规则：买大小，中奖则赚0.98，不中赔1。我采用每次2注的投注方式，每注100元，输注的以倍注的形式追投，赢注的回到原来投注的金额投注，模拟每轮买10次投注，购买100轮，判断平均输赢金额
（此算法根据博彩规则而写，博主不建议购买实践，奉劝大家，赌博有风险，十赌九诈）
import random
def absCheck(num1,num2):
temp=num1-num2
if temp>=0:
return '盈利:',abs(temp)
else:
return '亏损:',abs(temp)
def kaizero(monery,num1,num2):
print('大投注%d'%num1)
print('小投注%d'%num2)
print('开大')
print('大+%d'%(num1*0.98))
print('小-%d'%num2)
strr,yk=absCheck(num1*0.98,num2)
print(strr,yk)
# print('余额%d'%(monery+num1-num2))
return monery+(num1*0.98)-num2,num2*2
def kaione(monery,num1,num2):
print('大投注%d'%num1)
print('小投注%d'%num2)
print('开小')
print('大-%d'%num1)
print('小+%d'%(num2*0.98))
strr,yk=absCheck(num2*0.98,num1)
print(strr,yk)
# print('余额%d'%(monery+num2-num1))
return monery-num1+(num2*0.98),num1*2
def runner(runNum,monery,num1,num2):
i = 0
ii = 0
tempzero=num1
tempone=num2
count = 0
while True:
count+=1
print('##########%d次开始下注###########'%count)
j=random.randint(0,1)
# print(j)
if j==0:
i+=1
ii=0
monery,num2=kaizero(monery,num1,num2)
num1=tempzero
print('余额%d'%monery)
print('##########%d次结束下注###########'%count)
# print(num1,num2)
else:
ii+=1
i=0
monery,num1=kaione(monery,num1,num2)
num2=tempone
print('余额%d'%monery)
print('##########%d次结束下注###########'%count)
# print(num1,num2)
# if i==runNum or ii==runNum or monery <=0:
# print('%d次结束'%count)
# print('%d块余额'%monery)
# break
if count==runNum:
print('%d次结束'%count)
print('%d块余额'%monery)
return monery
break
if __name__ == '__main__':
runNum=10
monery=5000
num1=100
num2=100
list=[]
sum=0
for i in range(100):
res=runner(runNum,monery,num1,num2)
list.append(res)
for item in list:
print(item)
sum+=item
print('平均：%d'%(sum/len(list)-5000))
