Well, I know that it has already been analyzed and I haven't come to a particularly remarkable conclusion. Still, if I'm studying Python, I'll give it a try.
I can't do manual copying, so I decided to scrape it. The last data as of today (20/10/09) is the 1524th.
In LOTO6, 6 numbers and 1 bonus number will be drawn, but this time we will only play with these numbers. The numbers drawn will be 1-43 and will not be duplicated.
scloto6.py
import requests
from bs4 import BeautifulSoup
r = requests.get('http://hogehoge.com/loto6/data/list1/')
soup = BeautifulSoup(r.content, "html.parser")
numbers = soup.find_all('td', class_='w4')
i = 1
with open('index.txt','w') as f:
for number in numbers:
s = number.text.replace('\t', '')
s = s.replace('\n', '')
if len(s) == 1:
s = '0' + s
if (i % 6) != 0:
s = s + '\t'
else:
s = s + '\n'
f.write(s)
i += 1
A TAB-delimited text file like this will be generated.
index.txt
02 08 10 13 27 30
01 09 16 20 21 43
01 05 15 31 36 38
16 18 26 27 34 40
09 15 21 23 27 28
Using this as an input file, the following processing is performed.
--Calculate σ from the sum of each time --Count the number of occurrences of each number (1..43) and sort in descending order ――I will buy 5 units ――Two of the six numbers for one unit will be picked up from the top 10 appearances. --The remaining 4 numbers will be randomly picked up from 11..33 excluding the bottom 10 appearances. --6 If the sum of the selected numbers is out of the range of ± σ, start over.
Since the average of the total of each time up to the 1524th time is 131.94 and σ is 28.37, -σ to σ is in the range of 103.57 to 160.31.
| The number of data | Percentage | |
|---|---|---|
| -σ～σ | 1,042 | 68.37% |
| -2σ～2σ | 407 | 95.08% |
| -3σ～3σ | 73 | 99.87% |
In other words, the probability that the total value of the 6 numbers will fall between 104 and 160 is about 2/3. The lottery results such as [1, 2, 3, 4, 5, 6] are 21 in total and deviate from 3σ, so you can almost ignore them (≒ give up if that happens).
pyloto6.py
import math
import random
import numpy as np
appearance_count = {} #Number of occurrences of each number
for i in range(44):
appearance_count[i] = 0
sums = [] #Array of total values each time
with open('C:\\Python\\scloto6\\index.txt', 'r') as f:
lines = f.readlines()
for line in lines:
array = line.split('\t')
array_n = list(map(int, array)) #Since array is a str array, convert it to an int array
sums.append(sum(array_n))
for i in array_n:
appearance_count[i] = appearance_count[i] + 1
#Sort in descending order by appearance count
sorted_appearance_count = sorted(appearance_count.items(), key=lambda x:x[1], reverse=True)
avg = sum(sums) / len(sums)
print(f"AVG:{avg}")
sigma = np.std(sums) #standard deviation of sums
sigmalower = avg - sigma
sigmaupper = avg + sigma
print(f"σ:{sigma}({sigmalower}～{sigmaupper})")
for index in [0, 2, 4, 6, 8]:
while True:
a = []
a.append(sorted_appearance_count[index][0])
a.append(sorted_appearance_count[index + 1][0])
#The remaining 4 are random(In other words, there is no basis for pickup)
while len(a) < 6:
no = random.randint(10, 32)
value = sorted_appearance_count[no][0]
if not value in a:
a.append(value)
asum = sum(a)
if sigmalower <= asum and asum <= sigmaupper:
#The sum of the 6 numbers picked up is-Adopted in the range of σ to σ
break
print(f"{asum}{a}")
The output looks like this.
Execution result
AVG:131.93635170603673
σ:28.37024181798044(103.56610988805629～160.30659352401716)
123[6, 38, 22, 26, 11, 20]
128[10, 27, 39, 21, 26, 5]
108[37, 12, 2, 16, 21, 20]
126[24, 15, 39, 35, 5, 8]
129[19, 43, 25, 3, 23, 16]
Since it is 200 yen per unit, it is 1000 yen for 5 units, so I have 5 units, but I think I read that 3 units are the best for LOTO 6 somewhere (I forgot the basis ...)
Even if you visualize the appearance numbers for 1524 times, you can not see any tendency, so the physical machine lottery method may not have a strategy. If so, it has been done for 20 years, so it seems that the ancestors have already discovered it. The lottery days are twice every month and Thursday!
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