Cups problem: I built this table and can't see anything wrong in the 50% chance:

PRIZE IS IN CUP 1
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Choose 1, lift 2, change to 3 : loose
Choose 1, lift 3, change to 2 : loose
Choose 2, lift 3, change to 1 : win
Choose 3, lift 2, change to 1 : win

PRIZE IS IN CUP 2
-----------------
Choose 1, lift 3, change to 2 : win
Choose 2, lift 1, change to 3 : loose
Choose 2, lift 3, change to 1 : loose
Choose 3, lift 1, change to 2 : win

PRIZE IS IN CUP 3
-----------------
Choose 1, lift 2, change to 3 : win
Choose 2, lift 1, change to 3 : win
Choose 3, lift 2, change to 1 : loose
Choose 3, lift 1, change to 2 : loose

============================================
PRIZE IS IN CUP 1
-----------------
Choose 1, lift 2, stay on 1 : win
Choose 1, lift 3, stay on 1 : win
Choose 2, lift 3, stay on 2 : loose
Choose 3, lift 2, stay on 3 : loose

PRIZE IS IN CUP 2
-----------------
Choose 1, lift 3, stay on 1 : loose
Choose 2, lift 1, stay on 2 : win
Choose 2, lift 3, stay on 2 : win
Choose 3, lift 1, stay on 3 : loose

PRIZE IS IN CUP 3
-----------------
Choose 1, lift 2, stay on 1 : loose
Choose 2, lift 1, stay on 2 : loose
Choose 3, lift 2, stay on 3 : win
Choose 3, lift 1, stay on 3 : win

I guess (only guess) that the fact that an empty cup is removed does not influence the final selection which becomes a two-cup only game. It is the same as if the house said "Ok, we always take out one empty cup, let's play with only two cups, you make your choice, I ask you if you are sure, and you can keep your choice, or change it" The same would be (always guessing) with 1000 cups, I choose one, they take out 998 empty cups and let me choose between the remaining two. In my line of reasoning, the probability remains 50%. In yours, we would have an enormously high probability of a "bad" choice in the first selection, and then would more strongly recommend a change.

Intriguing, but I don't see any mistake in the above tables.

Un abrazo - Andrés