# Probability problem found in textbook

"Shown is the board for a simple dice game. You roll a die and move the same number of squares (for example if your first roll is a $3$, move to the $3$ square). If you land on an arrow's tail, you must move to the square where that arrow's head is. You win if you land on the "6" square or beyond (for example rolling a $5$ when you are on the $4$ square will win).

What is the probability of winning this game in EXACTLY $2$ moves?"

I'm having trouble understanding the wording and how to calculate the probability for this, could anyone help? Reading the sample solution didn't give me more hints on what exactly the rules for the game are.

-
Is the answer 5/36? –  Calvin Lin Jan 17 '13 at 22:24
No it supposed to be 5/9. –  user54609 Jan 17 '13 at 22:25
I didn't read the "win if land beyond 6 part". 5/9 makes sense. –  Calvin Lin Jan 17 '13 at 22:31
I don't understand "move the same number of squares". Do you mean "move to the square with the same number as your roll"? And what does "6 square or beyond" mean? –  David Mitra Jan 17 '13 at 22:32

If you win in exactly 2 moves, consider what the first move is.

If the first roll is 1, you need to roll a 5 or a 6.
If the first roll is 2, (you end up at 4) and need to next roll a 2, 3, 4, 5, or 6.
If the first roll is 3, you need to roll a 3, 4, 5 or 6.
If the first roll is 4, you need to roll a 2, 3, 4, 5 or 6.
If the first roll is 5, (you end up at 4) and need to next roll a 3, 4, 5 or 6.
If the first roll is 6, you already won, so can't win in exactly 2 moves.

Counting the cases, there are $2 + 5 + 4 + 5 + 4 = 20$ ways to win. Hence, the probability of winning in exactly 2 moves is $20 \times \frac {1}{6} \times \frac {1}{6} = \frac {5}{9}.$

-
Well this is confusing. Why do you need 5 or 6 to win after rolling 1? Of course this was the textbook answer, but the way I read "beyond" was "second roll larger than first". –  user54609 Jan 18 '13 at 12:26
My interpretation of beyond, is that the second roll gives you (more than) enough to get to square 6. (It has nothing to do with the compared values of the 2 rolls). After rolling a 1, you end up on the square 1. To get to 6 or beyond, you will need to roll 5 ($1+5 =6$) or 6 ($1+6 \geq6$). You can see that the rest of the values do not result in $1+x \geq 6$. –  Calvin Lin Jan 18 '13 at 14:36

Even though the initial wording is confusing, here is how I see the rules of the game as described.

\begin{array}{|c|c|c|} \hline \text{roll} & \text{start position}\rightarrow & 0 & 1 & 3 & 4 \\ \downarrow \\ \hline \\ 1 & &1 & 4 & 4 & 3 \\ 2 & &4 & 3 & 3 & \text{win} \\ 3 & &3 & 4 & \text{win} & \text{win} \\ 4 & &4 & 3 & \text{win} & \text{win} \\ 5 & &3 & \text{win} & \text{win} & \text{win}\\ 6 & &\text{win} &\text{win} &\text{win} &\text{win} \\ \hline \end{array}

Starting positions $2$, $5$ and $6$ are not possible due to the rules (slide to another position or have already won). And of course, Calvin's answer is the way to solve it, I just wanted to practice a table example here.

-