有没有一种方法可以用概率树而不是模拟来计算概率
我想在 python 中实现一个函数，它可以为我提供以下问题的确切概率（如果可能的话以分数形式）：
您有一个包含 8 个元素的列表，假设 l=[1,2,3,4,5 ,6,7,8]，然后你在该列表中连续取出 k 个数字，如果该数字是 1,2,3 或 4 则将其从列表中删除，否则你让该列表保持原样。每个数字出现的概率是相等的。
我想计算在这n次尝试中至少有数字1的概率以及有1和2的概率。
对于n=1，有1的概率是1/8
对于n=2，有1的概率是 1/8 + 4/8 * 1/8 +3/8 * 1/7= 27/112
有 1 和 2 是 2 * 1/8 * 1/7 = 1/28
等等..
但是，因为我无法将该概率形式化为公式。我尝试用算法来计算它，但它并不简单。
我能够模拟它以获得近似值，但我对此并不满意。
def amalsimu2(adapt,n=int(1e6)):
prob_to_find_both=0
prob_to_find_one=0
for _ in range(n):
l=[i for i in range(1,9)]
rec=[]
for _ in range(adapt):
draw=l[rd.randint(0,len(l)-1)]
rec.append(draw)
if draw<5:
l.remove(draw)
if 1 in rec:
prob_to_find_one+=1
if 2 in rec:
prob_to_find_both+=1
return [round(prob_to_find_one/n*100,3), round(n/prob_to_find_one,3)],[round(prob_to_find_both/n*100,3), round(n/prob_to_find_both,3)]
我对 python 中的树做了一些研究，但我不知道这是否是一个很好的处理方法。
如果您对如何公式化我想要计算的概率有任何想法，或者您对如何使用 python 算法进行处理有一个好主意，我将非常感激
I want to implement a function in python that gives me the exact probability (if possible in fraction form) of my following problem:
You have a list of 8 elements let's say l=[1,2,3,4,5,6,7,8], then you take succesively k number in that list, if the number is 1,2,3 or 4 then you take it off that list, otherwise you let that list as it is. The probability of having every number is equivalent.
I want to calculate the probability, within these n tries, to have at least the numer 1 and the probability to have 1 and 2.
For n=1 the probability to have 1 is 1/8
For n=2 the probability to have 1 is 1/8 + 4/8 * 1/8 +3/8 * 1/7= 27/112
to have 1 and 2 is 2 * 1/8 * 1/7 = 1/28
etc..
However as I can't formalize that probability into a formula. I tried to calculated it with an algorithm but it isn't simple.
I was able to simulate it in order to have an approximation but I'm not really satisfied with it.
def amalsimu2(adapt,n=int(1e6)):
prob_to_find_both=0
prob_to_find_one=0
for _ in range(n):
l=[i for i in range(1,9)]
rec=[]
for _ in range(adapt):
draw=l[rd.randint(0,len(l)-1)]
rec.append(draw)
if draw<5:
l.remove(draw)
if 1 in rec:
prob_to_find_one+=1
if 2 in rec:
prob_to_find_both+=1
return [round(prob_to_find_one/n*100,3), round(n/prob_to_find_one,3)],[round(prob_to_find_both/n*100,3), round(n/prob_to_find_both,3)]
I looked a little bit into tree in python but I don't know if it is a good way to process.
If you have any idea on how to formulize the probability I want to compute or if you have a good idea on how to process to make it with a python algorithm, I would really appreciate that
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它更像是概率图而不是树，但这取决于您如何定义状态（图中的节点）。下面的代码将“状态”定义为仍然可以选择的整数元组。最大的优点是只有 16 种可能的状态（取决于已经选择了
{1, 2, 3, 4}的子集），并且状态越少，速度就越快。缺点是可能的状态太少，限制了可以回答的问题。例如，没有状态记录是否曾经选择过 5。但我不知道你所说的“等”是什么意思，你的描述和你的代码都只询问了 prob_to_find_one 和 prob_to_find_both ，这里使用的状态足以回答那些。
输出开始如下：
请注意，虽然
Fraction类型会自动转换为最低项，但分子和分母会快速变大。这是精确的有理算术。It's more like a probability graph than a tree, but that depends on how you define a state (a node in the graph). The code below defines "a state" as the tuple of integers that can still be chosen. The great advantage is that there are only 16 possible states then (depending on which subset of
{1, 2, 3, 4}has already been picked), and the fewer the states the faster this goes.The disadvantage is that having so few possible states constrains questions that can be answered. For example, no state records whether a 5 has ever been picked. But I don't know what you meant by "etc.", and both your description and your code only asked about
prob_to_find_oneandprob_to_find_both, The states used here are sufficient to answer those.The output starts like so:
Note that while the
Fractiontype automatically converts to lowest terms, the numerators and denominators grow large quickly. This is exact rational arithmetic.