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Lottery Code Optimization (Not enough RAM)

3 vues (au cours des 30 derniers jours)
john
john le 20 Nov 2015
Commenté : Image Analyst le 28 Nov 2015
So I want to create a matrix with all the lottery combinations. 6 numbers between 1 and 49. This gives 13,983,816 possible combinations. From this matrix I will start reducing some combinations such as 6 consecutive numbers, all odd numbers, etc. to theoretically increase my chances. Anyway, I'm stuck at the phase of creating the actual matrix, I only have 16GB of RAM which is not enough. This is my code:
close all; clear all; clc
opt_a=zeros(0,49^6);
opt_b=zeros(0,49^6);
opt_c=zeros(0,49^6);
opt_d=zeros(0,49^6);
opt_e=zeros(0,49^6);
opt_f=zeros(0,49^6);
index=0;
for a=1:49
for b=1:49
for c=1:49
for d=1:49
for e=1:49
for f=1:49
index=index+1;
opt_a(index)=a;
opt_b(index)=b;
opt_c(index)=c;
opt_d(index)=d;
opt_e(index)=e;
opt_f(index)=f;
end
end
end
end
end
end
options=[opt_a' opt_b' opt_c' opt_d' opt_e' opt_f']
Is there anyway to make this code finish running with 16GB of RAM? If not, can I split the code somehow into parts and assemble the parts to create the matrix. Or maybe there already exists such matrix in a txt.file which I can import into Matlab?
Thanks
  3 commentaires
John D'Errico
John D'Errico le 21 Nov 2015
Modifié(e) : John D'Errico le 21 Nov 2015
How would not having all the odd numbers theoretically increase your chances? (Hint: it does not.) Similarly for consecutive numbers.
Second hint:
help nchoosek
Third hint:
A lottery ticket is a poor way to invest your money in general.
Walter Roberson
Walter Roberson le 21 Nov 2015
Image Analyst: you certainly can use 0 as a argument to zeros.
>> z = zeros(0,49^6)
z =
Empty matrix: 0-by-13841287201

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Réponses (1)

Walter Roberson
Walter Roberson le 20 Nov 2015
"From this matrix I will start reducing some combinations such as 6 consecutive numbers, all odd numbers, etc. to theoretically increase my chances"
To theoretically increase your chances of what? Certainly not your chances of winning: the combination
1 2 3 4 5 6
is exactly the same probability as the combination
14 17 32 38 41 47
unless the random number generator is biased.
I have a suspicion that you also need to read about The Gambler's Fallacy
  3 commentaires
Walter Roberson
Walter Roberson le 28 Nov 2015
See though this article about winning streaks
Image Analyst
Image Analyst le 28 Nov 2015
Interesting. I wonder if their coin was not fair/evenly weighted. What gave rise to the bias?
Also I wonder why multiple researchers over decades have all disproved hot hand and yet these guys got a contrary result. Supposedly all the others before them didn't know what the heck they were talking about???
And if the streak continues, then the historical mean will start to change and by definition they will eventually start to regress towards the mean.

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