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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.

Marconius
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Mery
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  • What are your thoughts on the problem? Typically, you will get better responses to questions by sharing what you have tried and where you are stuck. – Ben Sheller Oct 16 '15 at 19:40
  • Hi Mery! Welcome to MATH.SE! What have you tried? In which part of the solution have you had difficulties? Take a look at this page about how to use the platform. – Carlos H. Mendoza-Cardenas Oct 16 '15 at 19:41
  • I thought the following:
    • The probability That P1 wins on the first round is p1 = 1/6,
    • The probability That P2 wins on the second round is p2 = 0,
    • The probability That P3 wins on the third round is p3 = 5/6 * 1/6 * 1/6

    but I can't find a formula for n greater than 3 to add it for all the values of n and find the probability that I want. Thanks

     – Mery Oct 17 '15 at 10:01
  • I think p4 = 5/6 * 4/6 * 1/6 * 1/6 .. but i don't know p5, p6, ..., pn – Mery Oct 17 '15 at 10:16

3 Answers3

2

Let $p$ be the probability that the player having the roll wins the game.

The sequence of events is:

  • $A$ wins outright with probability $1/6$; or
  • control passes to one of the other $5$ players, $X$ say; then
  • $X$ wins eventually; or
  • $X$ does not win (prob. $1-p$), so one of the other five players (one of which is $A$) does

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:

  • the die is fair and hence all numbers $1$ to $6$ are equally likely
  • all players not having the roll have the same chances of winning at that point in the game -- at that point either the player having the roll wins or control passes back (with equal probabilities) to one of the other five players
Marconius
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  • I think the recurrence relationship is not quite right. When you write $(1-p)$, it is the probability of $A$ losing, whereas it should be that of $B$ losing. – true blue anil Oct 17 '15 at 17:45
  • @trueblueanil - this is the probability that player X loses, so as to give A a chance of winning after the first turn. Probability that control passes to X is $\frac{5}{6}$, and probability that A is the player to win if X does not is $\frac{1}{5}$. – Marconius Oct 17 '15 at 18:47
  • The point is you are using (1-p), which is the probability of A losing. I don't see how you can use it for the probability of X losing. – true blue anil Oct 17 '15 at 18:55
  • @trueblueanil - The probability that any player loses, given that the game continues to the point where they obtain the roll, is $1-p$. This is a conditional probability and is the same for all players by symmetry. What differs is the chance that each player has of ever getting the roll in the first place. Obviously, A is ahead of all other players in this regard. – Marconius Oct 17 '15 at 21:48
  • ok, +1 for explanation (and solution) – true blue anil Oct 18 '15 at 08:15
1

Here are some hints.

  • What is the probability that P1 wins on the first round?
  • What is the probability that P1 doesn't win on the first round?
  • What about winning on the second round? (Hint: Can he?)
  • What about winning on the third round? (Hint: This would mean that the first two players, whoever they are, didn't win.)
  • The fourth round?
  • The $n$th round?
  • On any round? (This is your answer.)
John
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  • Sorry but I understand that the probability that player 1 wins is 2/7. But I can't find any general formula for the nth round. Thanks – Mery Oct 17 '15 at 22:09
1

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 $\lim\limits_{n\to\infty} A^n x$)

JMoravitz
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