Monty Hall question, but varied. I haven’t seen the option that no matter what you have a 40% chance overall to win. You always pick 1 of 3 doors, then 1 is discarded & you pick 1 of 2. So don’t you always have a 2/5 chance of being right no matter what?
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Because there aren’t five doors, there are three. You choose one out of three doors, which makes you 2/3 likely to be wrong. This doesn’t change when one door is opened, so when you get to choose between your door and the remaining door, the chance you chose wrong remains at 2/3.
The percentages don’t multiply like that.
If you make a random 1/3 guess you have a 33% chance.
If you make a random 1/2 guess you have a 50% chance.
The Monty Hall problem just creates the illusion that “keeping” your first choice gives you the 50/50 chance when you really only had the 1/3.
You have 100 doors, pick 1.
1/100 chance of being right.
Discard 98, so 2 doors.
Do you think you should stick with your 1/100 pick?
_______
Your initial pick was a 1/3 chance. So even with 1 of the 3 discarded, it still is a 1/3 chance, not a 50/50 chance. If the doors were randomized and you had to pick again, then sure, but that’s not how it happens.
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I can see why you’re saying 2/5, but simply put that’s just not how probability works. Along with the other explanations here it might help to look at this demonstrated in a diagram. [This](https://brilliant-staff-media.s3-us-west-2.amazonaws.com/tiffany-wang/zWt5C1YlGm.png) is the best one I can see (from [this explanation](https://brilliant.org/wiki/monty-hall-problem/)).
As you can see from that, there are six possible combinations of choice & outcome: 3 where you stick and 3 where you change.
(One thing that’s hard to get your head around here is that probability depends on our *knowledge*. The location of the car doesn’t change, but opening a door changes our knowledge and our assessment of the probability.)
Essentially, the first door you pick is your bet that it is not the correct door. You get to know whether either of the other two doors would be the correct one, so you have 2/3 chance of being correct by choosing the first one to be the one that won’t be the correct door.
Another way of thinking of it: Monty will never open the correct door. If you’ve picked the correct door (1/3 of the cases) he can open either of the others; if you’ve picked the wrong door (which you have done 2/3 of the time), he’s forced to open the only remaining wrong door, and you should switch.
It’s 1/3 if you don’t switch, 2/3 if you do. It’s not intuitive but very easy to verify if you just work out the 3 possibilities in your head. If the correct door is B…
A: guess wrong door, switch to correct door.
B: guess right door, switch to wrong door.
C: guess wrong door, switch to correct door.
The key is you’ll never guess the wrong door and then switch to the other wrong door.
Not sure where you’re getting these numbers from.
1 in 3 and 2 in 5 aren’t the same. There are only 3 doors, so in the first stage you have a 1/3 chance of picking the correct door.
Because the host always opens a door with a goat, swapping will *always* change what you got. If you got a car you’ll always swap to a goat, if you got a goat you’ll always swap to a car. Therefore the probabilities are swapped and you now have a 2/3 chance of getting the car.
There’s no 50% probability or 2/5 probability at any point, the relevant probabilities are 1/3 and 2/3 and that’s it
Easiest way to think about the problem: Your first choice *constrains* Monty Hall. If you pick the prize (1/3 of the time), he can show you one of two goats. Switching always loses. If you pick a goat (2/3 of the time), he *has* to show you the other goat. Switching always wins.
It’s that simple.
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