Cards in a 5x5 grid- Probability of a diagonal of all hearts All the aces, 2’s, 3’s, 4’s, 5’s and 6’s, as well as the jack of diamonds are taken from a
regular deck of 52 playing cards, and then placed face up on a table in a 5 × 5 square
grid configuration randomly.

What is the probability that at least one of the diagonals in the array is all hearts?

I know that the probability of having all 5 hearts in a single row is 4!20!/24! but I am having trouble with this specific problem. Any help would be appreciated.
Since both diagonals can't be all hearts at the same time, the desired probability is simply twice the probability that one diagonal is all hearts, and thus

$$
2\cdot\frac{\binom65}{\binom{25}5}=2\cdot\frac{6!\cdot20!}{25!}=\frac2{8855}\;.
$$
There are two branches that must be considered. First, let's start with the center.  It has a 6/25 chance of being a heart.  Next, we'll traverse the diagonals.  The chance that the next spot we uncover is a heart is 5/24.  If it is a heart then we keep uncovering spots on that diagonal.  The chance that the next card is a heart is 4/23. If it is not a heart we still have a chance that the other diagonal is all hearts. So, there are two sub-branches here. If it is a heart then we have probabilities 3/22, and 2/21 until we uncover the entire diagonal.

There is a 19/23 chance that the third card uncovered is not a heart. If that occurs we switch to the other diagonal and require that all turns reveal a heart. We have probabilities 4/22, 3/21, 2/20, and 1/19 for this sub-branch. 

After the center, there is a 19/24 chance that card we turned over is not a heart.  If that is the case then we switch to the other diagonal. The probabilities there are 5/23, 4/22, 3/21, and 2/20 until the entire diagonal is exposed.

So we have

P(total) = 6/25*(5/24*(4/23*3/22*2/21 + 19/23*4/22*3/21*2/20*1/19) + 19/24*5/23*4/22*3/21*2/20)


Recombining, we have

P(total) = 6/25 * (5!*20!/24! + 5!*19!/24! + 19*5!*19!/24!)
         = 6!/25! * (20! + 19! + 19 * 19!)
         = 6!/25! * (20! + 20!/20 + (19 * 20!)/20)
         = 6!*20!/25! * (1 * 1/20 + 19/20)
         = (2 * 6! * 20!)/25!


This is the same value as answer generated by my original solution. The original solution made some assumptions not made here.