So how do you calculate the odds if you only have x spins? For example, how many times would it land on 1-12 6 times in a row out of 100 spins?

Question

You have a spinner with 38 numbers on it. The odds of it landing on 1-12 are .32. The odds of it landing on 1-12 multiple times in a row is .32^x where x equals the number of times. So how do you calculate the odds if you only have x spins? For example, how many times would it land on 1-12 6 times in a row out of 100 spins? Or out of 1000 spins? what is the way you set this problem up?

Answer 1

Yes, 12/38=0.32 approx. Let p=12/38=6/19.

Let q=1-p=13/19, so this is the probability of the spin landing on 13-38.

Let n be the number of spins and x be the number of consecutive spins in which the numbers 1-12 come up. The possible outcomes and probabilities can be listed:

The first x spins the numbers 1-12 come up, and the remaining n-x spins the numbers 13-38 come up.

The first spin, a number between 13-38 comes up, and the following x spins numbers between 1-12 come up, and the remaining n-x-1 spins the numbers 13-38 come up.

The first two spins both result in a number between 13-38, then x spins result in numbers between 1-12, and the remaining n-x-2 spins result in numbers 13-38.

And so on until the first n-x spins result in numbers 13-38 followed by x spins resulting in numbers 1-12.

In each case the combined probability is pˣqⁿ⁻ˣ.

If we sum these probabilities we get (n-x+1)pˣqⁿ⁻ˣ.

Example: p=⅓ (q=⅔), n=10, x=6: (10-6+1)(⅓⁶)(⅔⁴)=80/59049=0.001355 approx. This seems to be the probability of 6 consecutive successes in 10 trials, where the probability of success is ⅓. It happens that p=⅓ is the probability of throwing 1 or 2 on a die. So this example would be the probability of throwing 1 or 2 6 times in a row if the die was thrown 10 times.