(# of possible worlds in which we win after switching) / (all possible worlds in which we are shown a goat after the initial pick)
vs
(# of possible worlds in which we win without switching) / (all possible worlds in which we are shown a goat after the initial pick)
in this case 2/3 vs 1/3. We are only considering the cases where a goat has been shown so what happens when a goat isn't shown or what the host's intentions are when a goat is show is irrelevant.
1. We simulate a million trials where the contestant chooses one of three door, one of which has a car and two a goat
2. We simulate the host randomly choosing one of the remaining doors to open.
3. We discard all trials that resulted in a car. We are left with the number of trials that resulted in a goat and store that number in a variable 'total'
4. In the remaining trails we switch and reveal what was behind the door. We store the number of times we saw a car in a variable called 'wins'.
5. The probability of winning after switching is 'wins' / 'total'
It doesn't matter how many times the host shows a car because those trials are discarded.
If the host picks randomly and you discard rounds with a car, your odds switching are 50/50. If the host uses knowledge of where the car is to definitely reveal a goat, your odds switching are better.
I've written the simulation, and you're right. It is 50% when the host chooses randomly!
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.
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.
Seriously. Write the simulation.
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.
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
That is false 100% of the time given the parameters that the outcome of the host's choice is a goat (as the question states)
Edited to add:
I think there may be a different misunderstanding here...
Set aside, for the moment, whether the above implements the originally posed question or the variant. Imagine a gameshow that runs the above code, but the switch/stay code is moved below the "decision time!" line and the decision is made based on a prompt rather than hard coded. When you hit that prompt, "Monty" has already picked his door, and we've checked that it's not a car.
Do you expect your chance to win, on answering "switch" at that prompt, to depend on which line above is commented out?
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.
open_door = host_opts.sample
choice2 = [0, 1, 2] - [choice1, open_door] # switch
if open_door == car then