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