This is the original Marilyn's Goat problem - 珍珠湾ART

Appeared on the Parade Magazine, on September 9, 1990:

You're on a game show and have a choice of three doors. Behind one is a car and the other two goats. You choose #1. The host, who knows what's behind the doors, opens #3 and shows you a goat. He then asks you if you want to switch to #2. Should you?

Also Marilyn clarified later that the host always give the chance to switch.

It has dirty words. How could someone do this?

“主持人知情”与“主持人不知情”时的三门问题是两种不同的随机试验，但都可用古典概型来描写，并都可用条件概率来解决。然而，在电视节目中出现这一问题时，只能是在“主持人知情”的情况进行。否则，一旦出现（以１／３概率）"打开的门后有车" ，节目就无法进行下去了。所以，这三门问题应该在“主持人知情”的情况下求解。其解就是:要换。

yinyin知道让easy是好意，但还有一点看法。
若仅需 "知其然" ，而不需 "知其所以然" 的话，是只要记住这easy综合结论就行了。但若再遇一概率问题，恐怕不会马上就easy了。难得有这么一个许多朋友都感兴趣的概率题（上个月yinyin贴了一个概率题，一直没人做），不妨就此机会把概率论的基本概念和原理弄清楚。这两天来，大家的讨论是很有益的。
此外，数学上possibility这词近半个世纪来已另有定义，它是一种非可加集函数，常用于人工智能领域。最好不要把它用作probability的同义词。

<上个月yinyin贴了一个概率题，一直没人做），不妨就此机会把概率论的基本概念和原理弄清楚。这两天来，大家的讨论是很有益的。>

Do you mean the following problem?

如果不管你所选择的门后边是车还是山羊，主持人都要打开另一扇没车的门，那么你就应该换。这样就把得车的概率从１／３提升到２／３。如果主持人有利益关系（譬如，是老板，想尽量不让你赢车），不是总打开一山羊门后再问你换不换，而是虚虚实实，见机行事（例如，往往在你选对的时后才问你换不换，而当你选错的时候往往就直接公布答案），那就不一定是一个概率论问题了（心理学？）。一个引伸出来的概率论问题是：在第二种情况下，主持人（老板）以什么对他自己最有利的条件概率分布（在猜对的条件下及在猜错的条件下）来确定他给不给你第二次重猜机会？假定他应把这两个条件概率分布告诉你。

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I might misunderstand something, but this seems a trivial case (but again, I had thought Marilyn's Goat problem was trivial, and I was wrong). If the host's purpose is to reduce the player's chance of getting the car, he should know that whatever he does, he can not expect to reduce that chance below 1/3, because the player can always just stick to his choice. Now, if the host always declare the result right after the play pick the door, then the player's chance is the lowest possible value: 1/3, so this naive strategy of the host serves him best.

本贴由[HF:]最后编辑于：2007-2-24 7:38:31