1. If 1st = 1, then stop. If 1st = 0 then continual. 2. If 2nd = 0 then continual; if 2nd = 1 then continual too, because 50% chance to get 2/3. 3. A: If 2nd = 0 and 3rd = 0, then continual; B: if 2nd = 0 and 3rd = 1 or 2nd = 1 and 3rd = 0 then continual because the expected value = 3/8 > 1/3. C: If 2nd = 1 and 3rd = 1 then stop because the expected value = 5/8 < 2/3. 4. Keep doing on this way until you get more than ½. |
...if 2nd = 0 and 3rd = 1 or 2nd = 1 and 3rd = 0 then continual because the expected value = 3/8 > 1/3... why expected value = 3/8 here?
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赞同pf的结论, 但是理由粗略的想了一下,要是奖励没到1/2 继续玩,根据大数定理(strong law of large numbers), 最后出现超过或者等于1/2的概率为1. 要是在某刻的奖励m/n>1/2, 那么接下来的期望是m+K/n+2K 对于任何接下来的K次投币, 应该可以观察出这个期望总是小于m/n 的,所以应该停止 |
“要是在某刻的奖励m/n>1/2, 那么接下来的期望是m+K/n+2K 对于任何接下来的K次投币” 没这么简单。因为“接下来”,你还可以选择何时停止,而不是固定再投K次币。
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你有答案么说出来听听。 |
这是一个涉及随机游动的问题。只能说“在第一次出现正面时就打住”(得到的奖励为1);但当第一次出现反面时,就没有策略能保证达到“最优”(就象开区间(0, 1)中没有最大的数那样)。 |
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