Pi Digits High Low Game

Suppose you take a long number and for each digit if it is bigger then the previous one increase a counter by one. If it is less then the previous number reduce the counter by one. You keep a running total and graph that total. Noting when it passes 0. This total number will go up and down and can get to zero many times. If you play this game with random numbers in a million digits on average the number of times you will have crossed 0 is 1594.4 and the standard deviation of the number of times crossed 0 is 1207.3. Though as the number of times zero is crossed cannot be less then 0 this is a bit odd.

       
import random

numcrossed=[]
j=0
while j < 200:
	i = 0
	last=0
	total=0
	x=[]
	y=[]
	crossed=0
	while i < 1000000:
		ran= random.randint(0, 10)
		if ran==last:
			total=total
		elif ran>last:
			total=total+1
		else:
			total=total-1
		if total==0:
			x.append(i)
			y.append(total)
			crossed=crossed+1        
		i=i+1        
		last=ran
	numcrossed.append(crossed)
	j=j+1
       
 

If instead of random numbers the digits of pi are used. This is what the path of total counts looks like

       
file = open("pi1000000.txt", "r") 
#3.14159265358979323846264338327950 pi2.txt
x = []
y = []
text=file.read() 

pi = list(text)
total =0
i = 0
crossed=0
i=0

while i < len(pi):
	if pi[i]>pi[i-1]:
		#print(pi[i])
		total=total+1
	if pi[i]
 

Pi has a 0 total 657 times. Which is more than 51 out of 200 random million long sequences did in my tests. None of this means anything. Going up or down based on digits in a base ten number but i like these pattern sort of sequences.

E crosses 0 1725 times

sqrt2 crosses 0 1300 times

and with 2 million digits

The python code for visualisation is 
       
import numpy as np
import matplotlib.pyplot as plt

plt.scatter(x, y, alpha=0.5, color='green')
plt.title('Sqrt 2 High Low Game')
plt.show()