PE 121 Disc game prize fund （第一类斯特林）

小灰灰 2022-07-14 13:07 163阅读 0赞

#### Disc game prize fund ####

#### Problem 121 ####

#### A bag contains one red disc and one blue disc. In a game of chance a player takes a disc at random and its colour is noted. After each turn the disc is returned to the bag, an extra red disc is added, and another disc is taken at random. ####

#### The player pays £1 to play and wins if they have taken more blue discs than red discs at the end of the game. ####

#### If the game is played for four turns, the probability of a player winning is exactly 11/120, and so the maximum prize fund the banker should allocate for winning in this game would be £10 before they would expect to incur a loss. Note that any payout will be a whole number of pounds and also includes the original £1 paid to play the game, so in the example given the player actually wins £9. ####

#### Find the maximum prize fund that should be allocated to a single game in which fifteen turns are played. ####

#### 题意：一开始，包里装有一个红色碟子和一个蓝色碟子。在一场概率游戏中，每一轮玩家从包中取出一个碟子，记录下其颜色，随后将碟子放回包中，并在包中加入一个红色碟子，再进行下一轮。 ####

#### 玩家需要付￡1来玩这个游戏，如果他们在游戏结束时拿出的蓝色碟子数比红色碟子数更多，则获得胜利。 ####

#### 如果游戏进行4轮，那么玩家获胜的概率是11/120，因此游戏所设定的最高奖金为￡10，否则可能会造成亏损。注意奖金必须是整数，而且包含了玩家付出用于玩游戏的￡1，也就是说玩家实际上赢得的数额是￡9。 ####

#### 如果游戏进行15轮，求游戏所设定的最高奖金。 ####

#### 题解： ####

#### 我们很容易得知道抽了n遍之后抽到蓝色碟子的概率为1n\+1，P(n,k)表示抽了n次后，蓝色碟子的数量为k的概率。所以P(n,k)=1n\+1(nP(n−1,k)\+P(n−1,k−1)) 。其实这等于P(n,k)=S(n,k)(n\+1)!，其中S(n,k)是第一类斯特林数。 ####

### 拓展： ###

#### 第一类Stirling数s(p,k) ####

#### s(p,k)的一个的组合学解释是：将p个物体排成k个非空循环排列的方法数。 ####

#### s(p,k)的递推公式：s(p,k)=(p−1)∗s(p−1,k)\+s(p−1,k−1),1<=k<=p−1 ####

#### 边界条件：s(p,0)=0,p>=1s(p,p)=1,p>=0 ####