Ask The Wizard #401

Let's let:

• a=expected more shots assuming initial state or last shot was a miss.
• b=expected more shots assuming last shot was made.
• c=expected more shots assuming last two shots were made.

We can set up the following equations on going from state to state:

a = 1 + 0.01b + 0.99a
b = 1 + 0.01c + 0.99a
c = 1 + (1-p)a

We now have three equations and three unknowns so we can solve it. I prefer matrix algebra.

Without teaching a whole less on that, the solution can be expressed as the determ(A)/determ(B). The terms in the matrices are taken from the three equations above.

The answer to this ratio of determinants is 101010.

To answer the second question, the answer for any probability p and number n of consecutive successes is:

(1/p)^n + (1/p)^(n-1) + (1/p)^(n-2) + ... + (1/p)^2 + (1/p)^1

In the case of this problem, the general formula shows the answer as 100^3 + 100^2 + 100^1 = 1000000 + 10000 + 100 = 1010100

This question is asked and discussed in my forum at Wizard of Vegas.

This question is asked and discussed in my forum at Wizard of Vegas.

You didn't say the type of wheel, but here is the answer all three ways:

• Single Zero = 8.306669466
• Double Zero = 8.408797212
• Triple Zero = 8.509594851

The following table shows the probability of a first repeat at each spin for all three wheels.