If you are at a full 6-handed table, then you can approximate by multiplying by C(6,4)=15, so I'd guess the probability should be close to about 0.021 x 0.023 x 0.025 x 0.028 x 15 which gives very roughly about 200,000 to 1 against. I worked one thing out ages ago, but haven't double checked: if you have precisely trips and are playing HU, the chances the other player has precisely trips is 35486/C(47,5) ~ 0.023133906 which is better than 54912/C(52,5) ~ 0.021128451, so I have reason to believe that if two players are already dealt trips, the next player could have trips with a probability of about 0.025 or 0.026 and then the fourth player maybe between about 0.027 and 0.029 if I had to guess, but I'm a bit lazy to work out the calculation. The 5 card concept means swapping cards offers more outs to make a respectable hand, but it is down to the cards as in any game of poker. Last night I had a hand where it was trips over trips over trips over trips, if the odds of getting trips are 2.1% what would be the odds of 4 people getting trips at that table? It was 555/TTT/QQQ/KKK.

5 card draw hands