最近看到一个有趣的数题

tensaix2j

有两个信封， 里面各自有一个 real number（－infinity，＋infinity），（数值的来源可能是故意设定，也有可能是随机的。）

你会随机的选择其中一个信封，打开信封后，看到里面的数值， 然后，你可以选择换，或是不换。

如果你选择的信封里的数值高过另一个信封里的数值，你就算赢，
请问，要如何可以增加你赢的几率？

tjfoo

這好象叫two envelope problem還是exchange paradox的。
但是你是說任選兩個實數，好像是一個叫什麼necktie paradox的。

puangenlun

乍看之下像是monty hall problem
但是这个只有两个物件，不是三个

尝试用monty hall problem 的分析方法来算
得到的答案是一切都是天注定

Batultu

少过零就换？

tensaix2j

是从这个人发的贴转来的，（XKCD 漫画的作者）
https://plus.google.com/111588569124648292310/posts/dv9Fi45h91T

comment 里应该可以找到答案。

saturn85

回复 5# tensaix2j


   深。。

tensaix2j

抱歉， 两年前的帖，现在才公布答案。。。

The solution is pretty simple - before opening one of your envelopes, I pick my OWN random real number. Then, if the number in the envelop I open is less than the number I picked, I switch. If envelope number is greater than or equal to the number I picked, I stay with this envelope. (Note: I edited this to make my strategy clear!)

Why does this work? It's simple - look at all the possible cases. There's an equal chance that the numbers in your envelope are both below, or both above, the number I picked. And in either of those cases, either switching or staying still gives me a 50/50 chance of being right.

BUT, if I happen to pick a real number that is between your two numbers (and there must be always some very small chance of this happening, though this chance approaches 0 as the range of numbers increases out to infinity), then I have a 100% chance of being right by either switching or staying per my strategy. Therefore, it follows that overall, I have a greater than 50% chance of getting the bigger envelope, since some cases give me a 50% chance and other cases give me a 100% chance.

鸭王之王

tensaix2j 发表于 4-1-2013 10:19 PM
抱歉， 两年前的帖，现在才公布答案。。。

好久没有看到你了

tensaix2j

鸭王之王 发表于 5-1-2013 03:44 PM
好久没有看到你了

近来可好？

鸭王之王

tensaix2j 发表于 5-1-2013 05:19 PM
近来可好？

好，还在新坛混？？

tensaix2j

鸭王之王 发表于 5-1-2013 06:06 PM
好，还在新坛混？？

是的。 你呢？

鸭王之王

tensaix2j 发表于 5-1-2013 07:37 PM
是的。 你呢？

我？在浪费时间