Game probability

Question

I hope you could help me with this problem...

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.

I thin the probability of the Player 1 to win is $\frac{2}{7}$. But the question is: If player one has won, wich is the expected number of times he thrown the dice?

I try to do this: $p_n$ = probability that player 1 wins on the n round. $$p_1 = 1/6$$ $$p_2 = 0$$ $$p_3 = \frac{5}{6} \times \frac{1}{6} \times \frac{1}{6}$$

If I have a general formula for $p_n$ I would sum $$n \times p_n$$ for all value of $n$ and I have the expected number of rounds. But I can't find this $p_n$.

Thank you so much!!!

Answer 1

Note: I extended this answer and fixed a missing $z$ in equation $(\star)$ based on the comments.

For each player the probability of win changes depending on whether they are the one rolling. Pick any player. Let $P$ be the probability that the player wins given that they are rolling and $Q$ be the probability that the player wins if someone else is rolling.

If the player is rolling, then they win in that roll with probability $1/6$ and have to pass the dice with probability $5/6$ and become a non-rolling player. So $$P = 1/6 + 5/6 Q.$$ Similarly, $$ Q = 4/6 Q + 1/6 P.$$

Solving the equations gives $P = 2/7$ and $Q = 1/7$. So the probability that player 1 wins, as you said, is 2/7.

Now let $f(z)=\sum_{n=1}^\infty p_n z^n$, where $p_n$ is the probability that player 1 wins in $n$ rounds given that she is rolling; and $g(z)=\sum_{n=1}^\infty q_n z^n$, where $q_n$ is the probability that player 1 wins in $n$ rounds given that she is not rolling. Similar to the previous part:

\begin{align} f(z)&=1/6 z + 5/6 z g(z)\\ g(z)&=1/6 z f(z) + 4/6 z g(z),\qquad(\star) \end{align} which results in $$f(z)=\frac{2 z (3-2 z)}{(z+6) (6-5 z)}$$ (note that $f(1)=2/7$). The requested expected value is $f'(z)/f(z)$ at $z=1$ so it equals $$27/7.$$

If we are only interested in the number of throws of player 1, then the equations change as follows: \begin{align} f(z)=&1/6 z + 5/6 z g(z)\\ g(z)=&1/6 f(z) + 4/6 g(z), \end{align} Note that $z$ does not count the throws by the other players. So we get $$ f(z)=\frac{2z}{12-5z}\cdot $$ Now $f'(1)/f(1)=12/7$ is the expected number of throws of player 1, given that she wins.

Answer 2

Edit: As pointed out in a comment, only a part of this calculation is required to answer the question.

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Let's analyze the entire game under the condition that player $1$ wins.

If another player rolls, player $1$ would win with probability $\frac27$ if she gets the die and with probability $\frac17$ if another player gets the die, so, given that she wins, she gets the die with probability

$$ \frac{\frac15\cdot\frac27}{\frac15\cdot\frac27+\frac45\cdot\frac17}=\frac13\;. $$

If she rolls herself, she would win with probability $1$ if she rolls her own number and with probability $\frac17$ if she doesn't, so, given that she wins, she rolls her own number with probability

$$ \frac{\frac16\cdot1}{\frac16\cdot1+\frac56\cdot\frac17}=\frac7{12}\;. $$

Now we can set up equations for the expected number of rolls $x$ if she's rolling and $y$ if another player is rolling:

\begin{align} x&=1+\frac5{12}y\;,\\ y&=1+\frac13x+\frac23y\;. \end{align}

Substituting the first equation into the second yields $y=\frac{48}7$, and thus the desired expected length is $x=\frac{27}7$.

Here's Java code that confirms the result numerically.