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 .