<s>[INST] Probability Puzzle: Mutating Loaded Die Take an (initially) fair six-sided die (i.e. $P(x)=\frac{1}{6}$ for $x=1,…,6$) and roll it repeatedly.
After each roll, the die becomes loaded for the next roll depending on the number $y$ that was just rolled according to the following system:
$$P(y)=\frac{1}{y}$$
$$P(x)=\frac{1 - P(y)}{5} \text{, for } x \ne y$$
i.e. the probability that you roll that number again in the next roll is $\frac{1}{y}$ and the remaining numbers are of equal probability.
What is the probability that you roll a $6$ on your $n$th roll?

NB: This is not a homework or contest question, just an idea I had on a boring bus ride. Bonus points for calculating the probability of rolling the number $x$ on the $n$th roll.
 [/INST] The transition matrix is given by $$\mathcal P = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ \tfrac{1}{10} & \tfrac{1}{2} & \tfrac{1}{10} & \tfrac{1}{10} & \tfrac{1}{10} & \tfrac{1}{10} \\ \tfrac{2}{15} & \tfrac{2}{15} & \tfrac{1}{3} & \tfrac{2}{15} & \tfrac{2}{15} &  \tfrac{2}{15} \\ \tfrac{3}{20} &  \tfrac{3}{20} &  \tfrac{3}{20} & \tfrac{1}{4} &  \tfrac{3}{20} & \tfrac{3}{20} \\ \tfrac{4}{25} & \tfrac{4}{25} & \tfrac{4}{25} & \tfrac{4}{25} & \tfrac{1}{5} & \tfrac{4}{25} \\ \tfrac{1}{6} & \tfrac{1}{6} & \tfrac{1}{6} & \tfrac{1}{6} & \tfrac{1}{6} & \tfrac{1}{6} \end{bmatrix}.$$  It is fairly easy to get numerical values for the probability distribution of being in state $6$ after $n$ steps, but a closed form solution appears difficult.
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