Probability of winning is about 7.09e-5 or about 1 in 14102.
Number of ways to pick 37 numbers from 75 is: 75!/(37!*38!) = X
To win you have to get all 12 numbers in the pattern selected. The 25 selected numbers that don’t fall in the pattern are from a set of 63 incorrect numbers. So the number of those sets of 37 which match the 12 numbers in the pattern is 63!/(25!*38!) = Y.
Probability of winning is Y/X= 63!*37!/(25!*75!).
That number makes a lot more sense. May I do a partial hijack of my own thread and ask why the pattern makes a difference? The numbers are randomly selected on the card, so the fact that one number is next to another on your bingo card seems irrevelant. Obviously it isn’t. Another way to look at it is the pattern could be randomly selected as well. With all three sets of data random, how does your calculation makes sense?
I’m sorry for being so thick-headed about this, but I really (still) don’t get it. 
Having to have the 12 numbers in a pattern will definitely lower the chances. Say you’ve got a 3x3 bingo card
1 2 3
4 5 6
7 8 9
Some numbers (out of some group) are going to be randomly drawn; you have to get five of them.
In one case, you’ve got all kinds of ways to do it: you could get {1,2,3,4,5}, or {6,4,2,7,8}, and so on (126 different possibilities).
On the other hand, say you have to get a specific pattern, “T” for example. Then you’ve lost a lot of winning possibilities, the only winning five number combination for you now is {1,2,3,5,8}.
Every winner in the second case would be a winner in the first case, but not conversely. Case one is much more likely to happen.
I think I agree with Manlob on this version of the problem. And, Sethlob, I don’t think the pattern really matters, except that it’s 12 numbers.
I thought of it this way: Pick a bingo card at random. Now say the pattern is a Z. Look at your card, and you can identify exactly 12 numbers that you need to match. Now suppose that someone goes into the BINGO drum and colors all your balls blue (nyuk, nyuk). Now, as the game proceeds to draw 37 balls at random, what’s the probability that all 12 of the blue ones will come up?
Earlier, we (I) was figuring you had 24 numbers from which to choose 12 (like we had 24 blue balls in the drum). But with the pattern restriction, you really only are looking for 12 out of 12, and there’s only one combination for that. So you have C(12,12)*C(63,25)/C(75,37) = .000071 like Manlob said.