Inequality involving sums of fractions of products of binomial coefficients

Let $n\in\mathbb{N}$. For $0\le l\le n$ consider $$b_l:=4^{-l} \sum_{j=0}^l \frac{\binom{2 l}{2 j} \binom{n}{j}^2}{\binom{2 n}{2 j}}\text{.}$$ Do you know a technique how to prove that $$b_l\ge b_n\text{,\quad 0\le l\le n-1?}$$ Going through a long list of binomial identities I did not find epiphany.

Addition: Plot of $b_l$ for $n=20$.

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May be $b_l$ can be interpreted as probability of some event. – Norbert Aug 26 '12 at 0:09
precarious: Why/How do you know the result holds? – Did Aug 26 '12 at 0:16
@did Numerical experiments gave sufficient evidence. – precarious Aug 26 '12 at 0:30
@precarious I have a solution, I hope it is correct. – Seyhmus Güngören Aug 26 '12 at 2:44
@precarious before giving any answer do you have any simulation result for $l=10^6,...,10^7$? – Seyhmus Güngören Aug 26 '12 at 3:28

The sum $b_\ell$ is clearly hypergeometric: $$b_\ell = 4^{-\ell} \sum_{j=0}^\ell \frac{\binom{2\ell}{2 j} \binom{n}{j}^2}{\binom{2n}{2j}} = 4^{-\ell} \sum_{j=0}^\ell \frac{(-n)_j}{\left(\frac{1}{2}-n\right)_j} \frac{(-\ell)_j \left(\frac{1}{2}-\ell\right)_j}{j! \cdot j!} = 4^{-\ell} {}_3F_2\left(\left. \begin{array}{ccc} -\ell & \frac{1}{2} -\ell & -n \\ & 1 & \frac{1}{2}-n \end{array} \right| 1\right)$$ This representation allows to find $$b_n = 4^{-n} {}_3F_2\left(\left. \begin{array}{ccc} -n & \frac{1}{2} -n & -n \\ & 1 & \frac{1}{2}-n \end{array} \right| 1\right) = 4^{-n} {}_2F_1\left(\left. \begin{array}{cc} -n & -n \\ & 1 \end{array} \right| 1\right) = \frac{1}{4^n} \binom{2n}{n}$$

The above function allows to extend the sequence to $\ell > n$. This sequence is not decreasing for all $\ell \geqslant 0$, but does appear to decrease on the interval $0 \leqslant \ell\leqslant n$. Here is an example for $n=20$:

Now, to the probabilistic interpretation of the $b_\ell$. Suppose an urn contains $2n$ balls, $n$ white and $n$ blue. We sample $m$ balls without the replacement. The probability that the sample contains equal number of balls of different colors is $$p_m = \cases{\frac{\binom{n}{j} \binom{n}{j}}{\binom{2n}{2j}} & m=2j \\ 0 & m = 2j+1}$$ If the size of the sample follows a symmetric binomial distribution, the probability of getting sample with equal number of colors is: $$b_\ell = \sum_{j=0}^\ell \frac{\binom{n}{j} \binom{n}{j}}{\binom{2n}{2j}} \binom{2\ell}{2j} 4^{\ell}$$

I am not seeing how to establish the inequality though, but hope this helps.

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do you have results for $l=10^6,...,10^7$? – Seyhmus Güngören Aug 26 '12 at 3:50
@SeyhmusGüngören What results are you interested in? – Sasha Aug 26 '12 at 3:54
I am interested in $b_l$ as a function of $l$ when $l$ is big enough such as in the range $10^6$ to $10^7$ – Seyhmus Güngören Aug 26 '12 at 3:59
But $b_\ell$ is also a function of $n$. If you have access to Mathematica, use the expression in terms of the hypergeometric function I provided. – Sasha Aug 26 '12 at 4:04
I dont have it right now. My Matlab is poor about such kinda things. let $n$ be $10^7$ and $l\in[10^6,10^7]$ could you please? if you nevermind? – Seyhmus Güngören Aug 26 '12 at 4:09