I could do the same thing for Player 2 to find Player 2's best response for every possible choice of Player 1.

同理对于参与人2来说,可以算出参与人1不同策略的最佳对策

耶鲁公开课 - 博弈论课程节选

Just like it wasn't a crazy thing to say if we only had the guy in Michigan.

同理，如果只有这个密西根家伙，那么也不算疯狂。

耶鲁公开课 - 死亡课程节选

Absolutely the same thing is true on the market timing front.

市场时机选择同理

耶鲁公开课 - 金融市场课程节选

The argument that eliminates strategies 67 and above, or 68 upwards, that strategy just involves the first lesson of last time: do not choose a dominated strategy, admittedly weak here, but still.

剔除67以上或者68以上的过程,仅仅用到了上堂课所提到的第一个结论,即不要采用劣势策略,虽然这里是弱劣势,但同理

耶鲁公开课 - 博弈论课程节选

So the translation here is that whereas last week and in Scratch, the looping structures are yellow and have kind of this brace structure to them.

同理，大家回想一下上个星期,在Scratch软件中，循环结构都用黄色标记出来了,与此类似。

哈佛公开课 - 计算机科学课程节选

Again it's like anyone who does important things it's hard to judge the whole picture and on that I am sure he has created value for the economy.

同理，对于那些做着重要事情的人们,很难估测大局,不过我很肯定的是,他已经为经济创造了价值。

耶鲁公开课 - 金融市场课程节选

So I could at the same, I could do the same kind of analysis but we know I'll get the same answer.

我们能够同理继续分析参与人II的,但我们都知道答案是一样的

耶鲁公开课 - 博弈论课程节选

The idea is Player I directly contributes profits to the firm by working, as does Player II.

说明了参与人I努力工作对公司有利,同理可知参与人II

耶鲁公开课 - 博弈论课程节选

Now I could go through again and do exactly the same thing for Player II, but I'm not going to do that because everything's symmetric.

如果我照此计算同理可得参与人II的,我就不算了,因为都是对称的

耶鲁公开课 - 博弈论课程节选

So we're throwing away all of the strategies less than 5/4 for Player I and bigger than 6/4 for Player I, which is 1? for Player I and similarly for Player II.

这样我们又剔除了,参与人小于5/4大于6/4的策略,参与人I只有1?的区间,同理可证参与人II

耶鲁公开课 - 博弈论课程节选

So similarly, I would find that ?2 2 equals 1 plus B S1 and this is the best response of Player II, as it depends on Player I's choice of effort S1.

同理可得?2等于1+B*S1,?2是参与人II的最佳对策,因为它与参与人I的策略S1有关

耶鲁公开课 - 博弈论课程节选

Similarly, for Player II, the lowest Player I could ever do, is choose 0, in which case Player II would want to choose 1, so the strategies below 1 are never a best response for Player II.

同理,参与人I最低只能选0,此时参与人II应该会选1,小于1的策略不是参与人I的最佳对策

耶鲁公开课 - 博弈论课程节选

Now we can do the same for Player II, we can draw Player II's best response as it depends on the choices of Player I, but rather than go through any math, I already know what that line's going to look like.

同理也可以画出参与人II的图像,参与人II的最佳对策,取决于参与人I的策略,不用任何数学的方法,我就知道图像是什么样的了

耶鲁公开课 - 博弈论课程节选