Assuming Team A gets the ball first, and Team b second...
If the game goes to a 3rd possession, and assuming each team has a 40% chance of scoring and ending the game on a given drive (as a thought exercise... roll with it), then the chances of winning on a given drive are as follows:
3rd possession, Team A wins .4 (or game continues .6)
4th, B, .24 (which is .6*.4)
5th, A, .144
6th, B, .0864
7th, A, .05184
8th, B, .031104
Keep going far enough and sum each teams win probabilities on each of their possible drives (the sum of odd numbered drives for team A and the sum of even numbered drives for Team B), and you get 62.4% odds for Team A to win the game, or 37.6% for Team B to win the game.
Adjusting the assumed chance of scoring on a given drive, win probability changes as follows:
.25 chance of scoring on a given drive: A wins the game 55.5% of the time, B wins 44.5% of the time
.28, A wins 57%, B wins 43%
.3, A wins 58%, B wins 42%
.35, A wins 60.2%, B wins 39.8%
.4, A wins 62.4%, B wins 37.6%
.45, A wins 64.4%, B wins 35.6%
I think based on this and the assumption that the probability of scoring on a given drive is greater than .28, as the coach of Team B I would take my 43% chance of winning by going for 2 after scoring a TD on the 2nd possession rather than keep playing.