Monty Hall problem altered

Three doors, two goats and a car. 

The contestant does not pick a door. Instead, Monty hall picks two doors at random. Then Monty says, do you want these two doors or the one door? You take the two doors, then Monty opens the booby prize, and their is a 2/3 chance the other door is a car.

This seems like the more intuitive understanding. The game is commutative because there is always at least one goat in the set of two. So having Monty prematurely open that door, with foreknowledge, is moot. Thus, just switch the choices, put Monty first and the contestant has an obvious choice.

Change the game. You can pick any of the three doors per round, and the number of rounds is large, but finite.  Each door has a variable payout which was the log of some number. The number will be the proportion of times that door was picked.  So each time you leave one door unopened, its payoff increases as -log(k/total), k is your number of picks from the start, including one free pick, so k initially was 1 for each window. 

So in this game, if I keep picking one door repeatedly, its pay off drops as k/t goes toward one, but the other doors increase in payoff. 

What is my strategy? What is the best guess of the number total rounds you are allowed. I would let the one door build up in value, never open it until the final move.  If that is the case, then at least one of the others be selected twice as often. And the final at east twice as often as that. But it depends on my foe knowledge of allowed rounds of the game. What if i don't know it exactly, they tell me, 100 plus or minus three?