八面公平骰子连续N次不掷1/8的概率及掷骰游戏期望收益求解
Let's break down these two dice probability problems clearly and step by step:
First, let's start with the basics:
- A fair 8-sided die has 8 equally probable outcomes.
- The outcomes we want to avoid are 1 and 8, which make up 2 out of 8 total results.
- So the probability of not rolling a 1 or 8 in one single throw is:
(8 - 2)/8 = 6/8 = 3/4
Since each die roll is independent (one throw's result doesn't affect the next), the probability of avoiding 1 or 8 for N consecutive throws is just the single-throw probability raised to the Nth power:P(no 1/8 in N rolls) = (3/4)^N
We know V stands for the total number of dice rolls (values range from 1 to 6), and the player's score is a * V where a is a constant. To find the expected收益, we first calculate the expected value of V, then multiply by a.
Step 1: Calculate probabilities for each possible V
Let’s define p = probability of rolling 2-7 (game continues) = 3/4, and q = probability of rolling 1/8 (game ends immediately) = 1/4.
- V=1: Game ends on the first roll (rolled 1 or 8). Probability:
q = 1/4 - V=2: First roll is 2-7, second roll is 1/8. Probability:
p * q = (3/4)*(1/4) - V=3: First two rolls are 2-7, third roll is 1/8. Probability:
p² * q = (3/4)²*(1/4) - V=4: First three rolls are 2-7, fourth roll is 1/8. Probability:
p³ * q = (3/4)³*(1/4) - V=5: First four rolls are 2-7, fifth roll is 1/8. Probability:
p⁴ * q = (3/4)⁴*(1/4) - V=6: First five rolls are all 2-7 (game terminates after the 6th roll no matter what). Probability:
p⁵ = (3/4)⁵
Step 2: Compute the expected value E[V]
The expected value is the sum of each V multiplied by its probability:
E[V] = 1*(1/4) + 2*(3/4)(1/4) + 3*(3/4)²(1/4) + 4*(3/4)³(1/4) + 5*(3/4)⁴(1/4) + 6*(3/4)⁵
Calculating this out:
- Exact fractional result:
E[V] = 3367/1024 ≈ 3.288 - Decimal approximation:
E[V] ≈ 3.288
Step 3: Expected收益
Since the player's total score is a * V, the expected收益 is:Expected Profit = a * E[V] = a*(3367/1024) ≈ 3.288a
内容的提问来源于stack exchange，提问作者peterxz
