Probability problem involoving an N-sided die If two people roll an N-sided die turn by turn and the game is lost by the one who rolls a lower number than the previous roll, then what is the probability that the first player wins the game?
I think the answer is somehow related to the binomial process but, am not able to properly formulate the distribution. Any help. Thanks beforehand.
 A: "Binomial process"? Not really.
A solution to a closely related problem is available at this link:
http://cr4.globalspec.com/blogentry/21768/Roll-of-the-Dice-Newsletter-Challenge-January-2013
A: This can be done through a recursion.  I'll work it for $N=6$ to illustrate.
Note:  a prior version of this post switched winning/losing.  I've corrected accordingly.
Let $p_n$ denote the probability that a player will eventually win, assuming they just tossed an $n$ (without losing).    We note that $$p_n=\left( \frac {n-1}6\right)\times 1 +\frac 16\times \sum_{i=n}^6 (1-p_i)\implies $$
$$6p_n=n-1+\sum_{i=n}^6 (1-p_i)\implies 7p_n=n+\sum_{i=n+1}^{6} (1-p_i)$$
We can use this to determine all the $p_n$:
$p_6$:  We have $$7p_6=6\implies p_6=\frac 67$$
$p_5$:  We have $$7p_5=5+(1-p_6)\implies p_5=\frac {34}{49}$$
To approximate the rest, we get $$p_4=0.629737609\quad p_3=0.539775094$$  $$p_2=0.462664366\quad p_1=0.396569457$$
Barring arithmetic error (always possible), the answer then is $$P=\frac 16 \times \sum_{i=1}^6 p_i=\boxed {0.603430543}$$
Sanity check:  If the starting person tosses a $1$ then they can not win on the next toss.  Hence, the second person becomes the starting person...so we should have $P=1-p_1$, which is indeed the case here.
