機率問題....深夜睡不著隨便打 大家隨便看 沒足球 0rz 不過.....ㄟ.....有時候修個課會比較好懂，很多東西是不太直觀的 我給這串討論串提供一點新的想法(還是有人講過我沒看到?) 「手上每一張非拍賣師的牌打出去之後，剩下的手牌有拍賣師的機率就會越變越高」 妳可能會覺得"幹甲賽，她又沒抽新的牌，為什麼手牌有拍賣師的可能性會越變越大" 這裡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 懶人包請直接看作者幫你寫好的spreadsheet 包妳直接知道"下一張"打出拍賣師的機率有多高 (那是設牌組中只有一張拍賣師的情況下 多的請自己算... 而且越多張拍賣師，機率和直覺算差距越大) 至於這篇文章為什麼會提到monty hall呢 你要注意一個重點 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的概念就是每一張非拍賣師的牌被打出來，相當於沒有中獎的門會一直被打開 所以剩下沒打的那些手牌就更有可能是拍賣師， 不要被「他有沒有抽牌/他有沒有換門」給誤導了 而且這樣的機率會比妳想像中的高很多，完全違背沒學過的人的直覺 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. 兩張的狀況 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%. 打到這邊突然有睡意了 不要怪我沒翻譯 看到機率手癢而已 請各位幫忙補充給看不懂英文的版友看啦 0rz -- 不過....看討論，很多版友可能沒上課的話 一輩子都搞不清楚monty hall problem的問題在哪 其實outcome畫個樹狀圖就好了 我之前也曾經教同學教到快崩潰過 習慣之後妳永遠不會忘記條件機率的那個"條件"相當重要 --

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