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enter image description here Players $A$ and $B$ are playing a game with a $6$ faced dice, each player gets to throw the dice once, alternately . The game starts with both the players placed on $b1$ and the final winner will be the player who reaches $b5$ first. And one rule that is to be followed is, if there are less blocks ahead of you than the number you got on the dice then the throw would result in $0$ score i.e. the player would not move . Now, player $A$ throws the dice first and gets $1$ ,so he moves to $b2$ i.e. $1$ block forward . Now player $B$ throws dice and gets $5$ ,but as per the rule as there are only $4$ blocks ahead of Player $B$ (as he is on $b1$) so $B$ would not move , Now player $A$ throws the dice and the game continues...
Question : In the scenario given in the image where $A$ is on $b2$ and $B$ is on $b1$, find the probability that $A$ wins when $A$ throws first and also find the probability that $B$ wins when $B$ throws first.

My attempt: I see that the possible combinations $A$ can throw without considering a $0$ score are $3$ and $2$,$1$ and $1$,$2$ and $1$,$1$,$1$ and possible combinations $B$ can throw without considering a $0$ score are $4$ and $3$,$1$ and $1$,$3$ and $2$,$2$ and $1$,$1$,$2$, and $1$,$2$,$1$ and $2$,$1$,$1$ and $1$,$1$,$1$,$1$.
And I am stuck here ,I am not able to think ahead of this.

To me this question seems to be a snake and ladder game without any snake or ladder as such scene usually occurs in ending part of snake and ladder.

Also I am new here and it's my first question so I had be grateful for any suggestions .

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    $\begingroup$ As you've recognised, there are many different ways in which the game could play out. It might help to know what you're studying in order to give you a more useful answer. One approach would be to use Markov chains (with states representing the positions of $A$ and $B$). Another would be to try to find a function $P(x,y)$ which returns the probability of a player in block $b_x$ (whose turn it is) beating a player in block $b_y$. I would also suggest coding a simulation so you can sense check your answers. $\endgroup$
    – Chris Lewis
    Commented Jul 15 at 8:41
  • $\begingroup$ I should have tried my own hint about simulation first!! Turns out the answer is very simple, but I haven't been able to find a simple explanation yet :-). Perhaps knowing that there's a quick solution will help someone else find it. $\endgroup$
    – Chris Lewis
    Commented Jul 15 at 11:59

2 Answers 2

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The position along the stack if the length of the stack is smaller than 6 is a red-herring, a distraction.

Regardless of the position of the players, you have a $1/6$ chance of winning, and a $5/6$ chance of passing the turn to the next player. The game is thus equivalent to the simpler game in which the players alternate rolling a dice and winning if they roll $6$.

I don't wish to write down a solution to this simpler exercise and leave it to the reader.

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  • $\begingroup$ Very nice!! That's a perfect solution. I'll leave mine up for now (perhaps the long route is useful for some other questions around this one) but I think your answer is what they're looking for. $\endgroup$
    – Chris Lewis
    Commented Jul 15 at 12:59
  • $\begingroup$ I would have never figured it out without your first solution! $\endgroup$
    – Guillaume Dehaene
    Commented Jul 15 at 14:55
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TL;DR: there's a trick in this question, but I haven't found it yet! The player who rolls first wins with probability $\frac{6}{11}$, regardless of the starting positions.

There must be an obvious reason for this (induction works, but isn't very elegant), but so far all I have is a long version.

Let $P(x,y)$ be the probability that a player (whose turn it is) in block $b_x$ beats a player in block $b_y$. Let's say it's player $1$'s turn and they're trying to beat player $2$.

The specific values we're after are $P(2,1)$ and $P(1,2)$.

The player whose turn it is can either win, bust or move to another block. They win if they roll exactly $5-x$ (with probability $\frac16$). They bust (and don't move) if they roll a value greater than this, with probability $\frac{x+1}6$. They move to block $x+k$ if they roll $k<5-x$ (each of which occurs with probability $\frac16$).

If player $1$ doesn't immediately win, say their new position is $x'$ (we could have $x'=x$). Then it's player $2$'s turn; their probability of winning is $P(y,x')$ - and their probability of losing is $1-P(y,x')$.

Putting all of this together, $$P(x,y)=\color{Red}{\frac16}+\color{Blue}{\frac{x+1}{6}(1-P(y,x))}+\color{Green}{\frac16 \sum_{k=1}^{4-x} (1-P(y,x+k))}$$

The term in red corresponds to player $1$ rolling their required number immediately. The blue term corresponds to player $1$ rolling more than this; the green less.

Tidying this up a bit, we have $$6P(x,y)+(x+1)P(y,x)+\sum_{k=1}^{4-x} P(y,x+k) = 6$$

This is unpleasant but not impossible to solve; we have a linear system in the $P(x,y)$.

From here, we can either write this all out, or note that as a linear system, it has a unique solution, and that $P(x,y)=\frac{6}{11}$ for all $(x,y)$ works: \begin{align}6P(x,y)+(x+1)P(y,x)+\sum_{k=1}^{4-x} P(y,x+k) &= 6\times \frac{6}{11}+(x+1)\times \frac{6}{11}+\sum_{k=1}^{4-x} \frac{6}{11} \\ &=\frac{36}{11}+\frac{6x+6}{11}+\frac{24-6x}{11} \\ &=6\end{align}

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  • $\begingroup$ Great job in figuring out that there is a trick. See my answer if you want to find out why it works out this way. $\endgroup$
    – Guillaume Dehaene
    Commented Jul 15 at 12:11

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