Well, it does lend creedence to the idea that the wheel is random.
It may be "more random" than before, but it's demonstrably not completely random.
Here's an illustrative little puzzle: assume there are forty years (1963 through 2002 inclusive)
on the wheel. If it was truly random, how often would there be a week with at least one
(or more) duplicated years?
No math nerds or puzzle enthusiasts?
Spoiler alert: The solution is presented below, after a couple of hints. Stop reading now if you
still want to try to solve it.
Since we are assuming a random distribution, with each possible arrangement of five 10@10
shows drawn from a pool of forty different years being equally likely, then the probability of
a week with one or more duplicated years is simply the number of arrangements with duplicates
divided by the total number of different arrangements. For example, if there were only ten
different arrangements, and three of them contained duplicates, then the probability would
be 3/10 = .30, or a 30% chance that a randomly chosen week would contain duplicate years.
The actual numbers are different of course. What eases their calculation is first knowing that
the probabilty of a week with duplicates is equal to one minus the probability of a week
without duplicates, i.e., if we know the probability of one we can compute the other, and
vice versa. The second piece is the realization that it is easier to determine the number of
arrangements
without duplicates than the number of arrangements
with duplicates.
The total number arrangements is simply 40 to the fifth power (40^5), since we have a choice of
forty different years for each of five days. The number of arrangements
without duplicates is
calculated as follows: on the first day, we a choice of any of 40 different years. On the second
day, there are only 39 different years to choose from, as one was used on Monday. Similarly,
on the third day we have only 38 years for our choice, as one was used on Monday and a second
on Tuesday. We have 37 choices on the fourth day and finally 36 choices on the fifth day. The
total number of arrangements
without duplicates is 40 x 39 x 38 x 37 x 36.
So the probability we will have week without duplicates is the above number divided by 40^5,
which is .7711... or just over 77%. Thus the probability that there would be a week
withduplicates is 1 - .7711 or .2289 -- almost 23%. If the wheel was truly random, (and ignoring theme
sets), we could expect a week with one or more duplicated years twice every nine weeks, or just
a little less than once a month.