不要頭二獎, 祇要三獎, 抽獎機率, 求助

2007-11-13 6:55 pm

10人參與抽獎, 每人出資100元(不論選多少票)

每人可選1票至10票, 如已中獎, 餘票馬上作廢, 即每人只有一次中獎機會。

共抽3個獎

第一獎 1元

第二獎 1元

第三獎 998元

現已知其餘9人全部選取 2票,

請問第十人應選取多少票, 取得三獎的機率是最大

更新1:

Syschiu05, 9x2 + m = 18m 應該是 =18+m, 答案是 1, 亦應是錯, 因有命到抽三獎時, 中三獎的機會才只有十五分之一,遠低于平均概率 (十分之一)

更新2:

hahatse,分析合理, 思路易明, 但 (18/20+x)(16/18+x)(x/16+x) 此式有點不對徑, 我感覺應是: 16/(18+x) * 14/(16+x) * x/(14+x) 請指教

回答 (2)

HaHa

2007-11-13 9:08 pm

✔ 最佳答案

設X是第十位所投的票
總票數＝9x2+ｘ＝１８＋ｘ
P(第十位得三獎）＝P(非第十位得第一獎）xP(非第十位得第二獎）xP(第三位中票）
＝(18/20+x)(16/18+x)(x/16+x)
=288x/(20+x)(18+x)(16+x)
P(x=1)=0.0424
P(x=2)=0.0727
p(x=3)=0.094148
P(x=4)=0.109
P(x=5)=0.11925
P(x=6)=0.12587
P(x=7)=0.1288
P(x=8)=0.131868
P(x=9)=0.1324
P(x=10)=0.131868
因此，第十位應選取9票

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Chiu

2007-11-13 7:12 pm

Let the number of votes selected by the tenth person be m.
Then, the total number of votes = 9x2 + m = 18m

P(3rd prize) = P(another person other than the tenth one won the 1st prize) x
P(another person other than the tenth one and the 1st prize winner won the 2nd prize) x
P(the tenth person wins the 3rd prize)
= 2/18m x 2/(18m-2) x m/(18m-2-2)
[2 votes are removed after each draw because all votes of the winner will be void]
= (1/9m) [1/(9m-1)] [m/(18m-4)]
= 1/[18(9m-1)(9m-2)]
In order to maximize P(3rd prize), the denominator 18(9m-1)(9m-2) should be minimized.
Since 1<=m<=10, one vote should be selected in order to maximize P(3rd prize)

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收錄日期: 2021-04-18 23:48:16
原文連結 [永久失效]:
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