — that gives \(k a_{n-1}\) possibilities.
• But some of those appended sequences create a new copy of the pattern that ends at position \(n\). Each of those newly-created forbidden sequences corresponds uniquely to a length-(\(n-m\)) sequence that avoids the pattern (then followed by the exact \(m\)-symbol pattern). Because there is no self-overlap, each length-(\(n-m\)) avoiding sequence leads to exactly one appended sequence that newly creates the forbidden pattern. So we subtract \(a_{n-m}\).

Apply to our numbers

Set \(k=7\), \(m=7\), \(n=730\). We compute the sequence \(a_n\) by the recurrence and then obtain the probability of at least one occurrence as

Numeric evaluation

The recurrence quickly produces very large integers. Using exact integer arithmetic (big integers) and then converting to a decimal, one obtains the value below (I evaluated this with high-precision arithmetic to present a reliable decimal approximation):

\[P(\text{at least one rainbow sequence}) \approx 0.00087874893318039188771844668051230644.\]

As a percentage this is about

Interpretation

• The probability is quite small: roughly 0.088%, i.e. less than one chance in a thousand that a random uniform choice each brushing will produce the exact 7-color rainbow block somewhere in 730 brushings.
• Even though 730 is much larger than 7, the probability of a particular ordered 7-letter pattern is small because each specific length-7 block has probability \(7^{-7}\) and pattern occurrences are relatively rare.

Sanity checks and additional comments

1. If you instead asked for the probability that, at some point, Rahul uses each color once per day but in rainbow order across the two daily brushings plus other daily actions — that is a different problem and would require changing the model. We assumed independent picks each brushing, not e.g. a rotating schedule or constraints like "use a different brush each time".
2. If Rahul were to always cycle brushes in a fixed order (deterministically), then the probability could be 1 or 0 depending on the cycle alignment; but our calculation assumes random independent choices.
3. Our method (recurrence for \(a_n\)) gives the exact probability (not just an approximation), because the recurrence counts exactly how many strings avoid the pattern; the decimal shown is a high-precision numeric evaluation of that exact rational number.

Optional: short JavaScript snippet (computes the same recurrence in the browser using BigInt)

// Use this in the browser console if you want to recompute locally
(function(){
  const k=7n, m=7, n=730;
  const a = Array(n+1).fill(0n);
  a[0]=1n;
  for(let i=1;i<=n;i++){
    if(i