挑戰概率題

I haven't got a proof yet, but I believe the answer is way simpler than that.
My solution is P(win) = (m - n + 1)/(m+1).

I have checked by answer by simulation.

2010-06-08 22:36:38 補充：
希望被選為最佳答案！
圖片參考：http://i187.photobucket.com/albums/x22/cshung/7010053100463_1.png


圖片參考：http://i187.photobucket.com/albums/x22/cshung/7010053100463_2.png


圖片參考：http://i187.photobucket.com/albums/x22/cshung/7010053100463_3.png


我使用的simulation software就是普通的programming啊。代碼如下，請參巧

static void Simulate()
{
Random random = new Random ();
int failCount = 0;
int labCount = 100000;
for (int lab = 0; lab < labCount; lab++)
{
bool fail = false;
int bt = 0, wt = 0, bb = m, wb = n;
for (int i = 0; i < (m + n) && !fail; i++)
{
int chosen = random.Next(bb + wb) + 1;
if (chosen > bb)
{
wt++; wb--;
}
else
{
bt++; bb--;
}
if (wt > bt)
{
fail = true;
}
}
if (fail) { failCount++; }
}
Console.WriteLine((labCount - failCount) / (double)labCount);
}

2010-06-09 13:18:53 補充：
http://i187.photobucket.com/albums/x22/cshung/7010053100463_1.png
http://i187.photobucket.com/albums/x22/cshung/7010053100463_2.png
http://i187.photobucket.com/albums/x22/cshung/7010053100463_3.png

2010-06-13 08:46:36 補充：
I tried, not easy to solve directly.
If you expand that by substituting recursively, you will realize that it an iterated running sum of running domains.

若直接解Q(m,0) = 1,Q(m,m) = Q(m,m - 1),Q(m,n) = Q(m - 1,n) + Q(m,n - 1)則更好。

I agree with Andrew's answer.

2010-06-04 20:43:07 補充：
Please see the following by mathematical induction:

圖片參考：http://img64.imageshack.us/img64/3430/11366369.png


圖片參考：http://img43.imageshack.us/img43/2283/39450624.png


2010-06-04 20:43:23 補充：
http://img64.imageshack.us/img64/3430/11366369.png
http://img43.imageshack.us/img43/2283/39450624.png

2010-06-04 20:50:09 補充：
P(m,n) is speculated by Excel simulation which can be downloaded:
http://www.funp.net/574341
Only test with small numbers of m,n. Excel cannot accomodate larger numbers due to size limitation

2010-06-08 19:36:56 補充：
家浩君:本人之證法沒前設甚麼,純用數學歸納法,只是沒留意大將軍原題,篹改了M及N變為m及n.我是在證若M>=N時, P(M,N) 必然成立.所以沒辦法了解及解決你的疑問.

你好. 以下是我的答案. 希望可以看得清楚.


圖片參考：http://imgcld.yimg.com/8/n/HA00923483/o/701005310046313873416600.jpg


2010-06-01 13:41:33 補充：
若看不清楚, 可以長按[ctrl] + 將滑鼠上下移輪向前滾來放大(向後滾是縮小).

2010-06-02 10:59:30 補充：
Sorry, 有錯. 除了 P(m >= n) 和 P(在 x 次中, m >= n) 中的 m 和 n 外, 其他的 m 和 n 都是大楷的, 即是 M 和 N.

2010-06-02 17:28:21 補充：
Dear Andrew :

I think you answer has some problems.

Try m = 3 & n = 2, P(win) = 1/2

Will it be true for one draw, two draws, three draws, four draws and five draws?

Obviously, it is not true for one draw at least.

2010-06-07 08:17:04 補充：
各位, 我知道我的答案是錯的. 因為我的答案包括了白球先領先, 後打和或被反超前的可能性.

另, 我對 002 的答案有一點疑問 : 相信 002 對 m 和 n 的前設與我是一樣的, 即 m+n = x <= M+N

這樣, 就 002 的答案, 既然已經知道 m 和 n 的確實數量, 那麼求 P(m >= n) 的意義何在?

2010-06-09 13:53:50 補充：
差點搞到我頭都暈! 你們的答案應該是 :
P(黑球數目一直保持領先或打和白球數目) = (M + 1 - N) / (M + 1)

2010-06-14 22:39:59 補充：
好不容易找到以下相關的文章 :
http://www.cs.ccsu.edu/~gusev/AlgLin/Gusev_TR448.pdf

相信確實的證明可以在 Paul W. Purdom 的 "The Analysis of Algorithm" 中找到.