求解概率值

chenyu93

设抛掷硬币时，得到正面的概率为56%，得到反面的概率为44%，连续抛掷1500次，出现连续七次反面的概率是多少？

Czhang271828

其实问问 GPT 就可以了, 答案 $1-(1-0.56^7)^{1500-7+1}$.

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Question A coin is tossed 1500 times, and the probability of heads is 0.56. What is the probability that there exists 7 heads in a row?

Answer by $\mathsf{ChatGPT4}$

We can approach this by calculating the probability of the complement event, which is that there are no runs of $7$ consecutive heads.
Let $q=1-p=0.44$ be the probability of tails. We can use the formula for the probability of no runs of $k$ consecutive heads in $n$ tosses, which is:
\[P(\text{no $k$ consecutive heads in $n$ tosses})=(1-p^k)^{n-k+1}.\]
So, the probability of no runs of $7$ consecutive heads in $1500$ tosses is:
\begin{align*}
&P(\text{no $7$ consecutive heads in $1500$ tosses})\\
=\,&(1-0.56^7)^{1500-7+1}\\
=\,& 0.3917409259.
\end{align*}
Therefore, the probability that there is at least one run of $7$ consecutive heads in $1500$ tosses is:
\begin{align*}
&P(\text{at least $1$ run of $7$ consecutive heads in $1500$ tosses})\\
=\,&1-P(\text{no $7$ consecutive heads in $1500$ tosses})\\
=\,& 1-0.3917409259\\
=\,& 0.6082590741.
\end{align*}

Answer
\[P(\text{at least 1 run of 7 consecutive heads in } 1500 \text{ tosses}) \approx 0.6083.\]

kuing

知乎有相关问题：zhihu.com/question/46388875

kuing

Czhang271828 发表于 2023-5-31 12:45
其实问问 GPT 就可以了, 答案 $1-(1-0.56^7)^{1500-7+1}$.

-----------------------以下是解答过程----- ...

`P(\text{no $k$ consecutive heads in $n$ tosses})=(1-p^k)^{n-k+1}` 这是为啥呢？

chenyu93

Czhang271828 发表于 2023-5-31 12:45
其实问问 GPT 就可以了, 答案 $1-(1-0.56^7)^{1500-7+1}$.

-----------------------以下是解答过程----- ...

chenyu93

kuing 发表于 2023-5-31 14:56
`P(\text{no $k$ consecutive heads in $n$ tosses})=(1-p^k)^{n-k+1}` 这是为啥呢？

tommywong

Czhang271828 发表于 2023-5-31 12:45
其实问问 GPT 就可以了, 答案 $1-(1-0.56^7)^{1500-7+1}$.

-----------------------以下是解答过程----- ...

ChatGPT出名吹水同做錯數
$n=8$時$2p^7-p^8$
唔會係$1-(1-p^7)^{n-7+1}$

另外人哋係問緊連續七次反面，唔係正面

kuing

记 `a=0.44` 为反面概率，记 `P(n)` 为抛 `n` 次出现连续七次反面的概率。

显然当 `n<7` 时 `P(n)=0`，及 `P(7)=a^7`。

当 `n>7` 时，出现连续七次反面这个事件包含两种情况：

（1）前 `n-1` 次内已经出现连续七次反面；

（2）直至抛完最后一次才出现连续七次反面。

情况（1）概率就是 `P(n-1)`，至于情况（2），那就是前 `n-8` 次内都没有出现且最后 `8` 次为正反反反反反反反，因此概率为 `\bigl(1-P(n-8)\bigr)\cdot(1-a)a^7`。

综上，有
\[P(n)=P(n-1)+\bigl(1-P(n-8)\bigr)\cdot(1-a)a^7.\]

Czhang271828

$\mathsf{GPT}$ 确实问题一堆. 可以尝试把

1. M = Matrix([
2.     [0.56 ,0.56 ,0.56 ,0.56 ,0.56 ,0.56 ,0.56 ,0],
3.     [0.44 ,0    ,0    ,0    ,0    ,0    ,0    ,0],
4.     [0    ,0.44 ,0    ,0    ,0    ,0    ,0    ,0],
5.     [0    ,0    ,0.44 ,0    ,0    ,0    ,0    ,0],
6.     [0    ,0    ,0    ,0.44 ,0    ,0    ,0    ,0],
7.     [0    ,0    ,0    ,0    ,0.44 ,0    ,0    ,0],
8.     [0    ,0    ,0    ,0    ,0    ,0.44 ,0    ,0],
9.     [0    ,0    ,0    ,0    ,0    ,0    ,0.44 ,1]
10. ]);
11. V = Matrix([[1],[0],[0],[0],[0],[0],[0],[0]]);
12. Out = (M^1500)*V;
13. Out[-1]

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扔给 $\texttt{sage}$ 在线计算, 得到答案:

1. (0.933403093589771)

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kuing

按 8# 的递推方法，在 MMA 里用如下命令

1. a = 0.44;
2. Do[P[n] = 0, {n, 0, 6}]
3. P[7] = a^7;
4. Do[P[n] = P[n - 1] + (1 - P[n - 8]) (1 - a) a^7, {n, 8, 1500}]
5. P[1500]

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输出结果是 0.933403，不知会不会有数值误差……

hbghlyj

kuing 发表于 2023-5-31 10:05
按 8# 的递推方法，在 MMA 里用如下命令

输出结果是 0.933403，不知会不会有数值误差…… ...

In SageMath, (Python syntax) you need to initialize the list P with zeros using multiplication [0] * 7 and access the elements using square brackets P[n]. The rest of the code remains mostly unchanged.

1. a = 0.44
2. P = [0] * 7
3. P.append(a**7)
4. for n in range(8, 1501):
5.     P.append(P[n-1] + (1 - P[n-8]) * (1 - a) * a**7)
6. P[1500]

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In Matlab / Octave, indexing starts from 1 instead of 0, so the range in the for loop starts from 9 instead of 8. The initial list P is initialized with zeros using zeros function, and P(8) is assigned the value a^7 instead of using append. The rest of the code remains the same.

1. a = 0.44;
2. P = zeros(1, 1501);
3. P(8) = a^7;
5. for n = 9:1501
6.     P(n) = P(n-1) + (1 - P(n-8)) * (1 - a) * a^7;
7. end
9. P(1501)

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tommywong

#10同#11都冇錯，但#9唔知係乜

hbghlyj

Czhang271828 发表于 2023-5-31 09:25
$\mathsf{GPT}$ 确实问题一堆. 可以尝试把

扔给 $\texttt{sage}$ 在线计算, 得到答案.  ...

In Matlab / Octave, you can directly perform matrix multiplication using the * operator. The end keyword is used to refer to the last element of a vector or the last row/column of a matrix.

1. M = [0.56 ,0.56 ,0.56 ,0.56 ,0.56 ,0.56 ,0.56 ,0;
2.      0.44 ,0    ,0    ,0    ,0    ,0    ,0    ,0;
3.      0    ,0.44 ,0    ,0    ,0    ,0    ,0    ,0;
4.      0    ,0    ,0.44 ,0    ,0    ,0    ,0    ,0;
5.      0    ,0    ,0    ,0.44 ,0    ,0    ,0    ,0;
6.      0    ,0    ,0    ,0    ,0.44 ,0    ,0    ,0;
7.      0    ,0    ,0    ,0    ,0    ,0.44 ,0    ,0;
8.      0    ,0    ,0    ,0    ,0    ,0    ,0.44 ,1];
10. V = [1;0;0;0;0;0;0;0];
12. Out = M^1500 * V;
13. Out(end)

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In Mathematica, MatrixPower[M, n] is used to raise a matrix M to the power n, and Last[out] or out[[-1]] is equivalent to Out(end) in MATLAB, giving the last element of the vector out.

1. M = {{0.56, 0.56, 0.56, 0.56, 0.56, 0.56, 0.56, 0},
2.      {0.44, 0,    0,    0,    0,    0,    0,    0},
3.      {0,    0.44, 0,    0,    0,    0,    0,    0},
4.      {0,    0,    0.44, 0,    0,    0,    0,    0},
5.      {0,    0,    0,    0.44, 0,    0,    0,    0},
6.      {0,    0,    0,    0,    0.44, 0,    0,    0},
7.      {0,    0,    0,    0,    0,    0.44, 0,    0},
8.      {0,    0,    0,    0,    0,    0,    0.44, 1}};
10. V = {1, 0, 0, 0, 0, 0, 0, 0};
12. out = MatrixPower[M, 1500].V;
13. out[[-1]]

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战巡

搞这么复杂...就很简单的一个随机过程而已啊