If you kick the PAT's, you have a 50% chance to win in ot. If you go for the first 2 pt conversion, you have a just under 50% chance to win in regulation. Its already close to the same odds. Then you add in the chance of making the second one and winning in ot and the math isn't even close. This one is even easier to follow than the one Kirk did.
You are certainly entitled to believe whatever you wish, but you are not allowed to conflate your belief with logic. Again, you're making several conditional assumptions to arrive at your probabilities. Logically, the goal is to score the most points, and this is based on probabilities (expectations), not belief systems.
So, if the expectation of successfully kicking the PAT is 99%, the expectation for points is .99 (99% of 1 point). Similarly, the expectation of converting the 2 PAT is 47% (based on global averages), so the expectation of points is ,94 (47% of 2 points), so logically, the 1 PAT is better.
Now, we're not just talking about 1 TD, we're needing to score 2 TDs. So if we ignore the probabilities of recovering an onside kick and the probability of scoring the second TD, let's calculate the expectation of total points. Also, we need to assume that the first TD and second TD are independent events. Both TDs have to be scored, but we need to see how what was done for the PAT after the first TD affects the end result.
NOTE: The following is based on being 14 points down, not the Iowa/Purdue scenario. Obviously, there is no ability to score fractional points, but that is what is used to determine the highest point expectation, which is what logic is about.
Scenario 1
First 1 PAT successful = total point expectation is 7 + 6.99 = 13.99 (probability 99%)
First 1 PAT fail = total point expectation is 6 + 6 + (.47 * 2) = 12.94 (probability 1%) (MUST go for 2 on second TD in order to tie the game)
So, overall point expectation is (13.99 * .99) + (12.94 * .01) = 13.9795
Scenario 2
First 2 PAT successful = 8 + 6 + (.99 * 1) = 14.99 (probability 47%) (Only need 1 PAT after second TD to win game)
First 2 PAT fail = 6 + 6 + (.47 * 2) = 12.94 (probability 53%)
So, overall point expectation is (14.99 * .47) + (12.94 * .53) = 13.9035
So from a logical perspective, Scenario 1 is preferable before you have made the decision to go for 1 or 2 after the first touchdown. Obviously, if you decide to go for two
and are successful, you have increased your point expectation, just as you commented, but before that decision is made, that is not the highest expectation.
Lastly, I would argue (based on my own belief system) that Iowa's offense is not average, it is well below average. Thus the 47% assumption is too high for them, and if that is the case, it just makes it even more logical to kick the PAT after their score in the Purdue game because their likelihood of failure was even greater. And once they failed (as would be expected in most cases), it made it impossible for them to win the game in the time remaining.