Danny,
Very well analyzed. Note that in case the “killing theory” is correct (i.e. the host does not know where the prize is) switching is the best strategy only in a trivial sense – all the possible strategies are equivalent and give you a chance of exactly 0.5 of winning (given that you are not killed ).

Yair,

We will only consider games in which the candidate chose to switch doors (obviously this is also the correct strategy, but we don’t care, as we can simply ignore the other cases).

We need to distinguish between two hypotheses:
H0: The host never opens the prize door, all candidates get their 15 minutes of television fame.
H1: The host might open the prize door, in which case the candidate is executed.

WLOG we can assume the candidate’s original choice was door #1 (everything is symmetrical, so it shouldn’t really matter).

Now, there are 3 scenarios under H0, each w.p. 1/3:
1. The prize is behind door #1: so the host opens some random door 2 or 3, the candidate switches to the other one and loses.
2. The prize is behind door #2, so the host opens door #2 w.p. 1/2 and the candidate dies, or opens door #3 w.p. 1/2 and the candidate lives and wins.
3. the prize is behind door #3 – same as last.

All and all, under H0 we have:
candidate loses: 1/3
candidate wins: 1/3 * 1/2 * 2 = 1/3
candidate dies : 1/3 * 1/2 * 2 = 1/3

So under H0 and limiting ourselves to TV appearences only, we can expect to see about 1/2 of the candidates winning and the rest losing.

Under H1 we have also 3 scenarios, each w.p. 1/3:
1. The prize is behind door #1: so the host opens some random door 2 or 3, the candidate switches to the other one and loses.
2. The prize is behind door #2, so the host opens door #3 and and the candidate wins.
3. The prize is behind door #3 – same as last.

And all and all, under H1 we have:
candidate loses: 1/3
candidate wins: 1/3 * 2 = 2/3

So under H1 and limiting ourselves to TV appearences only, we can expect to see about 2/3 of the candidates winning and the rest losing.

Now we just need to apply the (weak) law of large numbers, compute the win proportion from the actual (virtually unlimited) television shows we have, and see which of H0 and H1 is more likely.