用python写了下理想情况下铁血打捞的情况

本人从池子刷新到现在，共进行了22次打捞，结果18次遇到3个1星，4次遇到2个1星和1个2星，遇到2星打捞成功0次，最终结果如图。

理想情况下指的是所有铁血的出现概率相等，例如有100只铁血，则每只出现的概率均为100%/100。在理想情况下刷出3只1星的概率是：(x*(x-1)*(x-2))/(y*(y-1)*(y-2))，x，y分别代表1星剩余数和总剩余数

代码如下：建议运行方式 python 文件路径\文件名.py>文件路径\输出文件名.txt

from random import randint

N,R,SR=71,28,1

ALL=N+R+SR

Rand_1,Rand_2,Rand_3=0,0,0

temp_msg=''

cost=0

def wave_fresh():

global N,R,SR,ALL

global Rand_1,Rand_2,Rand_3

global temp_msg

global cost

if ALL>=1:

Rand_1=randint(1,ALL)

if Rand_1<=N:

temp_msg +='\n第'+str(cost+1)+'轮 '+'1:☆ '

elif Rand_1<=N+R:

temp_msg += '\n第'+str(cost+1)+'轮 '+'1:☆☆ '

else:

temp_msg += '\n第'+str(cost+1)+'轮 '+'1:☆☆☆ '

if ALL>=2:

Rand_2=randint(1,ALL)

while Rand_2==Rand_1:

Rand_2=randint(1,ALL)

if Rand_2<=N:

temp_msg += '2:☆ '

elif Rand_2<=N+R:

temp_msg += '2:☆☆ '

else:

temp_msg += '2:☆☆☆ '

if ALL>=3:

Rand_3=randint(1,ALL)

while Rand_3==Rand_2 or Rand_3==Rand_1:

Rand_3=randint(1,ALL)

if Rand_3<=N:

temp_msg += '3:☆ '

elif Rand_3<=N+R:

temp_msg += '3:☆☆ '

else:

temp_msg += '3:☆☆☆ '

print(temp_msg)

temp_msg=''

def RST(x):

Result=randint(1,100)

if Result<=x:

return True

else:

return False

def capture(a,b,c):

global N,R,SR,ALL

global temp_result

global cost

wave=[a,b,c]

temp=0

Result=False

#temp=eval(input('选择捕获目标:'))

if a>b and a>c:

temp=1

elif b>a and b>c:

temp=2

else:

temp=3

if temp>0:

try:

if wave[temp-1]<=N:

Result = RST(100);

N-=1

elif wave[temp-1]<=N+R:

Result = RST(50)

if Result==True:

R-=1

else:

Result = RST(25)

if Result==True:

SR-=1

except:

Result=False

if Result==True:

print('捕获成功。')

else:

print('捕获失败。')

cost+=1

ALL = N + R + SR

print('残余 N:',N,' R:',R,' SR:',SR,' ALL:',ALL)

def main():

global ALL

global Rand_1,Rand_2,Rand_3

global temp_result

wave_fresh()

capture(Rand_1,Rand_2,Rand_3)

if ALL>0:

#cont=eval(input('是否继续(1/default):'))

cont=1

if cont==1:

Rand_1,Rand_2,Rand_3=0,0,0

return main()

print('\n花费:',cost)

main()