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third update added
Richard Stanley
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Here is a somewhat different way from Johan's of looking at this problem. At each stage of the walk, choose a number $x$ uniformly from $[0,1]$ and then walk either a distance $x$ to the right or $1-x$ to the left. This does not affect the probability of becoming negative since there is still a uniform probability of taking a step whose length belongs to the interval $[-1,1]$. However, it does have the property that after taking $n$ steps and choosing $0\leq x\leq 1$, the two possible locations following the next step are the same modulo 1. Hence the walk can be described as follows. Choose $n$ numbers $0\lt x_1\lt \cdots\lt x_n\lt 1$, a sequence $\epsilon=(\epsilon_1,\dots,\epsilon_n)$ of signs $\pm 1$, and a permutation $w$ of $1,2,\dots,n$. Let the location be $y_k$ after the $k$th step. If $\epsilon_k=1$ then step to the least real number $y_{k+1}\equiv x_{w(k+1)}$ (mod 1), $y_{k+1}>y_k$. If $\epsilon_k=-1$ then step to the greatest real number $y_{k+1}\equiv x_{w(k+1)}$ (mod 1), $y_{k+1}\lt y_k$. But the question of whether any $y_k$ is negative depends only on $\epsilon$ and $w$, not the choice of $x_1,\dots,x_n$. There are $2^n n!$ ways to choose $\epsilon$ and $w$. Is there a simple combinatorial argument that the number of choices such that each $y_k>0$ is $(2n-1)!!=1\cdot 3\cdot 5\cdots (2n-1)$? Then the probability of success is $(2n-1)!!/2^nn! = (2n)!/4^nn!^2$.

Here is a reformulation of the combinatorial result that needs a simple direct proof.

Let $f(n)$ be the number of pairs $(a_1a_2\cdots a_n, b_1b_2\cdots b_{n-1})$ such that (a) $a_1 a_2\cdots a_n$ is a permutation of $1,2,\dots, n$, (b) $b_i=0$ or $1$ if $a_i\lt a_{i+1}$, (c) $b_i=0$ or $-1$ if $a_i>a_{i+1}$, and (d) $b_1 +b_2+\cdots+b_j\geq 0$ for all $1\leq j\leq n-1$. Then $f(n)=(2n-1)!!$.

Update. The combinatorial result is proved bijectively by O. Bernardi, B. Duplantier, and P. Nadeau in Séminaire Lotharingien de Combinatoire, B63e (2010). In their citation [1] they use this result for the same purpose as above, i.e., to compute the probability $P_n$ (though they state the result a little differently).

Second update. The method above can be applied to the $[l,r]$ generalization mentioned by Lwins in his comment. By rescaling we may assume $l=-1$. If we are at $y$ sometime during the walk, choose a number $x$ uniformly from $[0,1]$. With probability 1/2 step from $y$ to $y+\frac{r-1}{2}+\frac{r+1}{2}x$. With probability 1/2 step from $y$ to $y-1-\frac{r+1}{2}x$. This gives a uniform probability of stepping from $y$ to a point in the interval $[y-1,y+r]$. It has the property that once $x$ is chosen, the value of $y$ is determined modulo $\frac{r+1}{2}$. Thus the walk can be described as follows: pick uniformly and independently $0\lt x_1\lt \cdots\lt x_n \lt \frac{r+1}{2}$, pick a permutation $w$ uniformly from the symmetric group $S_n$, and a sequence $\epsilon=(\epsilon_1,\dots,\epsilon_n)$ of independently distributed signs, with a probability of $\frac{r}{r+1}$ for a plus sign and $\frac{1}{r+1}$ for a minus sign. Go through the same procedure as above, working mod $\frac{r+1}{2}$ instead of mod 1. Again a proper walk (i.e., one which never becomes negative) depends only on $w$ and $\epsilon$, and we get the following result:

Theorem. The probability $P_n(r)$ that the walk is proper is given by $$ P_n(r) = \frac{1}{(1+r)^nn!}\sum r^{1+f(w,\beta)}, $$ summed over all pairs $w=a_1a_2\cdots a_n\in S_n$ and $\beta=(b_1,\dots, b_{n-1})\in \lbrace 0,\pm 1\rbrace^n$ satisfying the conditions (b) and (c) above, where $f(w,\beta)$ is the number of integers $1\leq i\leq n-1$ for which either $a_i\lt a_j$ and $b_i=0$, or $a_i\gt a_j$ and $b_i=1$.

For instance, $P_2(r)= (r+2r^2)/2(r+1)^2$ and $P_3(r) =(r+8r^2+6r^3)/6(r+1)^3$. I conjecture that the numerator $N_n(r)$ of $P_n(r)$ is just the polynomial $\sum B_{n,i}r^i$ defined by equation (4) of http://math.mit.edu/~rstan/pubs/pubfiles/29.pdf. This paper gives some additional information about the polynomials $\sum B_{n,i}r^i$. Much additional information can be found in the literature on Stirling permutations, e.g., Bona proves in http://wenku.baidu.com/view/dfa70012cc7931b765ce15e4.html that all zeros of this polynomial are real.

Third update. Alas, the conjecture in my second update is false. Unless there is an error in my code, the sequence of coefficients of $N_n(r)$ for $2\leq n\leq 7$ are $(1,2)$, $(1,8,6)$, $(1,25,55,24)$, $(1,69,361,394,120)$, $(1,176,1999,4416,3083,720)$, $(1,426,9836,41019,52193,26620,5040)$. It is easy to see why the leading coefficient of $N_n(r)$ is $n!$.

Richard Stanley
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