Probability of winning dice game Question: Two people roll a die with n faces in an alternative manner. The game is over when a face shows that is exactly one point lower than the previous roll, and that person loses the game. What is the probability of the first person losing the game?
My attempt: Let P be the probability. If the first roll is $1$ (with probability $\frac{1}{n}$), the conditional probability that the first person loses will be $1-P$ (the probability the second player loses). After that I have no idea how to proceed with the other possible first rolls.
 A: I will do the $n=3$ case first.  Let $P1, P2, P3$ be the probability that the next player wins on receiving the indicated roll.  The first player's winning probability is $P1$ as he is in the equivalent state.  We have $$P1=\frac 13\left((1-P1)+(1-P2)+(1-P3)\right)=1-\frac 13(P1+P2+P3)\\
P2=\frac 13((1-P2)+(1-P3))\\P3=\frac 13((1-P1)+(1-P3))$$
Alpha gives $$P1=\frac {34}{61},P2=\frac {25}{61},P3=\frac {22}{61}$$
We see $P2 \gt P3$, which is not surprising.  If I receive a $2$, I won't pass back a $1$ so my opponent will not be in the best place.  For larger $n$ you have to consider each face separately to be really right but I suspect you won't be far wrong to just consider $P,Q$, where $Q$ is the chance somebody who doesn't receive a $1$ will win and ignores the differences between the higher numbers. This gives
$$P=1-\frac 1n(P+(n-1)Q)\\Q=\frac 1n(1-P)+\frac {n-2}n(1-Q)$$
Which Alpha solves to give $$P=\frac {n+1}{2n+1},Q=\frac{n^2-n-1}{(n-1)(2n+1)}$$
A: Your general idea is right, but you need to consider all the possible rolls separately.  Let $p_n$ be the probability that the player who is about to roll loses, if his opponent has just rolled $n$, for $n=1,...6$.  Then, $$
p_1 = \sum_{k=1}^{6}{\frac{1-p_k}{6}}\\
p_n= \frac{1}{6}+ \sum_{k\ne n-1}{\frac{1-p_k}{6}},\space n=2,\dots,6
$$
That is, he loses if he rolls $n-1,$ and if he rolls $k\ne n-1,$ he loses if his opponent wins.  This gives you a system of linear equations to solve for $p_n.$  Once you have solved it, the probability that the first player loses is $p_1$ because that is the position he is in at the beginning of the game.  (I had this wrong before, but I have learned from Ross Millikan's answer.)   I also see that I made a mistake when I converted my equations to sympy code. 
EDIT I think I have my sympy code correct now, and here are the answers I get $$
p_1=46656/100777\\
p_2=54432/100777\\
p_3=55728/100777\\
p_4=55944/100777\\
p_5=55980/100777\\
p_6=55986/100777 
$$
A: This answer follows the same basic outline as the others given so far: separate the case that a player is passed a one from the other cases. Let $p=\Pr(\text{lose}\mid\text{previous roll was one})$ and $q=\Pr(\text{lose}\mid\text{previous roll wasn’t one})$. The probability that the first player loses is then $$\frac1n(1-p)+{n-1\over n}(1-q).$$ 
The players’ roles are reversed with each roll and there’s no way to lose on your roll if you were passed a one, so $$p = \frac1n(1-p)+\frac{n-1}n(1-q),$$ i.e., the same as that of the first player losing.
The recurrence for $q$ is a wee bit trickier. Regardless of which of the other $n-1$ possible other rolls you’ve been passed, there’s only one roll that will lose immediately. Otherwise, only $n-2$ of the possible numbers that you might have received allow you to pass a one to your opponent, therefore $$q = \frac1n + {n-2\over n(n-1)}(1-p) + \left(1-\frac1n-{n-2\over n(n-1)}\right)(1-q).$$ Solving this system for $p$ produces $$p = {n(n^2-2n+2)\over2n^3-3n^2+2n+1}.$$ For $n=6$, $p=156/337 \approx 0.4629$, which agrees pretty well with the other answers. As $n\to\infty$, $p\to\frac12$, which one might expect: the the contribution from being passed a one becomes minuscule compared to all of the other possibile rolls.
