--- name: dice-wildcard-probability category: research description: Calculate probability that, when rolling n dice (standard six-sided) with a designated wildcard face that can count as any other face, at least k dice show the same face. --- ## Description Calculate the probability that, when rolling n dice (standard six-sided) with a designated wildcard face (e.g., 1) that can count as any other face, at least k dice show the same face (using wildcards as needed). ## When to Use - Analyzing dice games with wildcards. - Probability homework or game design. ## Assumptions - Dice are fair and independent. - Wildcard face value is known (default: 1). - Other faces are 2..6 (5 regular faces). - We want probability of ≥ k matches for any face (including using wildcards). ## Method Overview 1. Let n = total dice, p_w = probability of wildcard = 1/6, p_r = 5/6 for regular. 2. Let w = number of wildcards observed; w ~ Binomial(n, p_w). 3. Given w wildcards, remaining r = n - w dice are regular, uniformly distributed over 5 faces. 4. Need probability that among r regular dice, some face appears at least t = k - w times (if t ≤ 0, condition already satisfied; if t > r, impossible). 5. For regular dice, compute P_cond(t, r) = Prob{max count ≥ t} using inclusion–exclusion over subsets of faces: \[ P_{\text{cond}}(t,r)=\sum_{s=1}^{5}(-1)^{s+1}\binom{5}{s} \sum_{y=st}^{r}\binom{r}{y}\left(\frac{s}{5}\right)^{y} \left(1-\frac{s}{5}\right)^{r-y} \frac{\displaystyle\binom{y-st+s-1}{s-1}}{s^{y}}. \] 6. Overall probability: \[ P(\ge k)=\sum_{w=0}^{n}\binom{n}{w}p_w^{w}(1-p_w)^{n-w} \times \begin{cases} 1 & t\le 0\\ 0 & t> r\\ P_{\text{cond}}(t,r) & \text{otherwise} \end{cases} \] where \(t = k - w\) and \(r = n - w\). ## Steps (for n=10, wildcard=1) 1. Precompute binomial probabilities for w=0..10. 2. For each w, compute t = k - w, r = 10 - w. 3. If t ≤ 0 → contribution = binom prob. 4. Else if t > r → contribution = 0. 5. Else compute P_cond(t,r) using the formula (can be implemented via a short script). 6. Sum contributions. ## Example Results (n=10) | k (≥ matches) | Probability | |---------------|-------------| | 3 | 0.9981245713 | | 4 | 0.9112991700 | | 5 | 0.6298112188 | | 6 | 0.2928175250 | | 7 | 0.0877220349 | | 8 | 0.0163763622 | | 9 | 0.0017599261 | |10 | 0.0000846093 | ## Implementation Tips - Use logarithms for large factorials if needed. - The inner sum over y can be limited to y from s*t to r. - For small n (≤20) direct computation is fine. - Verify with Monte Carlo simulation for sanity. ## Pitfalls - Forgetting that wildcard can be used for any face, not just a specific one. - Miscalculating the inclusion‑exclusion sign. - Overlooking the case t ≤ 0 (already satisfied). ## References - Derived using generating functions and inclusion–exclusion. - See also: “Probability of dice matches with wildcards” (standard technique). ## How to Extend - Change number of regular faces (e.g., different dice). - Change wildcard probability (e.g., multiple wildcard faces). - Compute probability of exactly k matches by subtracting P(≥k+1) from P(≥k).