Solution #20

The Star Tours ride consists of one section of 2 options, and three sections of 3 options.

After n rides, the probability of getting a particular option on the 2-option section each time is 1/2ⁿ. There are two options, so the probability of getting the same option every time is 2(1/2ⁿ) = 1/2^n-1. So the probability of having seen both of the two options after n rides is 1-1/2^n-1.

Next consider the sections with 3 options. Picking two particular options, the probability of getting one of them on a particular ride is 2/3, so the probability of getting one of them each ride for n rides is (2/3)ⁿ. There are three selections of two options, so it looks like we can triple this number to get the probability of getting some set of two for all n rides, yielding 2ⁿ/3^n-1.

However, this is not quite right. Let the three options be A, B, and C. When we pick A and B, and consider the possibility of getting a sequence of n rides that draws only from A and B, we include the possibility of getting only A, and also the possibility of getting only B. When we pick B and C, we include the possibility of only B, and the possibility of only C. This creates a double-counting of the possibility of only B. Each single-option run is thus double-counted, and its probability needs to be subtracted off.

After n rides, the probability of getting a particular option on a three-ride section each time is 1/3ⁿ. There are three options to be subtracted off, so the probability to be removed is  is 3(1/3ⁿ)=1/3^n-1.

Thus the probability of getting at most two options of the three in n rides is  2ⁿ/3^n-1 - 1/3^n-1. The probability of getting all three is then 1 minus this.

The four ride sections are independent, so to find the probability of getting all the options in each section, we just multiply the probabilities. This yields:

(1-1/2^n-1)(1 - 2ⁿ/3^n-1 + 1/3^n-1)³

Running the numbers, this yields a probability of 89.86% after 11 rides, and goes above 90% to 93.17% after 12 rides.

(In fact, Sophia managed to see all eleven segments after a mere 7 rides.)