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12/26/2004 08:22 AM by name withheld; Strategy:: try Myerson's Game Theory (p.65-67) | section 2.7 on Common Knowledge. [View full text and thread]
Ok, they are 7 men who are hung. If there were only 1, her wife of the adulterer would know it inmediately, since she would observe 0 adulteries knowing there has been one. If there were 2 adulteries, each of the wifes would observe 1 adultery. If there is no hung men on day 1 and I am one of the two wifes, I know my husband has been untruth, since the other woman is observing some other adultery. There is no other, therefore, it is my husband. Imagine there are 3. Each of the 3 wives will observe only 2. On the morning of day 1 there are no men hung. So a wife may think: ah, there are observing each other, tomorrow the both are hung. But, day 2 arrives and no men are hung. So I, as the third wife, realize that there is one more man who committed adultery, MY MAN. I can prove it in the evening and have him hung. The three of them think the same. So, by induction, the strategy is as follows. You, as a wife, wait as many days as adulteries you have observed. If on that morning they are all hung, your husband is innocent, but if they are not, your husband is one of them and they will be all hung tomorrow. The answer is: 7 husbands are hung. [Manage messages]
I guess all the men are hung. No one observes adultery, since no one is unfaithful after day 0. Since all wifes do not see after day 0 any adultery, but they know that there has been one. No men are hung on day 1 and the only men they [View full text and thread]
haha I'm pretty sure the answer is yes [View full text and thread]
11/12/2004 01:47 AM by SJ; Strategy | Hello,
I have come across a question that has temporarily stumped me:
There is a kingdom, with a number of couples living in the kingdom. On day 0, the king announces that some men in the kingdom have committed adultery, and that [View full text and thread]
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