The Monty Hall dilemma is a probability puzzle based on the American television game show Let's Make a Deal. Imagine you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats.
You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then asks you if you want to stick with your original pick, or swap to the other remaining door. What do you do?
Most people think that it doesn't make a difference and they tend to stick with their first pick. But it does make a difference. Even pigeons know that.
3 comment(s):
How can a random choice be better if there is suddenly one less option to choose from?
The explanation is ridiculous as to why showing that a third different door has a goat behind it.
Either remaining door has just as much information as each other, so without having extra info the choice is still random.
It would have been embarrassing if the contestant had chosen the door with the goat behind it with their first choice.
What then?
Just ask a pigeon :)
Anyway, look at it this way: when you made your choice, you had a 1 in 3 chance of picking the winning door.
After you made your choice, one door is opened. Now, if you make a choice, you have a 1 in *2* chance of picking the winning door.
So, by swapping, you go from a 1 in 3 chance of winning, to a 1 in 2 chance of winning. Therefore, the best strategy is to swap.
Of course, if you already picked the winning door from the start, then swapping is not a good idea. But you don't know if you picked the winning door. So you should swap anyway, because it will improve your odds.
If you were to do this experiment say 1000 times, you would see that you win 50% of the times. If you don't swap, you win 33% of the times. It's counter-intuitive, I know, but true regardless :)
Anonymous, I wish you'd left your name just to make sure it's not the same anonymous who was wrong in previous comments.
Anyway if it's the same anonymous, or a different one, still you are wrong.
I have the impression you didn't read the article because the article says your answer is precisely the wrong assumption everyone makes when given the chance to swap.
There is no 50/50 chance of winning after one door is removed from the equation.
By swaping you increase your chances of winning from 0.33 to 0.66, not from 0.33 to 0.50.
"At the very start, the contestant has a one in three chance of picking the right door. If that’s the case, they should stick. They also have a two in three chance of picking a goat door. In these situations, the host, not wanting to reveal the car, will always pick the other goat door. The final door hides the car, so the contestant should swap. This means that there are two trials when the contestant should swap for every one trial when they should stick. The best strategy is to always swap – that way they have a two in three chance of driving off, happy and goatless."
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