How do you calculate this against 2 opponents or more than 3 opponents, etc?
Let's first calculate the odds of 1 opponent having a four. I will then give the general formula for 'x' opponents.
There are 3 cards on the board and you have 4 cards. This means there are 45 cards left in the deck. There are therefore a total of C(45,4) different combinations of hands your opponent can hold. Chances of your opponent having a four can be calculated by substracting the chances of your opponent not having a four from 1. Chances of your opponent not having a four simply means there are now only 43 cards your opponent is allowed to draw (since there are 2 fours left). Total combinations of hands without a four is therefore C(43,4). So the odds of your opponent having a four is
1 - ( C(43,4) / C(45,4) ) = 0.171717...
As general formula we get:
1 - ( C(43,4x) / C(45,4x) ), where 'x' is the number of opponents.