Yes, _this time_. But how on earth are you evaluating probabilities without considering all the priors and/or doing multiple trials? Is this some new kind of mathematics you've invented?
The hidden assumption is that the odds that he opens a door containing a car is zero. That is what changes the odds in the second step. Without that they don't change.
Read your second question again. It says that the host shows a goat. Then asks if you would switch. Yes, you would switch.
I understand what you were trying to ask but you worded the second question wrong. You can't tell the listener that the host picks a goat. That defeats the purpose of the "randomness" which is meaningless since we know the host picks a door with a goat.
I found this on the internet and modified it so you can just change the commented out options to run it under the different scenarios.
car = wins = 0
many = 100000
actual = 0
many.times do
choice1 = rand(3)
car = rand(3)
#host_opts = [0, 1, 2] - [choice1, car] # host knows what is behind
host_opts = [0, 1, 2] - [choice1] # host doesn't know what is behind
#choice2 = [choice1] # don't switch
choice2 = [0, 1, 2] - [choice1, host_opts.first] # switch
if host_opts.first == car then
# Discard? Automatic win? It doesn't actually matter!
# It only changes the denominator.
else
# According to the puzzle, it is decision time!
wins += 1 if choice2.first == car
actual += 1
end
end
I believe it doesn't change the result, but this is technically Monty always picking "the lower (leftmost?) door you didn't pick" not "a random door you didn't pick", right?
Also, "it only changes the denominator" may be misleading, as it does not change the only denominator actually visible in the code. It changes the total numbers of wins and losses, but not the total numbers of wins and losses given that you saw a goat.
If the host picked at random and happened to get a goat, it is 1/2 vs 1/2. If the host picked a door he knew had a goat, it's 2/3 vs 1/3.
Again, write the simulation. It shouldn't take you 5 minutes.