Re: [機統] 基本機率事件 - 交大電子所乙組

看板Math作者cuttlefish (無聊ing ><^> .o O)時間7年前 (2016/12/27 20:28)推噓3(3推 0噓 6→)

留言9則, 4人參與討論串2/2 (看更多)

※ 引述《martine318 ()》之銘言： : Two gamblers play the game of "head or tails" . Each time a coin lands : heads up , then player A wins $1 from B ; otherwise (i.e. a coin lands : tails up) , player B wins $1 from A . Suppose that player A initially has : a dollars and player B has b dollars. : (b)If the game is not fair and on each play , player A wins $1 from B : with probability p , p is not 1/2 , 0 < p < 1 . What is the probability that : player A will be ruined? : 解答 : A贏1$的機率為p，B贏1$的機率為q = 1-p，則 : A輸光的機率 = Pa = [1 - (q/p)^a] / [1 - (q/p)^a+b)] : -------------------------------------------------------------- : (a)小題問A輸光的機率，答案是b/a+b，因為可以理解所以沒問這題 : (b)小題中想知道為何有(q/p)^a 以及(q/p)^a+b : 道底是套甚麼式子?? 謝謝!! P(i) = Prob{ A 最後輸掉(Ruin) | A有i元 } 記N = a+b, 所以由定義P(0)=1, P(N)=0. 且由機率可寫出遞迴式 P(i) = p*P(i+1) + q*P(i-1) ^^^^^^^^下一回A贏 ^^^^^^^^下一回A輸所以 P(i+1)-P(i) = (q/p) [P(i)-P(i-1)] = ... = (q/p)^i [P(1) - P(0)] 解P(1) 利用P(0)=1, P(N)=0 P(n) = [P(n) - P(n-1)] + [P(n-1) - P(n-2)] + ... + [P(1) - P(0)] + P(0) = [(q/p)^{n-1} + (q/p)^{n-2} + ... + 1] * [P(1) - 1] + 1 = [(q/p)^n - 1]*[P(1) - 1] / [(q/p) - 1] + 1 且 0 = P(N) = [(q/p)^N - 1]*[P(1) - 1] / [(q/p) - 1] + 1 ...(*) (*)可解出P(1) - 1, 而所求 = P(a) = 1 - [1 - (q/p)^a] / [1 - (q/p)^{a+b}] -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 97.99.68.240 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1482841685.A.D84.html

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yyc2008

12/27 22:48, , 1^F

12/27 22:48, 1^F

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yyc2008

12/27 22:49, , 2^F

12/27 22:49, 2^F

改一下的確寫錯了

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doom8199

12/27 23:12, , 3^F