The game goes this way:
There are 6 players, numbered 1 to 6.
Player 1 starts the game, he rolls a die with six faces. If the result ($x$) of rolling the die is 1 then Player 1 wins. Else the player number $x$ starts his turn. The game goes on and the Player $x$ rolls the die, if the result ($y$) is equal to $x$ then Player $x$ win, else it's the turn of Player $y$. And so on.
What is the probability of the Player 1 to win?
Thanks.
On
If you prefer using matrices, this could be described using a twelve-state markov chain with states ($P1_{\text{win}}, P2_{\text{win}},\dots, P1_{\text{turn}}, P2_{\text{turn}},\dots$) and the transition matrix:
$$\begin{bmatrix} 1&0&0&0&0&0&\frac{1}{6}&0&0&0&0&0\\ 0&1&0&0&0&0&0&\frac{1}{6}&0&0&0&0\\ 0&0&1&0&0&0&0&0&\frac{1}{6}&0&0&0\\ 0&0&0&1&0&0&0&0&0&\frac{1}{6}&0&0\\ 0&0&0&0&1&0&0&0&0&0&\frac{1}{6}&0\\ 0&0&0&0&0&1&0&0&0&0&0&\frac{1}{6}\\ 0&0&0&0&0&0&0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\ 0&0&0&0&0&0&\frac{1}{6}&0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\ 0&0&0&0&0&0&\frac{1}{6}&\frac{1}{6}&0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\ 0&0&0&0&0&0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&0&\frac{1}{6}&\frac{1}{6}\\ 0&0&0&0&0&0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&0&\frac{1}{6}\\ 0&0&0&0&0&0&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&0\end{bmatrix}$$
This is rather cumbersome however, and would be nice if we cared about additional statistics such as the probability of player two or player three winning.
Since we only are interested in whether or not player 1 wins, instead, let us simplify this to a four-state markov chain with states $P1_{\text{win}}, Other_\text{win}, P1_{\text{turn}}, Other_{\text{turn}}$ with the transition matrix:
$$\begin{bmatrix}1&0&\frac{1}{6}&0\\0&1&0&\frac{1}{6}\\0&0&0&\frac{1}{6}\\0&0&\frac{5}{6}&\frac{4}{6}\end{bmatrix}$$
(note that if it is a player other than player 1's turn, there is a $\frac{4}{6}$ chance it will be a player other than player 1's turn next turn as well, a $\frac{1}{6}$ chance that it will be player 1's turn, and a $\frac{1}{6}$ chance that player 1 loses (and one of the other players had won).
This is of the form $A=\left[\begin{array}{c|c}I&S\\\hline 0&R\end{array}\right]$
The limiting form becomes $\lim\limits_{n\to\infty} A^n = \left[\begin{array}{c|c}I&S(I-R)^{-1}\\\hline 0&0\end{array}\right]$
Since it begins with player 1's turn, our initial state vector $x$ will have a one in the third entry (corresponding to it being player 1's turn) and zeroes everywhere else.
Going through the necessary matrix operations will yield the answer. (namely, the first entry of the resulting vector of $\lim\limits_{n\to\infty} A^n x$)
On
Let $p$ be the probability that the player having the roll wins the game.
The sequence of events is:
From this the following equation is derived:
$$p=\tfrac{1}{6}+\tfrac{5}{6}(1-p)\tfrac{1}{5}=\tfrac{1}{3}-\tfrac{1}{6}p$$
which has the solution $p=\tfrac{2}{7}$.
So player 1 has a $\frac{2}{7}$ chance of winning.
Note that the solution relies on two forms of symmetry:
Here are some hints.