The game wheel in the TV quiz "What? Where? When?" is divided into 13 equal sectors. The sectors are numbered from 1 to 13. At the beginning of the game, each sector contains an envelope with a question. Players choose a random sector using a spinner with an arrow. If this sector has already been chosen before, there is no envelope in it, and then the next sector clockwise is played. If it is also empty, the next one, and so on, until a non-empty sector is found. Before the break, the players played six sectors.

a) What is more likely: that among the played sectors there is sector No. 1 or that among the played sectors there is sector No. 8?

b) Find the probability that as a result, the six sectors with numbers from 1 to 6 were played in a row.
b) Consider the sequence of sectors $n_{1}, n_{2}, \ldots, n_{6}$, which have fallen when spinning the top. The event $A=$ "sectors 1-6 have been played" is favorable to all sequences where $n_{k} \leq k$, and all permutations of each such sequence. For example, the sequence 1, 1, 2, 3, 3, 6 and all its possible permutations are suitable, etc.

The number of sequences favorable to $A$ does not depend on the number of unselected sectors, and therefore is the same for any circle with more than six sectors. It is convenient to count these sequences on a circle with seven sectors. Then, whatever the sequence of fallen sectors, there will be only one unplayed sector left, and the played sectors will be arranged consecutively. There are a total of $7^{6}$ sequences. The number of sequences in which sector №7 is not played is exactly 7 times less: $7^{5}$.

On the circle, there are a total of $13^{6}$ equally possible sequences. Therefore, the probability of event $A$ is ${ }^{7^{5}} / 13^{6}$.

## Answer

a) The probabilities are equal; b) $7^{5} / 13^{6} \approx 0.00348$.