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[转载] 抛硬币出现连续正面的概率 楼主| 发表于 2008-7-21 14:48:34 | 显示全部楼层
 由此我们可以得到 \$b(n)="round"({r^{-1}(r-1)}/{(t+1)r-2t}r^n)\$ 而抛硬币n次出现连续t次正面的概率为 \$p(n)=1-{"round"({r-1}/{(t+1)r-2t}r^{n+1})}/{2^n}; r>1&&r^{t+1}-2r^t+1=0\$ 楼主| 发表于 2008-8-25 09:44:34 | 显示全部楼层
 Now we could summarize how the formula of t-step Fibonacci sequence is got. In 9#, we got the characteristic polynomial of the Linear recurrence equation of the t-step Fibonacci sequence: \${x^{t+1}-2x^t+1}/{x-1}\$. According to link solutions of equation \$x^m=2-x^{-n}\$, we could use Rouché's Theorem to prove that there's only one root of polynomial \$x^{t+1}-2x^t+1\$ whose norm is greater than 1 and t-1 roots whose norms are less than 1. And obviously, the root with greatest norm is a real root greater than 1. Let's assuming that the real root greater than 1 is r, and besides from \$1/{z_0}=1\$ and \$1/{z_1}=r\$, there're another t-1 roots of polynomial \$x^{t+1}-2x^t+1\$. Let's assuming they're \$1/{z_2},1/{z_3},...,1/{z_t}\$. And it is obvious there're no multiple roots. So that we could write the n'th item of the t-step Fibonacii sequence in the form: \$u_1z_1^{-n}+u_2z_2^{-n}+...+u_{t}z_t^{-n}\$, and we got that \$u_s={z_s-1}/{2t-(t+1)z_s^{-1}}\$ Now let's assuming \$e_n = F_n^{(t)} - u_1 z_1^{-n} = u_2z_2^{-n}+...+u_{t}z_t^{-n}\$, so we know that the characteristic polynomial of \$e(n)\$ is \${x^{t+1}-2x^t+1}/{(x-1)(x-r)}\$. In 30#, a probability modeling is used to prove that \$|e(n)|<1/2\$. Let's assuming a man is randomly walking in an axis. He starts from a positive integer location in the axis and each time he random walks towards the positive direction by 1 or towards the negative direction by t with equivalent probability. And he will stop when he reached a point whose coordinate is no more than 1. We will give a score \$e(x)\$ to him when he finally stops at coordinate x. We know the probablity that he will finally stop is 1 when t is no less than 1. It is interesting that we could find that the expected score is \$e_n\$ when he starts from point whose coordinate is n. In 29# it is proved that the absolute values of all those final scores (\$e(-t+2)\$ to \$e(1)\$) are less than \$1/2\$, so we proved that \$|e_n|<1/2\$ for all \$n>=-t+2\$ So we know now that \$F_n^{(t)}\$ could be represented by \$"round"(u_1 z_1^{-n})\$, or \$F_n^{(t)}="round"({r-1}/{(t+1)r-2t}r^{n-1})\$ for any \$n>=-t+2\$, where r is t-bonacci constant. 发表于 2009-2-11 12:37:49 | 显示全部楼层
 高手,我想问一个问题,如果此个题100次趋于无穷大,请求出多大次数的连续正面的概率为最大? 楼主| 发表于 2009-2-13 08:21:15 | 显示全部楼层
 你这个是个不同的问题,很简单.连续出现t次正面的概率为\$1/{2^t}\$,所以出现一次的概率最大,为1/2 发表于 2009-3-9 17:14:38 | 显示全部楼层 whoops... i just did something silly... fortunately the posted comment can be edited... 发表于 2012-2-11 13:43:20 | 显示全部楼层
 其中\$z_1^{-1}\$我们可以通过取初试值为\$1/2\$然后用牛顿迭代法得到 记\$r=z_1^{-1}\$ 而递推式中其他各式绝对值都远远小于，所以我们得到，对于n充分大 \$b(n)="round"({r^{-1}(1-r)}/{2t-(t+1)r} r^{n})\$ mathe 发表于 2008-7-19 08:19 此公式非常漂亮。 请教mathe：此公式是只有正反两种情况下的计算方法，是否能推广为m种情况的计算公式？ 即：n　个　大　小　相　同　的　小　球　排　成　一　排，给　每　个　球　涂　上　m　种　颜　色　的　一　种，要　求　相　邻　颜　色　相　同　小　球　个　数　小　于　t　个，问　有　几　种　不　同　涂　法？ 发表于 2012-2-12 11:32:03 | 显示全部楼层
 上面叙述有点问题，其中相邻颜色是指定的一种颜色（m种颜色中一种）。 发表于 2012-2-14 21:55:12 | 显示全部楼层
 请教mathe：是否能推广为m种情况的计算公式？ 即：n　个　大　小　相　同　的　小　球　排　成　一　排，每　个　球　的　颜　色　为　m　种　颜　色　的　一　种，如　果　相　邻　小　球　颜　色　为　指　定　的　一　种　颜　色（m种颜色中一种）时，这　些　相　邻　小　球　个　数　小　于　t　个，问　有　几　种　不　同　排　列　方　法？ 发表于 2012-2-15 12:42:39 | 显示全部楼层
 我自己演算的16次抛硬币里面，无法出现连续2次正面的几率： 0.0049591064453125 请老大用公式演算一下

评分 gxqcn + 20 首贴奖励，欢迎常来。 发表于 2012-2-17 09:15:29 | 显示全部楼层
 我自己演算的16次抛硬币里面，无法出现连续2次正面的几率： 0.0049591064453125 请老大用公式演算一下 沉默的见证 发表于 2012-2-15 12:42 按老大计算公式，无法出现连续2次正面的几率=0.039428711 你的计算有问题

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