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We select 4 numbers out of 37 according to the rule of multiplication of successive actions:
1 number - 37 options
2 number - 36 options
3 number - 35 options
4 number - 34 options to choose from
N = 37*36*35*34 = 1'585'080 combinations of 4 numbers, excluding permutations.
Since the order of numbers in the group is not important for the bet (2,3,30,17 = 17,2,30,3), then the combinatorics combination formula:
We can choose 4 numbers out of 37 without repetitions With 4 37 = 37! / (4! * (37-4)!) = 66'045 ways.
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Alatissa wrote: I showed this formula of the Swiss Bernoulli four months ago, then a week ago.
But you ignore it.
And the fact that the binomial distribution, which you got into, was also completely useless to explain
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No school combinatorics will show the markers of entry. You can't defeat mathematics with the help of the same mathematics. It's like trying to pull yourself out of the swamp by your hair. Without a fulcrum it won't work anyway. And combinatorics will not and cannot give such a fulcrum. On the contrary, it is like a primary source, and everything else follows from it.DLK wrote: Let's simplify the task and consider only the cones on the field: 1/5, 2/6, 4/8, 5/9, etc.
It is necessary to establish monitoring and divide them into three groups according to loss: within the normal range, undershoot and overshoot.
Task: to determine the markers of entry into the game and the forecast mechanism so that in 500 spins the bets bring the balance into the plus.
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