Possibly wrong answer because I don't get a clear-cut answer to the final question (but I've checked my working and haven't found any errors):
[I have used a computer, but only as a slightly more convenient pocket calculator.]
John: “I have no face cards. All my five cards are factorials. Their (five cards) multiplication is also a factorial number.”
Factorials are 1, 2, 6. Product is factorial of <= 4 because no factor of 5.
Product could be 1 if all are 1, but there are only four aces.
Product could be 2 if John has AAAA2, or 6 if he has AAAA3,
or 24 if he has AA223. I think that's all the options.
Jose: “No face cards. Factorial of the first card a perfect square.
Only perfect-square factorial is 1, so first card is an ace.
Note that this means John doesn't have AAAA2 or AAAA3. So John
has AA223.
Sum of the factorial of the first card and the factorial of the second card is also a perfect square.
Perfect-square-minus-1 factorials (up to 10!) are 4!=24, 5!=120, 7!=5040. So Jose has A4..., A5..., A7...
Sum of the factorial of the first card and the factorial of the second card and the factorial of the third card is also a perfect square.
Sum of first two cards' factorials is 25, 121, or 5041. Factorials up to 10! that make a square on adding one of these: 25+4!=49, 121+6!=841. So Jose has A44.. or A56..
Sum of the factorial of the first card and the factorial of the second card and the factorial of the third card and the factorial of the fourth card is also a perfect square.
Sum of first three cards' factorials is 49 or 841. Factorials up to 10! that make a square on adding one of these: amusingly, 5! works for both. 49+5!=13^2 and 841+5!=31^2. So Jose has A445 or A565.
Sum of the factorial of the first card and the factorial of the second card and the factorial of the third card and the factorial of the fourth card and the factorial of the fifth card is also a perfect square.”
Sum of first four cards' factorials is 169 or 961. For each of these there is again a single factorial adding to them to make a square: 169+5!=289 and 961+6!=1681. So Jose has A4455 or A5656.
Ari: “No face cards. Sum of the factorials of each of my five cards is a 3 digit perfect cube.”
3-digit perfect cubes are 125, 216, 343, 512, 729. Of these, 125 is obviously only 1+1+1+2+120 (so AAA25) which is impossible because John and Jose have three aces between them. 729 is obviously only 1+1+1+6+720 (so AAA36) which is impossible for the same reason. We can get 216 as 44445 (and obviously not in any other way). 343 is impossible (taking two 120s leaves 103 which we clearly can't make with three things <= 24). 512 is impossible (taking four 120s leaves 32 which isn't a factorial). So Ari has 44445. This means that actually Jose has A5566, not A4455, because Ari has all the 4s.
Neal: “No face cards. None of my cards are prime numbers. My five cards add up to a number whose reverse is a factorial.”
Sum of these five cards is at most 50, so it must be 1, 2, 6, or 42. Obviously it can't be 1 or 2, but it also can't be 6 because the only way to make it is 1+1+1+1+2 and 2 is prime. So the sum is 42. The summands are chosen from 4,6,8,9. There actually aren't many ways to do this. We can't use 4 because 49999 only gives 40. We can have 69999 and clearly that's the only way to use 6. Otherwise we have only 8s and 9s and it must be 88899. So Neal has 69999 or 88899. [EDITED to add:] Oops, as pointed out in comments some of Neal's cards could also be 10s, which means there are more possibilities than I listed. Let's see. We still can't use 4 because Ari has all the 4s. We can't use two 6s because 66TTT is only 32; with one 6 we need 36 from the other four which means 9999 (which we already listed) or 899T or 88TT. Otherwise, all 8s, 9s and 10s; five 9s would be 45 so we need five {-1,0,+1} adding to -3 which means either ---00 (meaning 88899, already listed) or ----+ (meaning 8888T). So, finally, Neal has one of these hands: 69999 88899 6899T 688TT 8888T. Of these, the first and last beat Ari's four fours, and the others lose.
What should Ari do?
John has AA223: two pair. Jose has A5656, two pair. If aces aren't considered face cards, presumably Jose's hand beats John's, but in any case neither of them is going to be winning this hand. Ari has 44445, four of a kind. Neal's hand may be better than Ari's or worse; the better possibilities are considerably less likely than the worse ones. John and Jose both have losing hands, and know it, and know that everyone else knows it. They will fold. Ari doesn't know whether he or Neal has the better hand. Neal does know, but knows that Ari doesn't know.
Unless I am missing something, at this point we are out of the realm of pure logic and into that of probability and bidding strategy.
As already mentioned, there are way more possible deals where Neal loses than where Neal wins, so it seems like Ari should place a sizeable bet. As to what he should do after that if Neal doesn't fold, I don't know; it seems like it might depend on things like how much money each has available to bet, how much of a risk-taker Neal is, etc.
