PLEASE ANSWER FROM WHERE IT SAYS "Continuing from Part 1"
I ran the length 5 and p = 0.5 eight times on my PC. The probability came around 0.53
I ran the length 7 and p = 0.7 five times on my PC. The probability came around 0.507
So scenario 1 definitely has greater probability.
Here's the code if you want to try it out. The code will take a few seconds to compute, especially if its an older PC.
import numpy as np
from collections import Counter
def count(ls,n): #n here defines the length of runs
ls = ''.join(ls)
#ms = ls.split('0')
#ms = [i for i in ms if i]
#cnt = Counter(ms)
#print(cnt)
srch = '0' + ('1'*n) + '0'
if srch in ls:
return 1
else : return 0
expts = 100000
n = 100
run_length = 5 #change length of successive 1's (runs)
pr = 0.5 #change probability
ls = np.random.choice([0,1],(expts,n),p =(1-pr,pr))
fav = 0
for i in ls:
i = list(map(str,i))
a = count(i,run_length)
fav = fav + a
#print(a)
print(float(fav/expts))
#ls = [0,1,0,1,1,0,0,0,0,0,0,0,0,0,0]
#count(ls1)
Let me know if you want help figuring out how to change the parameters, I've commented what you'll need though.
Oh, and the commented counter lines are if you want to see how it runs, just remember to set the expts to around 100-1000 if you want to try to run them.
PLEASE ANSWER FROM WHERE IT SAYS "Continuing from Part 1" Consider a sequence of n Bernoulli...
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