Has anyone tried a progression on a double repeat over 35 spins?
Bet on just the numbers that repeated once, in order to catch a double repeat.
Bet a progression, so you profit on a win. The reset rule is 35 spins, or win once.
Tested 1 million trials. A trial is a set of 35 spins. I wanted to know how many trials had triples and how many triples were there.
This is the result showing that 32% of the time, 2 triples happen in 35 spins. Also note, that less than 4% of the time no triple shows up. Most importantly, 96% of the time at least 1 triple happens. All we need is one.
0 triple happened 36126 times (3.61%).
1 triple happened 178258 times (17.83%).
2 triple happened 324891 times (32.49%).
3 triple happened 286649 times (28.66%).
4 triple happened 134345 times (13.43%).
5 triple happened 34608 times (3.46%).
6 triple happened 4765 times (0.48%).
7 triple happened 349 times (0.03%).
8 triple happened 9 times (0.00%).
Python code if anyone wants to play around
import pandas as pd
import numpy as np
number_of_spins = 35
trials = 1000000
total = np.zeros(number_of_spins)
for x in range(0, trials):
spins = np.random.randint(1, 39, number_of_spins)
# count doubles, triples etc
unique_elements, counts = np.unique(spins, return_counts=True)
# get elements that had 3 or more repeats
result = unique_elements[counts >= 3]
# Count the number of elements
total[len(result)] += 1
for index, value in enumerate(total):
if value > 0:
percentage = (value / trials) * 100
print(f"{index} triple happened {int(round(value))} times ({percentage:.2f}%).")
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The next post will be an analysis of the duplicates to see what kind of bankroll would be needed to bet the duplicates on a realistic table limit of $5 min inside bet.