Comparing Math Logic of both ai tools

Prompt

Grid: 5 reels × 3 rows, 243 ways, left to right, 3+ of a kind pays. Every cell is drawn independently from its reel’s weights (this keeps the problem tractable). Symbol H1 H2 L1 L2 W (wild) S (scatter) Weight, reels 1 and 5 2 3 5 6 0 2 Weight, reels 2-4 2 3 5 6 2 2 Paytable (× total bet, per way): Symbol 3 of a kind 4 5 H1 0.5 2 10 H2 0.3 1 5 L1 0.1 0.4 2 L2 0.05 0.2 1 Rules: W substitutes for H1, H2, L1 and L2 but not for S. S pays nothing. Cascades: after a spin evaluates, every symbol in a winning way (wilds included) is removed. The remaining symbols drop down and new ones are drawn from the same reel weights. Repeat until no win occurs. Trigger: 3+ S on the initial grid of a spin awards 10 free spins (base game) or +5 spins (retrigger inside free spins). Free spins multiplier: starts at 1, goes up by 1 after every winning cascade step, and never resets during the feature. Each step’s win is multiplied by the multiplier current at that step. Cap: total win per base round (base spin plus feature) is capped at 5,000× bet. The round ends immediately at the cap. What to compute P(feature trigger) on a base spin. Hit frequency of the initial grid (exact, using per-reel symbol counts). Exact RTP of the base game without cascades, then with them. Expected value of the free spins feature, including retriggers. The effect of the 5,000× cap on total RTP. Total RTP to 4 decimals, then verified by Monte Carlo with a stated number of spins and confidence interval. Why it’s hard Cascades break the independence. After the first cascade, the surviving symbols are conditioned on not having been in a win, so they are no longer drawn from the original weights. You can’t just reuse the single-spin formula for later steps. The exact route is to track the grid state, or the per-reel count of each symbol, as a Markov chain. Wilds create shared-win dependence. Whether a W lands on reel 3 changes the ways count for several symbols at once, so you can’t sum symbols independently. You need the joint distribution of per-reel counts. The multiplier is a state variable. Win at step j is worth (current multiplier) × win, and the multiplier depends on how many cascade steps all earlier spins produced. So you need a DP over (spins remaining, multiplier), not a per-spin average. Retriggers make the feature length random, with the geometric-like structure we discussed. The cap is nonlinear. You can’t take the uncapped expected value and subtract a correction. You need the full payout distribution in the tail, or a simulation large enough to sample it well. Hints on the approach Per reel, the number of each symbol type in 3 cells is multinomial. The ways for a symbol on a reel are (symbol count + wild count), so you can compute the joint distribution reel by reel and combine across reels. For cascades, define the state as the 5 per-reel column contents after dropping, then evaluate the chain with memoization. The state space is large but sparse in practice, so Monte Carlo with a known conditional model is a good cross-check. For the multiplier, let V(s, m) be the expected remaining feature win with s spins left and multiplier m. The per-spin transition needs the distribution of the number of cascade steps and the win accumulated at each step weighted by (m + j − 1), which is why the cascade distribution from the earlier bullets feeds directly into it. Sanity check: total RTP should equal base-without-feature + P(trigger) × E[feature], and both pieces must be computed under the same cap rule.