1樓：小嚴
根據 @靈劍 的解法，寫了個程式實現，算出來9張牌全部翻開的概率大約是0.010325，和@趙明的結果很接近。
具體來說，翻完10次後，0張牌正面朝上的概率約為0.001999，1張牌0.017885，2張牌0.
071108，3張牌0.164837，4張牌0.245569，5張牌0.
243642，6張牌0.160522，7張牌0.067765，8張牌0.
016343，9張牌0.010325
static void Main(string args)
curStates = intermediateStates;
intermediateStates = new double[10Print output
Console.WriteLine(curStates[9Flip:
// For given state, calculate the next state after one flip
// Update resultStates accordingly
// curStates: array storing initial probabilities
// resultStates: array to store the result probabilities
// numberOfUpCards: Number of cards facing up(current state)
static void Flip(double curStates, double resultStates, int numberOfUpCards)
int totalCase = 0;
// could flip 1, 2 or 3 cards at a time, so (1/3) for each case
double eachFlipRate = 1.0 / 3;
// Flip 1 card
totalCase = 9; // A(9,1)
// 1 card facing up -> down
if (numberOfUpCards > 0resultStates[numberOfUpCards - 1] +=
curRate * eachFlipRate * numberOfUpCards / totalCase;
} // 1 card facing down -> up
if (numberOfDownCards > 0resultStates[numberOfUpCards + 1] +=
curRate * eachFlipRate * numberOfDownCards / totalCaseFlip 2 cards
totalCase = 9 * 8; // A(9,2)
// 2 cards facing up -> down
// numberOfUpCards * (numberOfUpCards - 1) cases
if (numberOfUpCards > 1resultStates[numberOfUpCards - 2] +=
curRate * eachFlipRate * numberOfUpCards * (numberOfUpCards - 1) / totalCase;
} // 1 card facing up -> down, 1 card facing down -> up
if (numberOfUpCards > 0 && numberOfDownCards > 0resultStates[numberOfUpCards] +=
curRate * eachFlipRate * numberOfUpCards * numberOfDownCards * 2 / totalCase;
} // 2 cards facing down -> up
// numberOfDownCards * (numberOfDownCards - 1) cases
if (numberOfDownCards > 1resultStates[numberOfUpCards + 2] +=
curRate * eachFlipRate * numberOfDownCards * (numberOfDownCards - 1) / totalCaseFlip 3
totalCase = 9 * 8 * 7; // A(9,3)
// 3 cards facing up -> down
// numberOfUpCards * (numberOfUpCards-1) * (numberOfUpCards-2)
if (numberOfUpCards > 2resultStates[numberOfUpCards - 3] +=
curRate * eachFlipRate * numberOfUpCards * (numberOfUpCards - 1) * (numberOfUpCards - 2) / totalCase;
} // 2 cards facing up -> down, 1 card facing down -> up = -1 up
// numberOfUpCards * (numberOfUpCards-1) * numberOfDownCards * A(3,1)
if (numberOfUpCards > 1 && numberOfDownCards > 0resultStates[numberOfUpCards - 1] +=
curRate * eachFlipRate * numberOfUpCards * (numberOfUpCards - 1) * numberOfDownCards * 3 / totalCase;
} // 1 card facing up -> down, 2 cards facing down -> up = +1 up
// numberOfUpCards * numberOfDownCards * (numberOfDownCards - 1) * A(3,1)
if (numberOfUpCards > 0 && numberOfDownCards > 1resultStates[numberOfUpCards + 1] +=
curRate * eachFlipRate * numberOfUpCards * numberOfDownCards * (numberOfDownCards - 1) * 3 / totalCase;
} // 3 cards facing down -> up
// numberOfDownCards * (numberOfDownCards - 1) * (numberOfDownCards - 2)
if (numberOfDownCards > 2resultStates[numberOfUpCards + 3] +=
curRate * eachFlipRate * numberOfDownCards * (numberOfDownCards - 1) * (numberOfDownCards - 2) / totalCase
2樓：InvsGhost
每次隨機翻開1-3張.那麼單次過程中每張牌被翻的概率P是：
單次概率P，重複N次，發生奇數次概率是 f(p,n)=[ 1 - (1-2p)^n ] / 2；
所以應該就是 ( f(P,10) )^9
大家都沒答。。心裡沒底（好久不看概論了）
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