this post was submitted on 07 Dec 2023
504 points (87.2% liked)
Asklemmy
43786 readers
1195 users here now
A loosely moderated place to ask open-ended questions
Search asklemmy π
If your post meets the following criteria, it's welcome here!
- Open-ended question
- Not offensive: at this point, we do not have the bandwidth to moderate overtly political discussions. Assume best intent and be excellent to each other.
- Not regarding using or support for Lemmy: context, see the list of support communities and tools for finding communities below
- Not ad nauseam inducing: please make sure it is a question that would be new to most members
- An actual topic of discussion
Looking for support?
Looking for a community?
- Lemmyverse: community search
- sub.rehab: maps old subreddits to fediverse options, marks official as such
- [email protected]: a community for finding communities
~Icon~ ~by~ ~@Double_[email protected]~
founded 5 years ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
You're not really arguing against the whole crowd there, a lot of people (wrongly) hold the same opinion. The problem is thinking of the door swap as an independent event when it's not; the result is directly related to the original choice of door. If we label the doors A, B, and C and put the prize behind door A, here's the possible options:
Two out of three times swapping wins.
Edit: I see you added a table to your comment, but you're miscounting pretty badly there. You're giving double weight to initial choice being correct.
It is technically true that when you pick A the presenter can open either B or C, but then you need to account for that in your odds; it's 50% either way so the win/loss rate is halved. In other words:
As shown, including which door the presenter opens does not affect the odds. When sticking, you win (16.5% + 16.5% = 33%) and lose (33% + 33% = 66%), when swapping you win (33% + 33% = 66%) and lose (16.5% + 16.5% = 33%).
Dude you can't argue mathematics
Hell yeah I can. Why the fuck is Jimmy buying 35 watermelons on an average day???
Mathematics bro
Forget the 35 watermelons, why does he have 5 cakes at school?
That's mathematically correct!
Here is an alternative Piped link(s):
That's mathematically correct!
Piped is a privacy-respecting open-source alternative frontend to YouTube.
I'm open-source; check me out at GitHub.
That's not an opinion, that's an incorrect statement.
You can have an opinion and still be factually wrong I suppose
I was stubborn about this for so long, and I'm still not entirely sure I understand it, but here is a perspective that made me doubt my belief.
Imagine the Monty Hall Problem, but with 100 doors and only one grand prize. You pick one; it obviously has a 1/100 chance of being a grand prize. Then Monty reveals 98 doors without grand prizes in them such that the only doors left are the one you chose and one that Monty left unopened. Monty obviously arranged for one of those two doors to have the grand prize behind it. The "choice to switch" is really just a second round of the game, ~~but with a 1/2 chance of winning~~ (wrong, your odds change only if you "participate" in round two).
If you stick with your door, you are relying on your initial 1/100 chance of winning. If you switch, you are getting the ~~1/2~~ odds of the "second round".
Apparently with three doors, switching gives you a 2/3 chance of winning, but I don't understand the math of how to get that answer and I wouldn't be able to calculate the odds of the 100 door version. I just know intuitivey that switching is better.
With 100 doors swapping wins 99 out of 100 times; the only time you lose is when your initial door (1 in 100) contained the prize.
With 100 doors switching should give you a 99% win rate.
You're essentially concentrating the entire thing into this one vs not this one, and when you initially chose there was a 99% chance it was not this one.
After Monty opens all the other doors, the odds that the right answer is not this one is still 99% except that now the entirety of not this one is represented by that single other door. The Grand Prize has nowhere else to be, and the odds that you picked it first is still only 1%.
So, to bring it back down, with three doors, the odds that the right answer is not this one 66%, and we end up exactly where we expected to be.
There's four total outcomes of the problem:
Scenario 1: you originally pick the winning door (1/3) and don't switch (1/2), therefore winning. Probability = 1/6
Scenario 2: you originally pick the winning door (1/3) and did switch (1/2), therefore losing. Probability = 1/6
Scenario 3: you originally pick a losing door (2/3) and don't switch (1/2), therefore losing. Probability = 1/3
Scenario 4: you originally pick a losing door (2/3) and do switch (1/2), therefore winning. Probability = 1/3
Now consider scenarios 1 and 3 together, these two are when you don't switch. P(S1) is 1/6 and P(S3) is 1/3, meaning S3 is twice as likely than S1. So if you don't switch, you are twice as likely to lose. And now consider scenarios 2 and 4 together. P(S4) is 1/3 and P(S2) is 1/6, meaning if you switch you are twice as likely to win than to lose.
You can also consider this problem in terms of conditional probability like this:
P(win as long as no switch) = P(win and no switch) / P(no switch) = P(S1)/(1/2) = (1/6)/(1/2) = 2/6 = 1/3
P(win as long as switch) = P(win and switch) / P(switch) = P(S4)/(1/2) = (1/3)/(1/2) = 2/3
P(win as long as switch) > P(win as long as no switch)
I have no clue what this actually is about. But I always remember watching Deal or No Deal and thinking "If it was down to two boxes, Β£1 and Β£250K, I would absolutely swap my box." There is no way I would believe that all along I've been holding the 250K box. In my mind it makes more sense that I'm holding the Β£1 box and I need to swap.
The "second round" of the game is always just, "flip your odds of winning if you swap". That's all it is.
Monty will always open the proper doors to ensure this happens every time. Did you pick the winning door in the first round? Monty will eliminate all other doors but leave one of the losers. Did you pick a losing door in the first round? Monty will eliminate all the other losers and only leave the winner. It's always the opposite of what you picked. Therefore, if you swap, you will simply get the opposite odds of the first round.
100 doors to pick from, only 1 winner? 1/100 chance to win if you just picked at random and ended it there. Now Monty offers a swap. Without the swap, you have 99 different ways to lose this. But with the swap, all 99 of those ways become winners, because Monty will always swap the opposite with you.
But by staying on your door you're still making a choice relying on that Β½ chance...
No, by staying on your door you're relying on the 99/100 chance of originally picking the wrong door.
This is worded better than what I said. The second round isn't 1/2 because the door you initially picked was 1/100.
Not if I bribe the people in charge of putting the prize behind the door it won't
Only if Monty Hall didn't know where the prize is
Say there are 100 doors, you choose one, then 98 are knocked out randomly (likely including the prize) - Now each of the 2 doors has the same chance of winning, so there is no reason to change
But starting with 100 doors and a knowledgeable Monty Hall, once you've chosen a door, the only reason Monty Hall leaves your door alone is because you chose it, whether it is the 1/100 winner, or one of the 99/100 losers
Either you chose the right door the first time (1/100 chance) or the other door has the prize behind it - those are the only options - the other door literally represents the 99/100 other doors in a single choice
There's a flaw in this problem, which is the fact Monty Hall didn't consider the possibility I may have a gun pointed to his head
Do you have a Monty Hall problem, or does Monty Hall have a you problem?
You can't have opinion on maths
Yes you csn, for example: math is what scientists do while on meth