Debunking Monty Hall problem fallacy
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- Опубліковано 17 вер 2024
- I did a realistic simulation of Monty Hall problem and proved that switching choice never influences the outcome. In this simulation I acknowledge a very simple fact, that most of the "high IQ-ers" somehow always overlook. Any sane person will always change initial choice, if it was revealed to be a goat by the host. Therefore, first choice does not matter. It never influences the end outcome.
In this game you always choose between 2 doors, not 3. Therefore the chance of win is always 50%, no matter if you switch your choice or stick with original one (if it wasn't revealed as goat).
Source: github.com/Sci...
I've argued this so many times. It's basically the same thing as guessing whether two coins being flipped will come up both the same (regardless of side) or one of each (regardless of which coin is which), hoping for the opposite of my choice, then flipping one coin first before asking if I want to change.
Example:
I guess outcome. (Pick a door)
Options 1 and 2 - First coin comes up heads or tails (reveal an empty door).
Option 3 - Outcome of second coin reduced to either match or doesn't match. (Your door is revealed).
Three steps, three options, 50/50 result.
Yep. Coin flip is a very good analogy!
Except that isn't the case. It should become quite obvious if we increase the number of non-winning doors. If you have to pick from, let's say, 1000 doors it should be quite clear that your door can't have a 1/2 chance of winning.
It's a shame that you programmed it in a very complicated way just to end violating the rules of the game. In Monty Hall, the host takes care of always revealing a goat, but it has to be from the doors that the player did not pick, which means that as long as the player starts picking a goat, the revealed goat will be the second one, so switching necessarily leads to the car.
In contrast, here you put as if when the player has picked a goat, the host could decide to reveal that same goat, and if it happens, then the player makes a random selection between the other two doors that remain closed, with only 1/2 chance to pick which has the car, because in the other 1/2 he would end picking which has the other goat.
That means that not in all the games that the contestant has picked a goat he will manage to get the car by switching, and that's why you end getting less wins for switching than there should be.
I "violated the rules" because the rules are wrong and makes 0 practical sense.
@@ScienceDiscoverer
SD: I will debunked MHP. First, I change the rule.