Bayes in roulette?


New Member
If 2 people go to the casino, 1 of them is blind, the other is not.
When arriving at the table, the sighted person sees that in pockets 0 32 15 19 4 21 2 25 17 34 6 27 there are particles that could obstruct the entry of the ball, the blind man does not see that.
The 2 players decide to play and write down what happened in the roulette. The blind man will play a dozen, the 1st dozen. The other player would also play 12 numbers, but making guesses he would play the next adjacent 6 numbers on the cylinder , the 6 numbers 26 3 35 12 28 7 and 13 36 11 30 8 23(12 numbers)
By scoring and playing 1000 balls results are: the blind have tied and the other player has won. Over 1000 balls, the dozen chosen by the blind man came out 334 times, which he did not lose. And the 12 numbers chosen by the sighted player came out 361 times so they came out 28 times more than the 12/36 payment rate and 36 times more than the 12/37 average. For the blind player, he simply analyzes 1000 balls where a sector of roulette has come out based on the +2.5 standard deviations. But, for the sighted player, the sector played and what happened and we could say that the probability of choosing 12 numbers in advance and achieving +2.5 standard deviations happens by chance in 1 test of 1000 each more than 30 tests of 1000. That is to say that it has had a very strong streak, or its prediction obeys some previous subjective knowledge. There is a saying that says: "He who does not know is like he who does not see" The blind will go the other day to play another dozen or the same, the other player will apply the same prediction, and its result will have a very high probability of repeating itself.
I think this kind of example has never been raised before. We can use subjectivity, or maybe it's something from Bayes giving A and B some of the assumptions and events in the example.
What do you think?