How to calculate winning odds with opposed 2d6 dice rolls

Hello. As a part of a game i play, i am interested in what the odds are of opposed dice rolls.

In the game, both players can control wizards of certain levels, from level 1 to 4. Each time they cast a spell, they roll 2 6 sided dice (2d6 for short), add those together and add their wizard level. They need to get a certain threshold to cast their spell. A double 1 always fails, a double 6 is always successful. An opposing player can try to dispel a spell that was just cast. To do that, they also roll 2d6, add those together and add their wizard level to it, if they beat their opponent the spell is dispelled. A double 1 always fails, a double 6 is always successful when casting or dispelling, no matter what the other player rolls. They need to beat their opponent to win, so lets say if a lvl 4 rolls 2 2s, for a total of 8, the dispelling wizard needs at least a 5 on his 2d6 for a 9 in total.

I would like to know what the chances are of a lvl 1 wizard dispelling a lvl 4 wizard, when the lvl 4 wizard is casting a spell that needs at least a 7+ to cast. I would like to know the same odds for a lvl 2 wizard, for a lvl 3 wizard and a lvl 4 wizard trying to dispel. I want to show that because of that 2d6 rolls tend to aggregate to the middle, that lvl 1 wizards are very poor at dispelling higher level wizards. I can't figure out the exact maths myself, if someone could help that would be much appreciated.