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The answer depends on the design of the spinner: how many sides and what numbers on them.
If the probability of an event will occur is p, then the probability that it will occur in n trials is pn.(That's p raised to the n power). So if you bet on 12 numbers, then (38-12=26) numbers are empty. The probability of the ball landing on one of these empty numbers is 26/38. So (26/38)^11 = 0.01538, which is about 1.538 % or a 1 in 65 chance.
If a five color spinner with equal sections of red blue green yellow and orange is spun six times, the probability of getting no reds in all six spins is 26.2%. The probability of no red on one spin is 4 out of 5, or 0.8 The probability of no red in six spins is 0.86.
The answer depends on the shape of the spinner and the numbers on it.
The chances of one of these numbers NOT appearing on one spin are 29 out of 37 ie 78.4%, so for 40 spins it is (29/37)^40 which is a remote 0.006%!