Okay. In that case it is best for the contestant to switch the first time, since surely he can do better than 1/100. Then we are playing the classical monty hall problem with the following modifications:
1. The car is not uniformly distributed between doors; it is behind the first with probability 2/200, the second 99/200, the third 99/200.
2. The contestant is forced to select door 2.
3. Then the host opens door 1 or 3, whichever has a goat behind it. If both have a goat the host chooses door 1 or 3 with a strategy which minimizes the chance that the contestant wins, instead of randomly as in the classical problem.
4. Then the contestant opens one of the two remaining closed doors of their choice
This is a fairly simple game to solve. The contestant has 4 pure strategies: Open door 1 if possible, else door 2, Open door 2, open door 3 if possible, else door 2, and open the other door. Call these strategies 1, 2, 3, and 4, respectively. The host has two pure strategies: Open door 1, or open door 3 (since he only has a choice when both are goats). Here's a payoff table, where the payoff is that of the contestant and a win is +1, a loss -1. The payoff for the host is the negative.
1 2 3 4
Door 1 .01 -.01 -.01 .01
Door 3 -.98 -.01 .98 .01
Clearly it can never be better to choose strategies 1 or 2, so we just consider 3 and 4. Then it is never better for the host to choose to open door 3 if he has the choice, so then we are left with just Door 1, and strategies 3 and 4. Then of course the contestant should always switch, strategy 4.
So we got lucky and there was a pure-strategy equilibrium. The host should always open Door 1 if he has a choice*, and the contestant should always switch*. The contestant's expected payoff is than .01, or equivalently, he has a 101/200 = 50.5% chance of winning.
*Technically, the host could also open door 3 when he has a choice with probability up to and including 2/99, and then the contestant would be fine opening door 2 when the host opens door 3 if P(D3) = 2/99. The expected payoff remains the same of course.