what are the odds that in 10 consecutive rolls with 2 dice a 4 or 10 is rolled before a 7?

You have to solve this in steps,

So first roll, it's either a 4/10 or 7 or none of them. There are 6 combinations for 4 and 10 (1 and 3, 2 and 2, 3 and 1, 4 and 6, 5 and 5 and 6 and 4) and there's 6 combinations to throw a 7 1+ 6 2+5 3+4 4+3 5+2 and 1+6 .

So the chance is 6/36 for the first roll, and there's a 24/36 change you get to throw a second time, in which case yo have another 6/36 chance of rolling 4 or 10 (or 7) for a total of 6/36+6/36*(24/36) repeat.

You could probably make a nice formula out of it too, but in too lazy to think of one right now. Probably something like 6/36*n+6/36*24/36*n-1 or something.

Where n is 10 in this case.