概率最优解骰子游戏问题：寻求最优决策策略技术问询

云上老周

2026-5-19

Let's work through this dice game problem to figure out Agnishom's optimal strategy for maximizing his win probability. Here's the breakdown:

Step 1: Map Satvik's Probability Distribution

Since Satvik can only roll once, we first define the probability of each possible sum (2-12) from two standard dice, plus the probability that Agnishom wins if he holds a given sum ( S ):

Sum 2: Probability ( \frac{1}{36} ), Win probability (Satvik rolls lower) = 0
Sum 3: Probability ( \frac{2}{36} ), Win probability = ( \frac{1}{36} )
Sum 4: Probability ( \frac{3}{36} ), Win probability = ( \frac{1+2}{36} = \frac{3}{36} )
Sum 5: Probability ( \frac{4}{36} ), Win probability = ( \frac{1+2+3}{36} = \frac{6}{36} )
Sum 6: Probability ( \frac{5}{36} ), Win probability = ( \frac{1+2+3+4}{36} = \frac{10}{36} )
Sum 7: Probability ( \frac{6}{36} ), Win probability = ( \frac{1+2+3+4+5}{36} = \frac{15}{36} \approx 0.4167 )
Sum 8: Probability ( \frac{5}{36} ), Win probability = ( \frac{1+2+3+4+5+6}{36} = \frac{21}{36} \approx 0.5833 )
Sum 9: Probability ( \frac{4}{36} ), Win probability = ( \frac{21+5}{36} = \frac{26}{36} \approx 0.7222 )
Sum 10: Probability ( \frac{3}{36} ), Win probability = ( \frac{26+4}{36} = \frac{30}{36} \approx 0.8333 )
Sum 11: Probability ( \frac{2}{36} ), Win probability = ( \frac{30+3}{36} = \frac{33}{36} \approx 0.9167 )
Sum 12: Probability ( \frac{1}{36} ), Win probability = ( \frac{33+2}{36} = \frac{35}{36} \approx 0.9722 )

Step 2: Calculate Expected Win Probability of Re-Rolling

If Agnishom re-rolls, his expected win probability is the weighted average of win probabilities for all possible sums. Calculating this:
[
\text{Expected Win (Re-Roll)} = \frac{1}{36^2} \left( 01 + 12 + 33 + 64 + 105 + 156 + 215 + 264 + 303 + 332 + 35*1 \right)
]
The numerator simplifies to 575, so:
[
\text{Expected Win (Re-Roll)} = \frac{575}{1296} \approx 0.4437
]

Step 3: Decide Hold vs. Re-Roll for Each Sum

For each initial sum ( S ), compare the win probability of holding ( S ) against the expected win probability of re-rolling:

Hold the roll if ( P_{\text{win}}(S) > \text{Expected Win (Re-Roll)} ): This applies to sums 8, 9, 10, 11, 12 (their win probabilities are all above ~0.4437)
Re-roll if ( P_{\text{win}}(S) < \text{Expected Win (Re-Roll)} ): This applies to sums 2-7 (even sum 7's ~0.4167 win rate is lower than the re-roll expectation)

Step 4: Validate the Strategy's Total Win Rate

Using this strategy, Agnishom's total win probability is:
[
\left( \sum_{S=8}^{12} P(S) * P_{\text{win}}(S) \right) + \left( P(\text{sum 2-7}) * \text{Expected Win (Re-Roll)} \right)
]
This calculates to ~0.5675, which is higher than any alternative (e.g., holding sum 7 would drop the win rate to ~0.5629).

Final Optimal Strategy

Keep the initial roll if the sum is 8, 9, 10, 11, or 12.
Re-roll if the initial sum is 2, 3, 4, 5, 6, or 7.

内容的提问来源于stack exchange，提问作者user517784