Hijack - need help with a different card game problem

[Fenris78]

Since I am taking a break from poker atm due to my last downswing I have turned my interest towards another card game that I have been playing quite a lot since my youth. I still play this game a lot with my friends, especially on the weekend while drinking, and it's similar to the game called "euchre" as far as I know, but is only played with 24 cards. Now I want to clear some misconceptions we have about "correct" play using the same mathematical methods I use for poker problems.

My problem now is the following:

I have a certain 6card hand, which I can only loose if one of my 3 opponents has 2 certain cards together in his hand, all the other cards do not matter. So I need to know, how to calculate the probability that none of my 3 opponents has excatly these 2 cards in his hand.

My thinking was the following:

After I know my 6 card hand there are 18 cards unseen distributed among my 3 opponents. There are (6 out of 18 ) = 18564 possibilties to make up a 6 card hand from the 18 remaining cards.

The probability the first of my opps has the two cards he needs to beat me in his hand is the number of combinations that contain exactly those two cards over the total number os possible combinations.

(4 out of 18 )/(6 out of 18 ) = 3060/18564 = 16.5%

So the probability that the first played does NOT have these two cards is 83.5%, which means the probability that no one has those two cards together in his hand should be 0.835^3

[1-(3060/18564]^3] = 58.3%

This means we would win the game more often than we lose it.

Could anyone please confirm this to me or do I have a serious error in my thinking somewhere?