求两枚骰子点数绝对差≤2的概率（非枚举公式法）

Great question! Let’s solve this without listing every possible outcome—here’s a clean, formula-driven breakdown:

1. Total Possible Outcomes #

When rolling two 6-sided dice, the total number of unique outcomes is 6 * 6 = 36—this will be our denominator for the probability calculation.

2. Use Complementary Probability (The Simpler Path) #

Instead of counting valid outcomes directly, let’s first calculate the probability of the opposite event: |X-Y| > 2 (i.e., the absolute difference is 3, 4, or 5). We’ll subtract this from 1 to get our desired result.

Derive the count of invalid outcomes (|X-Y| > 2) #

For two dice with values X (1-6) and Y (1-6):

• When X=1, Y must be 4, 5, 6 → 3 outcomes
• When X=2, Y must be 5, 6 → 2 outcomes
• When X=3, Y must be 6 → 1 outcome
• By symmetry, X=4 mirrors X=3 (Y=1 → 1 outcome), X=5 mirrors X=2 (Y=1, 2 → 2 outcomes), X=6 mirrors X=1 (Y=1, 2, 3 → 3 outcomes)

Summing these: 3+2+1+1+2+3 = 12 invalid outcomes.

Calculate the desired probability #

The number of valid outcomes (where |X-Y| ≤2) is total outcomes minus invalid ones: 36 - 12 = 24.

Probability = 24/36 = 2/3.

3. Generalized Formula (For Any n-sided Die and t ≤ n-1) #

If you want to apply this to an n-sided die and find P(|X-Y| ≤ t):

1. Total outcomes: n²
2. Invalid outcomes (|X-Y| >t) = (n-t-1)(n-t) (derived from summing symmetric invalid cases and simplifying the arithmetic series)
3. Valid outcomes = n² - (n-t-1)(n-t)
4. Probability = [n² - (n-t-1)(n-t)] / n²

For our problem, plug in n=6, t=2:
[36 - (6-2-1)(6-2)] / 36 = [36 - 3*4]/36 = 24/36 = 2/3

This formula works for any n and t, so you never have to list out sample spaces again!

内容的提问来源于stack exchange，提问作者Syk