抛硬币出现连续正面的概率

mathe

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 $p(n)=1-{"round"({r-1}/{(t+1)r-2t}r^{n+1})}/{2^n}; r>1&&r^{t+1}-2r^t+1=0$

mathe

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.

zhangjianquan

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?

mathe

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

shangwen_ren

whoops... i just did something silly... fortunately the posted comment can be edited...

AAAAA

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？

AAAAA

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）。

AAAAA

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？

沉默的见证

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

AAAAA

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