"VERY, VERY often a complete street hits in a short period of tme (like 12 spins)"
Ken
Such a statement ought to be a challenge for every reader with a basic knowledge of probabilities. I do not have a maths degree; but I am able to play with my cheap electronic calculator with 12 digits. No. 1 might be absent in the first 12 spins: (36/37)^12 = 0.71979582014. By subtracting this result from 1 I find the probability, that this number comes within this 12 spins: 0.28020417986.
The chance that all three numbers in street 1-3 come in that period: 0.28020417986^3 = 0.02200005813.
By subtracting this result from 1 I find the probability, that the street is not complete: 0.97799994187.
There are twelve streets. So the chance that no street comes with all three numbers within 12 spins must be 0.97799994187^12 = 0.76571287166.
A street might not complete itself in spins 2-13, in spins 3-13 .....One visit might last 105 spins. 105 = 93+12.
0.76571287166^93 = 0.00000000001. My cheap calculator has reached its limit. Are we losing every time within the 105 spin? The probability of 0.00000000001 suggests, that it does not happen "VERY, VERY often"! ACCORDING TO THIS KEN IS RIGHT!
Ken
Such a statement ought to be a challenge for every reader with a basic knowledge of probabilities. I do not have a maths degree; but I am able to play with my cheap electronic calculator with 12 digits. No. 1 might be absent in the first 12 spins: (36/37)^12 = 0.71979582014. By subtracting this result from 1 I find the probability, that this number comes within this 12 spins: 0.28020417986.
The chance that all three numbers in street 1-3 come in that period: 0.28020417986^3 = 0.02200005813.
By subtracting this result from 1 I find the probability, that the street is not complete: 0.97799994187.
There are twelve streets. So the chance that no street comes with all three numbers within 12 spins must be 0.97799994187^12 = 0.76571287166.
A street might not complete itself in spins 2-13, in spins 3-13 .....One visit might last 105 spins. 105 = 93+12.
0.76571287166^93 = 0.00000000001. My cheap calculator has reached its limit. Are we losing every time within the 105 spin? The probability of 0.00000000001 suggests, that it does not happen "VERY, VERY often"! ACCORDING TO THIS KEN IS RIGHT!