js520 is extremely close, tough he made a very slight mistake (when introducing the factorials) that alters the result in ~2%. He also considered only 8 villains when the OP asks for 9 villains. His calculations do handle the case where the two fives are out there in the villain's hands. I'll reexplain, with the calculations fixed:
First, out of the 52 cards of the deck, we know 5 of them, and 2 of them are fives. This means that there's 47 unseen cards, and 2 of them are fives. We're going to deal a total of 18 cards (9 villains, 2 cards each), and we want to know the chance that one or more of them are fives.
Logically, what we want to calculate is: (Note how if two of the cards are fives, the logical condition still holds true).
P(Card1 is a five OR Card2 is a five OR ... OR Card18 is a five)
To simplify the calculation, we apply a double negation and distribute one of the negations according to De Morgan's Laws, the other just as a normal probability negation. After doing this, we obtain the equivalent expression:
1-P(Card1 is not a five AND Card2 is not a five AND ... AND Card18 is not a five).
Now this is easier to calculate. The probability that Card1 is not a five is 45/47. The probability that Card2 is not a five is 44/46. Etc Etc. The probability that Card18 is not a five is 28/30. The probability that a series of mutually exclusive events all happen is done by multiplying the chance of all of them. So the final expression to calculate is:
Here's the calculation on WolframAlpha.
. The chance that one or more cards are fives is 62,44%. With 8 villains the chance is 56,98% (http://www.wolframalpha.com/input/?i=1-%2845%2F47%29*%2844%2F46%29*%2843%2F45%29*%2842%2F44%29*%2841%2F43%29*%2840%2F42%29*%2839%2F41%29*%2838%2F40%29*%2837%2F39%29*%2836%2F38%29*%2835%2F37%29*%2834%2F36%29*%2833%2F35%29*%2832%2F34%29*%2831%2F33%29*%2830%2F32%29) (js520 missed the last fraction of this expression).
BTW, this calculation is useless anyway because if you have a high pair, you should raise, and any non-donk opponent will almost always fold hands with a 5 except pocket pairs, connectors and gappers, and which ones of those he does fold depends on his style, skill and the size of your raise, so it is a completely different problem.
PS: You can't use tricks like the rule of 2 or approximating ORs by doing sums in this calculation, because the calculation is long and too many error accumulates.