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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.

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"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

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    Good find. (+1) Note, however, that the problem there is slightly different...There "the player who does not improve from the previous roll loses." Here throwing the same number keeps the game going. Presumably the method can be adapted. – 2017-01-17
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    @lulu true. the method is slightly different here. Anyways, good find. By the way, I think quasi is trying to strictly monitor me for plagiarism, isnt it!? – 2017-01-17
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    @lulu -- I missed the slight variation. I edited my reply to reflect your observation. – 2017-01-17
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    @vidyarthi -- You did cross the ethical line quite a few times in the last week. Have you now reformed? – 2017-01-17
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    @quasi yes, slightly! – 2017-01-17
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    @vidyarthi -- CooI. I like many of the same problems that seem to interest you. As long as it's "fair play", it's all good. – 2017-01-17
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    @quasi by the way, are you a professor or something of the sort? – 2017-01-17
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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.

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    I think you are calculating for a six sided die, isnt it? – 2017-01-17
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    Yes, I missed that part. Of course the recursion works the same way for any $N$. Don't immediately see how to get a simple closed formula, but it might well be possible. – 2017-01-17
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    Might help to notice that the denominator of $p_i$ is $7^{7-i}$, presumably one replaces $7$ with $N+1$ in general. For the case $N=6$ the numerator is $46656$ which doesn't instantly suggest a formula to me, but which might to someone else. – 2017-01-17
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    Shouldn't it be $7p_n = 1 + \sum_{n+1 \leq i \leq 6} (1 - p_i)$? – 2017-01-17
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    @FabioSomenzi Oh, of course. Thanks. I implemented it correctly so it's just a typo in that line, which I will now correct. – 2017-01-17