>Suppose 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 says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Does this make sense? Seems like the odds would be 50/50 to me.
>>25071913
Think of it like this:
When you pick a door you have a 2 in 3 chance of it being a goat. The host revealing the a goat does not change that probability and so staying at the same door will land you a free goat 2/3rds of the time.
Also, since the host reveals the goat, you know that you cannot land on the same prize if you switch doors now that there is only one of each in play. This means that if you pick a door (2/3 chance of being a goat) then switch doors (which guarantees you switch prizes) you will find that you end up with the car two out of three times.
Statistical projections are representations of reality and not reality
The goat is where it is regardless of the door you pick
You always have the same "chance"
>>25072261
No that's stupid.
Just think of it with 100 doors instead of 3.
If the host opens the 98 doors and lets you re choose and you stick with your first option, you're dumb as a brick.
But what if I want the goat? What use is a car to a NEET?
>>25072316
this desu
At least I can pretend the goat is my friend.
https://en.wikipedia.org/wiki/Monty_Hall_problem
1 thing that's always bothered me about this: zonks. If there's 1 $0 door, 1 $500, and 1 $3,000 door, and the $500 door is revealed first, is changing doors still worth it? Wouldn't there also be a 2/3 chance of leaving with nothing? 4/3 total: how does that work?
>>25071913
ok dickhead
I give you a bowl of 500 sweets, 499 are poisoned, if you choose the non-poisoned one you will recieve $1,000,000
I'll let you pick one to start with, and then throw out 498 because they're poisoned.
Wanna stay with the one you originally picked?
>>25072300
If and only if he's revealing new information by explicitly leaving only doors which meet the following two conditions:
>The door originally chosen, regardless of rightness or wrongness
>The correct door
Because then he is telling you new information.
If you are actually asking the question: "What is the chance that 1/100 is 1/100 if he takes away 98/100?", the answer is still 1/100. The answer only changes if Monty is assumed to reveal new information.
>>25072347
>and the $500 door is revealed first
What if they selected the $500 door first? Is their choice invalid now?
>>25072358
Yea because it's 50-50 between my one and the other one.
>>25072397
Talk about bait
In essence, you're being offered to swap to two doors. Imagine 1000000 doors, would you not switch if the host opened 999998 other doors with goats in them?
>>25072372
fuck off with your what if
they obviously didn't pick the $500 door in his example, or else that wouldn't have been the one the host opens.
if you pick the $500 one, unknowingly, and the $0 door was opened, you still wouldn't know what was behind your door, only that you've made money.
>>25072463
"Here's 2 options, one does this, one does something else."
That would be a 50-50 chance if you randomly chose between them.
>HE DIDN'T BELIEVE MY CLEARLY PHENOMENAL OPINION, MUST BE BAIT
That or you're wrong.
>>25071913
I love this one, out of all the bullshit people post, no one outs themselves as ignorant more easily than this. You have a statistically superior chance to win by switching doors. Everyone gets caught up in thinking there are 2 left which means 50/50, it doesn't work like that. The act of being "able" to reveal the goat is what masks the fact it then changes to 2/3. Some people cant get this through their head if they stare at it all day and think. This problem has even stumped some of the greatest thinkers before being entirely spelled out for them.
>>25072537
>Monty Hall Problem
>Opinions
Cmon
>>25072537
>if you randomly chose between them
But you didn't. You chose randomly between three doors, then had new information given to you.
>>25072537
You understand that the problem in OP (Monty Hall) has been mathematically proven and verified that switching doors increases your odds?
It's not even up for debate. Sure, you might struggle with the concept because you're retarded, but you're objectively wrong.
>>25072362
http://betterexplained.com/articles/understanding-the-monty-hall-problem/
Understanding The Game Filter
Let's see why removing doors makes switching attractive. Instead of the regular game, imagine this variant:
- There are 100 doors to pick from in the beginning
- You pick one door
- Monty looks at the 99 others, finds the goats, and opens all but 1
Do you stick with your original door (1/100), or the other door, which was filtered from 99? (Try this in the simulator game; use 10 doors instead of 100).
It's a bit clearer: Monty is taking a set of 99 choices and improving them by removing 98 goats. When he's done, he has the top door out of 99 for you to pick.
Your decision: Do you want a random door out of 100 (initial guess) or the best door out of 99? Said another way, do you want 1 random chance or the best of 99 random chances?
We're starting to see why Monty's actions help us. He's letting us choose between a generic, random choice and a curated, filtered choice. Filtered is better.
But... but... shouldn't two choices mean a 50-50 chance?
>tldr:
THE CONFUSION RESULTS FROM PEOPLE NOT UNDERSTANDING THAT MONTY IS REVEALING NEW INFORMATION, WHICH IS IMPLICIT (AND ACTUALLY VAGUE AND NOT AT ALL GUARANTEED) BY THE PHRASING OF THE GAME.
THE BIGGEST RETARDS ARE ACTUALLY THE BANDWAGONERS WHO JUMP ON THE "IT'S SIMPLE STATISTICS, BROS!" WHICH IS NOT WHAT THE PROBLEM IS ABOUT AT ALL. IT'S ABOUT INFORMATION; ONLY TRIVIALLY ABOUT MATH.
IF YOU ARE ONE OF THE "IT'S SIMPLE STATISTICS BROS!" PEOPLE, CONGRATS, YOUR BRAIN IS INCAPABLE OF TRACING THE LEAP OF LOGIC YOU MADE UNDER PRESSURE CONFORM BY RECOGNIZING WHAT YOU PERCEIVED TO BE THE "SMARTER" SOLUTION. YOU ARE ACTUALLY AN IDIOT! ENJOY.
This always used to fuck with me, it took me awhile to understand, but there's a video around that basically illustrates the same question but with 100 doors instead of 3, say you pick door 42, the host eliminates 98 of the doors leaving door 63 and the door you picked, door 42, which is more likely, the door you picked, or the random door that wasn't eliminated by the host?
>>25072549
This. It's the Monty Hall problem. We were presented this question in my math class freshman year of high school. It sounds confusing, but it's really just that your chance of guessing correctly the first time is lesser than your chance of guessing correctly with fewer options.
https://www.youtube.com/watch?v=mhlc7peGlGg
>>25072608
it becomes so obvious this way, yet with only 1 option eliminated it's still so unintuitive
Okay. Here's a twist:
Regular Monty Hall procedure, 3 doors 2 goats 1 car and so on. Now switch out the player for some random person who wasn't paying attention to what was going on. He must pick the final door. What's the probability that the door he picks will contain the car? Then answer why the fuck the probability changes depending on who's playing at a particular moment.
>>25071913
It is cause the host know's where the goat is and will never eliminate it. So if it is in those 2 doors it will never be eliminated
If the host chose randomly then yes it would be 50/50. But of course 1/3 of the time the goat would be chosen by the host
>>25072742
>probability that the door he picks will contain the car
50-50. The random person will choose randomly from the 2/3 door and the 1/3 door. Making it 50-50. Half the time he will pick the one where he is more likely to win, half the time he will pick the one where he is less likely to win
>why the fuck the probability changes depending on who's playing at a particular moment
Because the original person has the knowledge of which is the 1/3 chance and which door is the 2/3 chance
>>25072742
It doesn't matter who was playing. The act of switching to a door that was revealed by the host to be more statistically profitable is what matters.
>>25071913
>Seems like the odds would be 50/50 to me.
the chances of picking the wrong door first time are 2/3
if you pick the wrong door and he shows you the other wrong door, then switching means a win
Of course there is the possibility that he wouldn't ask you to switch if you picked the wrong door because normies are sly pieces of shit. So really this is up to puree chance.
>>25072742
Nicely worded. Subtle enough to seem genuine, but still built around a highly flawed premise.
The probabilities don't change, but the question you're asking did.The Monty Hall problem isn't about a participant's chance of winning. It asks what the chances are for switching versus staying. Regardless of who's playing, switching gives a 2/3 chance of winning, staying give you a 1/3 chance. A random schmuck pulled in at the end who chooses to switch doors will have a 2/3 chance at the car, a mathematician who has intently followed every step of the game and chooses to stay will have a 1/3 chance.
>>25072358
That's different though, if you pick one and he throws away all but one because they are poisoned, the one left is obviously not poisoned or he would of tossed it out
>>25072742
The entire problem changes that way. The reason that the probability is 2/3 in the original problem is because you have an initial 2/3 chance of picking the wrong door, and therefore the same 2/3 chance of picking the right door if you change your choice.
I'm a bit confused about your scenario. If it were one guy who chose 1 out of 3 doors and he had a 2/3 chance of picking the wrong door, and Monty opened one of the other two remaining doors, then it would be better for him to switch. If he subbed in a friend at that moment and asked which door he should choose, and his friend knew that he had previously chosen 1 of 3 doors, then he should still switch.
If it's just a random guy given the option to choose one of 2 doors with no background knowledge, it's still better for him to switch from the original guy's choice given the situation, but since he doesn't know what the guy's previous choice was then he can't make that decision. He has a 50% chance because he doesn't know what door the other guy chose before him.
Another way to explain with only 3 doors is this.
You have a 1/3 chance to pick the car.
Once you pick, say door 1, host has 2 doors to pick from.
The situations that could happen are as such
2 has a goat, 3 does not
2 does not have a goat, 3 does
both 2 and 3 have goats
Now we have to find the probability of each outcome happening.
both 2 and 3 having goats means 1 does not have a goat, so we can find that probability with
1/3
the others are joint probabilities.
to find the probability that 2 does not have a goat, given 3 does
firstly, there are only 3 situations that can ever happen. either 1, 2, or 3 do not have a goat. order does not matter, so each has a 1/3 chance.
Then we can use the formula
P(A|B) = P(AnB)/P(B)
= (1/2)*(2/3)/(2/3)
=1/2
The same is true for 3 not having a goat given 2 does
This means you have a 1/2 chance of getting the car if you switch and 1/3 if you don't. It is simple stats, because the whole information thing (kolmogorove eq) quantifies the new information, and is one of the first things you learn in a stats class.
>>25072261
>>25072742
>2015
>not being Bayesian
>>25073115
>"It is simple stats"
>distribution does not sum to 1
bad news
>>25071913
>Probability
>Testable
>Definable
>Real
Lmao
>>25073160
The sum of events that can happen does add to one. There are only three events. 1 has the car, 2 has the car, and 3 has the car. Each has a probability of 1/3. That adds to one. Your choices and the chances you have have no bearing on that.
>>25073216
>This means you have a 1/2 chance of getting the car if you switch and 1/3 if you don't.
These should add to 1. It's 2/3 if you switch.
No, I'm not going to locate your error for you because your post was too informal and I've no idea what refers to what.
no its not to your advantage to switch your choice. the odds are always the same. this is like trying to predict a roulette
>>25073338
If Monty revealed a goat before you made a choice, it would be 50/50. Since you did, and he can't open the door you chose, you have constrained his choice. This gives you more information. In the 2/3 chance that you originally picked a goat, he has no choice at all.
>>25072608
This guy gets it.
While it's basis is in statistics, it's more probability and information theory based, like error correction and decoding.
Put it this way, say get a random string of bits, but in this string is one bit that is on and all the others are off. However, you don't know which bit is the bit that is on and which ones aren't.
Now, let's say this string of bits is of arbitrary size n. Your initial pick has probability of 1/n of being correct. Now Monty reveals n - 2 bits that are off leaving you with your bit with initial probability 1/n and one other door that has optimally been chosen out n - 1 bits.
Essentially, Monty has filtered out irrelevant noise leaving you with your random choice and a filtered choice. Obviously you're going to want the choice that has been filtered and is therefore optimal
Basically, and correct me if I'm wrong, you're now left with a bit that is either on or off, probability of 1/2, or a bit that has a probability of 1/n of being off since it was not chosen optimally, but the other bit was. You want the one that has the higher probability of being correct.
So, technically, it is still 50/50 if you choose the other bit, but that has a higher probability of being the correct choice than the bit you initially chose because it was chosen at random.
>>25073513
No. The probability that you find the 1 bit, over all choices you have left, has to sum to 1. (It's a probability distribution: probability the 1 bit is in a particular location. Distributions have to sum to 1 over the entire space. The space is now constrained to the choices you have left.)
It's 1/n if you stay, and (n-1)/n if you switch.
Also, don't be fooled into thinking the original 1/n probability has to remain fixed. It doesn't. It just so happens in this case the new information doesn't change it.
Consider this: once there's only one choice left (you examine all but one possibility), you know where the 1 bit is. That means every location has probability 0 or 1 - it's known.
>>25071913
50/50 when you only consider the two doors, but this game is based on a 1 in 3 door chance so you have to include the third door from the start.
>>25071913
you need to think of it as being done 3 times, only then will it make sense
If it is played three times, in two out of three scenarios it will work - 60% chance that youre better off if you change
