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First, let us assume that the car is behind door no. 1. We can
do this without reducing the validity of our proof, because if
the car were behind door no. 2, we only had to exchange all
occurrences of "door 1" with "door 2" and vice versa, and
the proof would still hold.
The candidate has three choices of doors. Because he has no additional information, he randomly selects one. The possibility (I'll write p from here on like real statisticians do) to choose each of the doors 1, 2, or 3 is 1/3 each:
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| Candidate chooses | p | |||||
|---|---|---|---|---|---|---|
| Door 1 | 1/3 | |||||
| Door 2 | 1/3 | |||||
| Door 3 | 1/3 | |||||
| Sum | 1 | |||||
|
Going on from this table, we have to split the case depending on the door opened by the host. Since we assume that the car is behind door no. 1, the host has a choice if and only if the candidate selects the first door - because otherwise there is only one "goat door" left! We assume that if the host has a choice, he will randomly select the door to open.
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| Candidate chooses | Host opens | p | ||||
| Door 1 | Door 2 | 1/6 | ||||
| Door 1 | Door 3 | 1/6 | ||||
| Door 2 | Door 3 | 1/3 | ||||
| Door 3 | Door 2 | 1/3 | ||||
| Sum | 1 | |||||
|
Let us now look at a candidate who always sticks to his original choice no matter what happens:
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| Candidate chooses | Host opens | final choice | win | p | ||
| Door 1 | Door 2 | Door 1 | yes | 1/6 | ||
| Door 1 | Door 3 | Door 1 | yes | 1/6 | ||
| Door 2 | Door 3 | Door 2 | no | 1/3 | ||
| Door 3 | Door 2 | Door 3 | no | 1/3 | ||
| Sum | 1 | |||||
| Sum of cases where candidate wins | 1/3 | |||||
|
Now it is almost obvious, but to go all the way here's another table for a candidate who always switches to the other door whenever he gets the chance:
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| Candidate chooses | Host opens | final choice | win | p | ||
| Door 1 | Door 2 | Door 3 | no | 1/6 | ||
| Door 1 | Door 3 | Door 2 | no | 1/6 | ||
| Door 2 | Door 3 | Door 1 | yes | 1/3 | ||
| Door 3 | Door 2 | Door 1 | yes | 1/3 | ||
| Sum | 1 | |||||
| Sum of cases where candidate wins | 2/3 | |||||
What I did here is a very simple proof because I just enumerated all possible cases and added the probabilities. But a proof it is nonetheless, and thus I can boldly write qed. (Which is mathematician-speak and stands for "quod erat demonstrandum", "which had to be proved".)
If that is not sufficient for you, there's still
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