机率问题....深夜睡不着随便打 大家随便看 没足球 0rz 不过.....ㄟ.....有时候修个课会比较好懂，很多东西是不太直观的 我给这串讨论串提供一点新的想法(还是有人讲过我没看到?) 「手上每一张非拍卖师的牌打出去之後，剩下的手牌有拍卖师的机率就会越变越高」 你可能会觉得"干甲赛，她又没抽新的牌，为什麽手牌有拍卖师的可能性会越变越大" 这里就是要用功的部分啦 http://iam.yellingontheinternet.com/2013/09/01/ hearthstone-probabilities-and-the-monty-hall-effect/ http://goo.gl/kIurmn 懒人包请直接看作者帮你写好的spreadsheet 包你直接知道"下一张"打出拍卖师的机率有多高 (那是设牌组中只有一张拍卖师的情况下 多的请自己算... 而且越多张拍卖师，机率和直觉算差距越大) 至於这篇文章为什麽会提到monty hall呢 你要注意一个重点 Monty Hall, however, tells you differently. Even though she only has 5 cards in her hand next turn, she’s been selecting out non-Polymorph cards and playing them all game. Just as Monty selected out non-winning doors and removed the pool, making the remaining doors of the ones he could have chosen more likely to be winners, Jaina has been casting non-Polymorph cards from her hand, making the ones she’s left in her hand more likely to be Polymorphs. monty hall的概念就是每一张非拍卖师的牌被打出来，相当於没有中奖的门会一直被打开 所以剩下没打的那些手牌就更有可能是拍卖师， 不要被「他有没有抽牌/他有没有换门」给误导了 而且这样的机率会比你想像中的高很多，完全违背没学过的人的直觉 For now consider the simplest case, where nothing had been cast so far this game that Jaina would have been likely to Polymorph. Applying the logic from above, there’s a 50% chance that the Poly started the game in the bottom 15 cards in the deck, and that probability has not been changed by any subsequent events. There must, then, be a 50% chance that it’s among the 5 cards in her hand. Quite a significant difference from the 25% that seemed completely intuitive before considering this effect. 两张的状况 The other complexity is what I alluded to at the start: the arithmetic is more complicated with two Polymorphs in the deck (but as I said, the logic is unchanged). To work the same example with two Polymorphs: The naive estimate is that out of 20C2 (20 choose 2, referring to combinations) places the Polys could be, 15C2 choices have them in the bottom 15, so the chance of having one in hand is 1-(15C2 / 20C2), which evaluates to 17/38, or around 45%. The Monty-corrected estimate would be that out of 30C2 possible placements at the start of the game, 15C2 have them undrawn in the first 15 cards, so the chance of having one in hand is 1-(15C2 / 30C2), which evaluates to 22/29, or around 76%. 打到这边突然有睡意了 不要怪我没翻译 看到机率手痒而已 请各位帮忙补充给看不懂英文的版友看啦 0rz -- 不过....看讨论，很多版友可能没上课的话 一辈子都搞不清楚monty hall problem的问题在哪 其实outcome画个树状图就好了 我之前也曾经教同学教到快崩溃过 习惯之後你永远不会忘记条件机率的那个"条件"相当重要 --

※ 发信站: 批踢踢实业坊(ptt.cc), 来自: 220.134.44.75 ※ 文章网址: http://webptt.com/cn.aspx?n=bbs/Hearthstone/M.1404765452.A.C9C.html