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Consider $2\times n$-matrices with entries 0, 1 or $-1$, such that the number of zeroes in both rows is the same. Let $P_n$ be the probability that the first non negative element of both rows is a zero (remember, conditioned to the fact that the number of zeroes of both rows is the same). I believe that $\lim P_n = \frac{1}{4}$, but I have difficulties to actually prove it.

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If the number of zeroes in both rows is $n$, $P_n=1$. If the number of zeroes in both rows is $0$ and all the $\pm1$ are chosen randomly and i.i.d. with probability $p$ for $+1$, $P_n=p^2$. Please add the missing hypotheses. – Did Apr 13 '13 at 10:27
@Did I think the number of zeros are random too, in that case you cant say if we have n zeros or 0 zeros or something like that – Lrrr Apr 13 '13 at 10:37
Ali Amiri is right. The number $n$ is fixed but the number of zeroes of the rows is not (we only require that both numbers are equal). In other words, we consider that each matrix with same number of zeroes (whatever this number is) in both rows has the same probability.The number $0\leq P_n\leq 1$ refers to the probability that such a matrix has a 0 as first non-negative element in the first row, and also a 0 as first non-negative element of the second row. – Antonio Apr 13 '13 at 12:14

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