Odds for Roullette Results

If you divide the wheel (American) into 4 9-number quadrants, along with the 0 and 00, what are the probabilities of a single quadrant not appearing for 16 or more spins?

Would it be correct to assume the probabilities are half that on a European wheel?

Thanks.

Quote: sharkbyte

If you divide the wheel (American) into 4 9-number quadrants, along with the 0 and 00, what are the probabilities of a single quadrant not appearing for 16 or more spins?

Would it be correct to assume the probabilities are half that on a European wheel?

Thanks.

There are 38 numbers, so if you divide them into 4 groups, there will be two extra left. The probability that none of given nine numbers hits in 16 spins is (29/38)^16 = 0.01324. On a single zero wheel, the probability of not hitting any of nine numbers in 16 spins is (28/37)^16 = 0.01157

Quote: weaselman

There are 38 numbers, so if you divide them into 4 groups, there will be two extra left. The probability that none of given nine numbers hits in 16 spins is (29/38)^16 = 0.01324. On a single zero wheel, the probability of not hitting any of nine numbers in 16 spins is (28/37)^16 = 0.01157

Forgive me if I ask this wrong, but what would be the expect high & low standard deviations for these?

Is that deviation affected by quantity? It would seem the fluctuation would be greater for 100 spins versus 1M spins.

Finally, as number of spins increase (100 up towards 1M), should the percentage seem to always be moving closer to the "ideal" (0.01324)?

Thanks.