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Suppose we start at the number zero, and repeatedly roll a fair $6$-sided die, adding the resulting number to the total. We win if we reach the total $k$, and lose if we overshoot, for example by reaching a total of $k-1$ and then rolling a number greater than $1$. What is the probability to win?

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    Well, there's a simple recursion ($P_n=\frac 16 \times \left(P_{n-6}+\cdots P_{n-1}\right)$). Were you looking for the limit as $k\to \infty$ or something like that? – 2017-01-06
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    k is a fixed number. I thought a dynamic programming solution but I am not sure it's the fastest and if it's correct at all. – 2017-01-06
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    Ok, but then the recursion works. It's linear so you can solve it in closed form (though the characteristic polynomial has degree $6$ so the closed form isn't very nice). – 2017-01-06
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    [This question](http://math.stackexchange.com/questions/978036/what-are-the-odds-of-hitting-exactly-100-rolling-a-fair-die) was technically about the case $k=100$ but the answers discuss the general case. The limiting probability is $2/7.$ – 2017-01-06
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    How does one get such a recursion? – 2017-01-06
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    @Kaynex In order to reach $k$ exactly one has to do exactly one of the following: for some $i\in [1,\cdots,6]$ you must reach $k-i$ and then throw an $i$. The recursion just restates that. – 2017-01-06
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    @Kaynex $P_n$ is your probability of hitting $n$ if you start from $0.$ On your first roll you are equally likely to roll $1,2,3,4,5,6.$ If you start by rolling $k$ your conditional probability of hitting $n$ is $P_{n-k},$ the probability of hitting $n-k$ with your subsequent rolls. $P_n$ is the average of those six conditional probabilities. – 2017-01-06
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    @lulu Yes, that's another way to look at it. Reels the mind. – 2017-01-06

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