Do you switch? Or do you stick with your original guess? As the parent mentioned, switching gives you a 99/100 chance of being correct, staying gives you a mere 1/100 chance.
Keep in mind I made my choice before we started, and it hasn't changed.
Do you want to change your guess to 437, or stick with 2?
The key to this problem is the behavior of the game host as he will never reveal the door with the car. So, after he reveals the contents of one of the doors, if you choose not to switch, you still only have a 1/3 chance of having chosen the door with the car.
On the other hand, if you choose to switch, 2/3 of the time you will have chosen incorrectly initially, and under the switching strategy, those 2/3ds of the time you are guaranteed to win the car. Hence, you have a 2/3ds chance of winning under the switching strategy, vs a 1/3d chance of winning under the "staying" strategy.
Hope this clarifies things.
Look at it this way: 1/3 of the time, you pick the right door up front. If you switch, you lose.
2/3 of the time, you didn't pick the right door. If you switch, you win.
So switch.
The key observation is that the problem states that Monty Hall opens an empty door. It's not clear whether Monty knows which doors are empty, but it doesn't actually matter, since the problem states that he will always open an empty door.
As others have said, your choice as a contestant is whether to take one door (the one you initially pointed at), or to take the other two doors (at least one of which is obviously empty). Stated that way, it's pretty obvious that taking 2 doors is better than taking 1 door.
The location of the prize and the player choice of door are independent events. Monty's choice is dependent on whether the player choice and the prize location are the same. Monty will never open the door the player picked. Monty will never reveal the prize. Those rules generate exploitable information for the player.
Here is the complete probability tree, and two possible strategies of play:
Prize behind door A; P = 1/3 ALWAYS STAY strategy:
Player picks door A; P = 1/3 * 1/3 = 1/9 P(WIN) = 1/18 + 1/18 + 1/18 + 1/18 + 1/18 + 1/18
Monty shows Door B; P = 1/3 * 1/3 * 1/2 = 1/18 = 6/18
Player stays with A; WIN = 1/3
Player switches to C; LOSS P(LOSS) = 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9
Monty shows Door C; P = 1/3 * 1/3 * 1/2 = 1/18 = 6/9
Player stays with A; WIN = 2/3
Player switches to B; LOSS
Player picks door B; P = 1/3 * 1/3 = 1/9 ALWAYS SWITCH strategy:
Monty shows Door C; P = 1/3 * 1/3 * 1/1 = 1/9 P(WIN) = 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9
Player stays with B; LOSS = 6/9
Player switches to A; WIN = 2/3
Player picks door C; P = 1/3 * 1/3 = 1/9 P(LOSS) = 1/18 + 1/18 + 1/18 + 1/18 + 1/18 + 1/18
Monty shows Door B; P = 1/3 * 1/3 * 1/1 = 1/9 = 6/18
Player stays with C; LOSS = 1/3
Player switches to A; WIN
Prize behind door B; P = 1/3
Player picks door A; P = 1/3 * 1/3 = 1/9
Monty shows Door C; P = 1/3 * 1/3 * 1/1 = 1/9
Player stays with A; LOSS
Player switches to B; WIN
Player picks door B; P = 1/3 * 1/3 = 1/9
Monty shows Door A; P = 1/3 * 1/3 * 1/2 = 1/18
Player stays with B; WIN
Player switches to C; LOSS
Monty shows Door C; P = 1/3 * 1/3 * 1/2 = 1/18
Player stays with B; WIN
Player switches to A; LOSS
Player picks door C; P = 1/3 * 1/3 = 1/9
Monty shows Door A; P = 1/3 * 1/3 * 1/1 = 1/9
Player stays with C; LOSS
Player switches to B; WIN
Prize behind door C; P = 1/3
Player picks door A; P = 1/3 * 1/3 = 1/9
Monty shows Door B; P = 1/3 * 1/3 * 1/1 = 1/9
Player stays with A; LOSS
Player switches to C; WIN
Player picks door B; P = 1/3 * 1/3 = 1/9
Monty shows Door A; P = 1/3 * 1/3 * 1/1 = 1/9
Player stays with B; LOSS
Player switches to C; WIN
Player picks door C; P = 1/3 * 1/3 = 1/9
Monty shows Door A; P = 1/3 * 1/3 * 1/2 = 1/18
Player stays with C; WIN
Player switches to B; LOSS
Monty shows Door B; P = 1/3 * 1/3 * 1/2 = 1/18
Player stays with C; WIN
Player switches to A; LOSS
[1] https://en.wikipedia.org/wiki/Monty_Hall_problem#Simple_solu...
That said, I made a simulation of it where it does the following: 1. Assign random locations behind doors, 2. Player randomly chooses door, 3. Host excludes door (random if both are goats, goat if one is a car), 4. Player either switches, or doesn't switch based on what's chosen for that simulation (the full set gets one or the other)
After running 50,000,000 then dividing success/tries, my number is nearly exactly the same between the two. Is my simulation working incorrectly?