Because when you first picked 27, it was 1 out of 100 choices. Then I tell you that you either got it right, or it’s this other number. None of the others are correct, only 27 or 44.
So you think your 1/100 choice was better than the one I’m giving you now? On average, you’ll be right 1% of the time if you don’t switch. If you do switch, you’ll be correct 99% of the time.
Another way to think of it is: you choose 27 or you choose ALL of the other 99 numbers knowing that I’ll tell you that 98 of them are wrong and you’ll be left with the correct one out of that batch. One of those clearly has better odds, no?
In this example, there were 100 choices in the beginning, and later you reduced to 2 choices. Clearly an advantage. Does the same apply to the 3 door problem?
Let’s take this question in another angle. Instead of 3, there are only 2 doors. I am to choose one out of 2, which has a prize. After I choose one, you show me a third door which is empty. Now, should I change my option?
Yes, it’s the same concept. The same math/logic behind it doesn’t change. You’re choosing 1/3 or you are choosing 2/3 and I’ll tell you which of the two is incorrect. It’s just easier to visualize with 100 doors instead.
I’m not sure I’m following the other angle…there are 3 correct possibilities at the start but I can only choose 2? Or there are 2 possibilities and then you introduce a 3rd door that is never correct?
Do you know the third door is never correct? Because then the probability doesn’t change.
Scenario 1: You chose 1/2 at first with a 50% chance of being correct, I introduce a 3rd door (but it isn’t a legit possibility), so the actual choice for you is still 50/50 (between doors 1 and 2)
Scenario 2: If you think it’s possible that 3 could be correct (but it actually never is) then, no, you wouldn’t want to switch. By staying with your first choice has a 50% chance of winning, by switching it only has a 33% chance. But there’s no way to know this ahead of time (because as soon as you know you shouldn’t switch bc 3 is the wrong door, then you’re back in scenario 1)
Scenario 3: For completeness, let’s say the 3rd door can be correct sometimes. Then it doesn’t matter if you switch or not. It’s a 33% chance of winning either way. If there is a chance it can be correct, then your first choice doesn’t matter at all and the second choice is the ‘real’ choice bc that’s the only time you’re able to choose from all real possibilities.
The only reason that the Monty Hall problem changes probability in the second choice is because you are provided more information before the switch (that the opened door is absolutely not the one with the prize)
In scenario 1, legit or not, you said the chance is still 50-50. In other scenarios also you shouldn’t change or it wouldn’t matter. That’s what I say, just in the opposite direction. But the problem of probability depends on the wordings and phrases, which means I may not have understood the ques well.
Another angle: You explained the Monty Hall problem at the end that the probability changes because in second choice we have more information. So you are implying that the initial 1/3 probability of the now-open door adds to the door we did not choose - making the switch advisable.
Here I also say the probability does change from initial 1/3, but to 1/2-1/2 for each remaining doors; why should the probability be poured to the unselected single door?
In the original the possibilities for a prize behind the doors 1,2,3 are:
A) YNN
B) NYN
C) NNY
In (A) -
A.1 you choose door 1 and then stay, you win
A.2 you choose door 1 and switch, you lose
A.3 you choose door 2 and stay, you lose
A.4 You choose door 2 and switch, you win
A.5 you choose door 3 and stay, you lose
A.6 you choose door 3 and switch, you win
By staying, you lose in 2 of 3 cases (A.3 and A.5)
This isn’t a trick or anything, the math is pretty clear and you can actually write out all the scenarios and count it up yourself. It’s just a little counterintuitive because we aren’t used to thinking in terms of conditional probabilities this way.
Another way to think about it is the probability of losing. If the contestant loses, it means that they picked correctly on their first choice and then swapped. This will happen 1/3rd of the games, because there is a 1 in 3 chance of picking correctly the first time. So, if you have a 1/3rd chance of losing by swapping, then it follows that you have a 2/3rds chance of winning by swapping (choosing incorrectly at the start and then switching to the correct door)
Because when you first picked 27, it was 1 out of 100 choices. Then I tell you that you either got it right, or it’s this other number. None of the others are correct, only 27 or 44.
So you think your 1/100 choice was better than the one I’m giving you now? On average, you’ll be right 1% of the time if you don’t switch. If you do switch, you’ll be correct 99% of the time.
Another way to think of it is: you choose 27 or you choose ALL of the other 99 numbers knowing that I’ll tell you that 98 of them are wrong and you’ll be left with the correct one out of that batch. One of those clearly has better odds, no?
In this example, there were 100 choices in the beginning, and later you reduced to 2 choices. Clearly an advantage. Does the same apply to the 3 door problem?
Let’s take this question in another angle. Instead of 3, there are only 2 doors. I am to choose one out of 2, which has a prize. After I choose one, you show me a third door which is empty. Now, should I change my option?
Yes, it’s the same concept. The same math/logic behind it doesn’t change. You’re choosing 1/3 or you are choosing 2/3 and I’ll tell you which of the two is incorrect. It’s just easier to visualize with 100 doors instead.
I’m not sure I’m following the other angle…there are 3 correct possibilities at the start but I can only choose 2? Or there are 2 possibilities and then you introduce a 3rd door that is never correct?
Yes that one. Similar to the one you did with 100 doors, just in opposite direction.
Do you know the third door is never correct? Because then the probability doesn’t change.
Scenario 1: You chose 1/2 at first with a 50% chance of being correct, I introduce a 3rd door (but it isn’t a legit possibility), so the actual choice for you is still 50/50 (between doors 1 and 2)
Scenario 2: If you think it’s possible that 3 could be correct (but it actually never is) then, no, you wouldn’t want to switch. By staying with your first choice has a 50% chance of winning, by switching it only has a 33% chance. But there’s no way to know this ahead of time (because as soon as you know you shouldn’t switch bc 3 is the wrong door, then you’re back in scenario 1)
Scenario 3: For completeness, let’s say the 3rd door can be correct sometimes. Then it doesn’t matter if you switch or not. It’s a 33% chance of winning either way. If there is a chance it can be correct, then your first choice doesn’t matter at all and the second choice is the ‘real’ choice bc that’s the only time you’re able to choose from all real possibilities.
The only reason that the Monty Hall problem changes probability in the second choice is because you are provided more information before the switch (that the opened door is absolutely not the one with the prize)
In scenario 1, legit or not, you said the chance is still 50-50. In other scenarios also you shouldn’t change or it wouldn’t matter. That’s what I say, just in the opposite direction. But the problem of probability depends on the wordings and phrases, which means I may not have understood the ques well.
Another angle: You explained the Monty Hall problem at the end that the probability changes because in second choice we have more information. So you are implying that the initial 1/3 probability of the now-open door adds to the door we did not choose - making the switch advisable. Here I also say the probability does change from initial 1/3, but to 1/2-1/2 for each remaining doors; why should the probability be poured to the unselected single door?
In the original the possibilities for a prize behind the doors 1,2,3 are:
A) YNN B) NYN C) NNY
In (A) - A.1 you choose door 1 and then stay, you win A.2 you choose door 1 and switch, you lose A.3 you choose door 2 and stay, you lose A.4 You choose door 2 and switch, you win A.5 you choose door 3 and stay, you lose A.6 you choose door 3 and switch, you win
By staying, you lose in 2 of 3 cases (A.3 and A.5)
By switching you only lose in 1 case (A.2)
It works out for (B) and © the same way. You have a 2/3rds chance of winning if you switch and a 1/3rd chance of winning if you don’t.
This isn’t a trick or anything, the math is pretty clear and you can actually write out all the scenarios and count it up yourself. It’s just a little counterintuitive because we aren’t used to thinking in terms of conditional probabilities this way.
Another way to think about it is the probability of losing. If the contestant loses, it means that they picked correctly on their first choice and then swapped. This will happen 1/3rd of the games, because there is a 1 in 3 chance of picking correctly the first time. So, if you have a 1/3rd chance of losing by swapping, then it follows that you have a 2/3rds chance of winning by swapping (choosing incorrectly at the start and then switching to the correct door)