1
赌博（装备锻造）必定破产
source link: https://changkun.de/blog/posts/%E8%B5%8C%E5%8D%9A%E5%BF%85%E5%AE%9A%E7%A0%B4%E4%BA%A7/
Go to the source link to view the article. You can view the picture content, updated content and better typesetting reading experience. If the link is broken, please click the button below to view the snapshot at that time.
赌博（装备锻造）必定破产
Published at:
2011-04-06
|
Reading: 338 words ~1min
|
PV/UV: 4/4
问题：某傻逼有本金a元，决心再赢b元停止赌博。设这个傻逼每局赢的概率是p=1/2，每局输赢都是一块钱，傻逼输光就不赌了，求傻逼输光的概率q(a)。
解：用A表示这个傻逼第一局赢，用B_k表示甲有本金k元时最后输光。由题意，q(0)=1，q(a+b)=0,并且
=P(B_k)
=P(A)P(B_k|A)+P(非A)P(B_k|非A)
=P(B_k+1)/2+P(B_k-1)/2
=q(k+1)/2+q(k-1)/2
于是有2q(k)=q(k+1)+q(k-1)，从而
q(k+1)-q(k)=q(k)-q(k-1)=…=q(1)-q(0)=q(1)-1.
上式两边对k=n-1,n-2,…,0求和后得到
q(n)-1=n[q(1)-1].
取n=a+b,得到
0-1=(a+b)[q(1)-1],q(1)-1=-1/(a+b).
由q(n)-1=n[q(1)-1]得到
q(a)=1+a(q(1)-1)=1-a/(a+b)=b/(a+b).
上式说明，当a本金有限，则贪心b越大，输光的概率越大。如果一直赌下去(b->infinity)，必定输光。
Recommend
About Joyk
Aggregate valuable and interesting links.
Joyk means Joy of geeK