This was exactly the type of problem that Pascal and Fermat discussed in length via a series of letters. Pascal thought it was easy. Fermat didn't get it.
Basically you count up the possible outcomes where one person wins and the number where the other person wins and then split the pot using that ratio.
They will keep rolling until they get a 2 or 6. Both 2 and 6 have the same probability of happening... so we can ignore how many times they have to roll before getting one of them.
If you imagine they kept rolling even once someone had one.. you get the following options
[2,2] [2,6] [6,2] [6,6]
In all cases except [6,6] player 1 wins. So player 1 has 3/4 chance.. so should get 3/4 of the pot... i.e. \$75 and player 2 gets \$25
see @drhab's answer for the clever way to work it out :)