Hello there !
I think I found a way of mixing two Markov chains in order to predict the next combination.
First consider these probabilities :
P1(i,j) = P( ball j is extracted at combination n+1, given that the ball i was extracted at combination n)
and
P2(i,j) = P(ball j is extracted after ball i in the same combination)
Now, consider the last combination :
C : a1, a2, a3, a4, a5, a6
and you want to predict the next one.
For this, consider X the probabilistic distribution of numbers at combination n :
| 1 2 3 ... a1... a2 ... a3 ... a4 ... a5 ... a6 ... 49 |
X = | |
| 0 0 0 ... 1/6.. 1/6...1/6....1/6....1/6....1/6.... 0 |
I consider p(a1)=p(a2)=...=p(a6) because they all contribute equally to the next outcome.
Now, using the first markov chain, you get the following distribution :
X(n+1) = X * P1
Now let's say b1, b2, b3, b4, b5, b6 are the numbers to predict.
And now take
b1 = the number with the maximum probability in the distribution of X(n+1).
This is the first number in the combination we try to find out.
Now, using the second Markov chain, we get the other 5 numbers : b2, b3, b4, b5, b6.
First, let's say Y = the distribution of the combination at moment n. Then, we have Y(1) : b1=100% the initial distribution.
Then you calculate Y(2) = Y(1) * P2 and choose b2 as the maximum from the resulting distribution.
To find b3, you follow the same algorithm : Y(3) = Y(2) * P2, and so on, until you come up with all the six numbers.
And this is all. I know it's an heuristic algorithm, where you only consider the local optimum, but it may work.
I would like to know your suggestions
Lucas
I think I found a way of mixing two Markov chains in order to predict the next combination.
First consider these probabilities :
P1(i,j) = P( ball j is extracted at combination n+1, given that the ball i was extracted at combination n)
and
P2(i,j) = P(ball j is extracted after ball i in the same combination)
Now, consider the last combination :
C : a1, a2, a3, a4, a5, a6
and you want to predict the next one.
For this, consider X the probabilistic distribution of numbers at combination n :
| 1 2 3 ... a1... a2 ... a3 ... a4 ... a5 ... a6 ... 49 |
X = | |
| 0 0 0 ... 1/6.. 1/6...1/6....1/6....1/6....1/6.... 0 |
I consider p(a1)=p(a2)=...=p(a6) because they all contribute equally to the next outcome.
Now, using the first markov chain, you get the following distribution :
X(n+1) = X * P1
Now let's say b1, b2, b3, b4, b5, b6 are the numbers to predict.
And now take
b1 = the number with the maximum probability in the distribution of X(n+1).
This is the first number in the combination we try to find out.
Now, using the second Markov chain, we get the other 5 numbers : b2, b3, b4, b5, b6.
First, let's say Y = the distribution of the combination at moment n. Then, we have Y(1) : b1=100% the initial distribution.
Then you calculate Y(2) = Y(1) * P2 and choose b2 as the maximum from the resulting distribution.
To find b3, you follow the same algorithm : Y(3) = Y(2) * P2, and so on, until you come up with all the six numbers.
And this is all. I know it's an heuristic algorithm, where you only consider the local optimum, but it may work.
I would like to know your suggestions
Lucas