M: This is a hypothetical question. I don't want you to do any of the math or anything like this. Just help me walk through how I would set up the problem. Assume there is a six-by-six grid and that a random number generator is being used. There's a binary choice for the tiles that drop down into the grid, so either black or white. As they fall and settle, once they have settled, if three or more, either horizontally or vertically, are the same color, they are removed. Everything shuffles down and new tiles replace the blank spots, and this continues until the board completely settles. There are no blank spots or matches of three or more. What I'm trying to figure out how to set up in terms of calculations is figuring out the probability of what's going on here. I haven't even got a good language for expressing what I'm trying to find, so I apologize, and I could use some help with that too. Okay, here's an exact element. If you follow those rules from a completely blank grid, replacing every blank with a tile and then removing them and reshuffling them in the order I described and using the rules I described, what would be the probability of getting a... Okay, calculate, if you can for me, the probability of not getting any matches on that first iteration. And then don't calculate, but either explain to me how you would go about finding the probability of getting x matches in a row as you followed those rules, or explain to me why the calculations become so complex that it's virtually uncalculable.

G: Got it—let’s set this up cleanly so you (or code) can compute what you want, and also see why later steps blow up in complexity.

What we’re modeling (terms)

Board: 6×6 grid.
Tiles: i.i.d. fair coin flips (black/white) when generated.
Match rule: Any horizontal or vertical run of length ≥ 3 of a single color is cleared.
Settle: Gravity drops tiles to fill gaps; then new i.i.d. tiles spawn at the top for empty slots.
Cascade step: One “detect → clear → drop → refill” cycle.
Stable board: No runs ≥ 3 anywhere.

A) Probability of no matches on the very first fill

You want ( \Pr(\text{first 6×6 fill is stable}) ).

This is a pure counting problem: [ \Pr(\text{stable}) ;=; \frac{#{\text{6×6 black/white boards with no ≥3 runs H or V}}}{2^{36}}. ]

An exact, efficient way to count the numerator is a transfer-matrix / dynamic-programming approach over columns.

How to set it up (no heavy math, just the plan)

Column validity (vertical constraint):
Enumerate all 6-bit columns (there are (2^6=64)). Keep only those with no vertical run ≥ 3 (i.e., they don’t contain 000 or 111 as a length-3 substring anywhere). Call this set (C).
State you carry across columns (horizontal constraint):
A horizontal run of 3 can only appear when you place a column given the previous two columns.
The minimal information you must remember per row as you sweep left→right is the row’s last two colors. For one row there are 4 possible 2-bit memories: {00, 01, 10, 11}. For 6 rows, a naïve state would be (4^6 = 4096) possibilities.

However, we can compress the state by tracking, for each row, the current color and run length (1 or 2) along the row so far. That’s still 2×2 = 4 options per row, giving the same (4^6) upper bound—but in practice many states are unreachable and get pruned.
Transitions (no horizontal triples):
When you append a new column (col \in C):
- Check that (col) itself has no vertical triples (already guaranteed by membership in (C)).
- Update each row’s run length using the new bit. Reject the transition if any row’s run length would become 3.
- If valid, move to the new aggregate state.
DP sweep across 6 columns:
- Initialize with an “empty” state before column 1.
- For columns 1→6, accumulate counts over all valid transitions.
- The total number of stable boards is the sum of counts in any state after column 6.

This runs quickly on a laptop and gives an exact count; divide by (2^{36}) for the probability of “no matches on the first iteration.”

Tiny sanity check (optional intuition builder):
In 1×6 (a single row), the probability there’s no run of 3 equals the number of binary length-6 strings avoiding 000 and 111, divided by (2^6). That 1-D count can be done with a small 2-state automaton tracking current run length. The 2-D plan above is the same idea, but applied across columns with vertical filters.

B) “x matches in a row” (cascade length) from the blank start

You asked for how to set up the calculation (not to do it). Here’s the conceptual plan and where it gets hard.

Exact formulation (why it explodes)

Each board configuration is a node in a (huge) Markov chain.
A transition is: detect all ≥3 runs → clear them simultaneously → gravity drop → refill top cells with i.i.d. fair tiles. If the result is stable, the chain absorbs; otherwise you continue.
Cascade length (X) is the number of clearing rounds until absorption. You want (\Pr(X = x)) when the initial state is “blank then fully filled with i.i.d. tiles.”

Why exact closed-form is “virtually uncalculable”

The state space is size (2^{36}) just for filled boards, plus intermediate “with gaps” configurations during the settle step.
After a clear, correlations are induced by gravity: whole columns compress and then refill from the top with fresh i.i.d. bits only for the exposed positions. The positions that were not cleared keep their previous colors, so you don’t return to independent bits everywhere.
The distribution after one round is thus highly dependent on the exact cleared geometry, which branches combinatorially. You can, in principle, do it with exact dynamic programming over equivalence classes of column profiles, but the number of classes grows quickly and becomes unwieldy.

Practical ways people compute (\Pr(X=x))

Monte Carlo simulation (recommended):
Implement the exact rules, run a large number of trials from the blank-fill start, and record cascade lengths. Converges fast and is easy to validate.
Structured DP with compression (researchy, still heavy):
- Use a transfer-matrix over column profiles that include: the current column’s bits, per-row horizontal run length (1–2), and a compact encoding of gaps after clears during the settle step.
- Enumerate all clear-patterns consistent with those profiles, apply gravity deterministically, then inject i.i.d. bits only into exposed top slots, yielding new profile distributions.
- This can, in principle, produce exact (\Pr(X = x)) for small boards (6×6 is borderline) but requires careful state design to avoid state-space blow-up.
Mean-field / branching approximations (quick but approximate):
Treat the formation of new triples after a clear as a kind of branching process driven by local overlaps, ignoring long-range correlations. This gives decent intuition for tail behavior of cascades but won’t be exact.

Quick recipe you can hand to code (summary)

First-fill “no matches” probability:
- Build the set (C) of 6-bit columns without vertical triples.
- DP across 6 columns, carrying per-row (color, runlength∈{1,2}). For each next column in (C), allow transition only if no row hits runlength 3.
- Sum counts after column 6 and divide by (2^{36}).
Cascade length distribution from blank:
- Simulation: implement detect→clear→drop→refill exactly; tally cascade lengths.
- Exact (if you insist): design a compressed column-profile state that survives clears and gravity, and build a (huge) transfer process over those profiles; iterate until absorption while tracking length-k paths.

Small but important omissions (noted explicitly)

No diagonals: Only horizontal/vertical triples are considered.
No wraparound: Edges don’t connect.
Simultaneous clears: I assumed all qualifying triples clear at once each step (the usual rule). If you meant sequential clears, the Markov dynamics change substantially.

If you want, I can turn the “first-fill probability” plan into compact pseudocode or a tiny (readable) DP outline so you—or a friend—can implement it directly.