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?
Yes. Imagine a 1,000,000 doors with 999,999 goats and one Lamborghini.
You choose a door. The host removes 999,998 of the doors and asks, with that shit-eating grin of his, "Would you like to switch?"
Yes. You switch and probably win the Lamborghini.
Yes. The advice is to switch. There was a 2/3 chance of one of the doors you didn't pick having a car behind it. Those odds are now reduced down to a single door. Meanwhile there was only a 1/3 chance that you had selected the single door with the car. Go with the door that has twice the odds.
Three men stand before me. One is OP, who I want to kill.
I point my uzi at one of the three men. Another of the three men says "I'm not OP," and proves it by not sucking cock.
Now, should I switch my target? My chance of killing OP has switched from 1/3 to 1/2, so of course I should switch.
this used to drive me nuts for hours. at first I too was under the impression that, since the game show is garunteed to remove a goat regardless of which door you pick, it's always 50/50. but this isn't the case.
the fact that he removes a goat door /after/ you choose a door is actually significant and plays into the probability. you chose a door while there was a 1/3 chance of you choosing the car. now odds have it that you picked a goat instead of the car (2/3 chance). now you've got a finger on a door that is most likely a goat. since the host CANNOT open the door you chose, he will reveal what is most likely the other goat (you've most likely chose a goat). now that the other goat is out of play, the probability of the door you chose initially to be a car remains 1/3, but the probability of switching and getting the car becomes 1/2.
1/2 is always better odds than 1/3. this was experimented and proven to be the case many times.
To better explain the problem, Monty Hall ALWAYS opens a goat door before giving you a chance to switch.
When you make your choice, you have a 1/3 chance of being right, and the other doors had a combined 2/3 chance of being right. That extra 1/3 doesn't go away when Monty opens a goat door because he ALWAYS opens a goat door.
So you initial choice retains its 1/3 chance of being right, while the other door has a 2/3 chance of being right. You should switch for better odds of getting the car.
>the fact that he removes a goat door /after/ you choose a door is actually significant and plays into the probability.
again, are you really this stupid?
If you laid your finger on the door with the car, the host could open a goat door
If you laid your finger on the door with the goat, the host could still open a goat door
It literally makes no difference
I am familiar with the Monty Hall Problem. It's invalid.
>1. You are presented with three doors
>2. There is an equal chance that the prize is behind any of them
>3. You pick a door
>4. Another door is opened, but it is never the one with the prize behind it
>5. You are asked to switch or not
>6. The reveal.
It has already invalidated itself by point 4. There is not a 1/3 chance of any given door having the prize behind it because by default the door that is revealed never has the prize behind it. That is contradictory to the second term. Either one of two possibilities must exist:
A. There was never any possibility that it was behind the superfluous door, meaning that you always had 50/50 chances of getting it right, which the Monty Hall Problem does not conclude. This is how the real world works given those parameters.
B. We accept the warping of probability based on the initial statements and work the problem within the given parameters. It says that you should switch doors because fiat has declared a set of conditions that do not naturally exist within a real life experiment, only within simulation.
In my opinion the Monty Hall Problem demonstrates a lack of critical thinking ability among mathematicians and scientists. It creates fictional conditions for a simulation that could not exist within the real world.
you're the stupid one m8.
1. you pick a door. 2/3 chance of goat. host can't open your door.
2. host reveals goat.
3. last door left untouched is 1/2 chance of car, while first door picked is 1/3 chance of car.
Google dude. find the experiments. be proven wrong. I dare you.
>A. There was never any possibility that it was behind the superfluous door, meaning that you always had 50/50 chances of getting it right
a 50/50 chance from choosing among 3 doors? Quit being an idiot. It only becomes a 50/50 chance after one door is removed.
>B. Word soup, the post
The example isn't any different. The goats in this equation are made fungible.
The chance of you picking a car is 1:1000000
The chance of you picking a goat is 999999:1000000
The chance of you picking that SPECIFIC goat, however, is equal to the chance of you picking the car, 1:1000000
By removing 999999 goats, you don't alter the balance of probabilities between the car and that specific goat. It's still a 50/50 chance between the two doors.
All right, let's break it down even more Barney-style, then. For our purposes here, the car is always behind Door 1, but "we don't know that" as the player.
100% Choose to switch no matter what:
>Pick Door 1, Monty opens Door 2 or 3, switch to 3 or 2, GOAT.
>Pick Door 2, Monty opens Door 3, switch to 1, CAR.
>Pick Door 3, Monty opens Door 2, switch to 1, CAR.
100% Choose to stay no matter what:
>Pick Door 1, Monty opens 2 or 3, stay, CAR.
>Pick Door 2, Monty opens 3, stay, GOAT.
>Pick Door 3, Monty opens 2, stay, GOAT.
Switching is 2/3 chance to win, staying is 1/3 chance to win.
Imagine that one goat is red, one goat is blue, and one door has a car,
You pick a door, then the host opens up the door with the red goat behind it. What are the odds that the door you picked has the blue goat?
>Pick Door 1, Monty opens Door 2 switch to 3, GOAT.
>Pick Door 1, Monty opens Door 3 switch to 2, GOAT.
>Pick Door 2, Monty opens Door 3, switch to 1, CAR.
>Pick Door 3, Monty opens Door 2, switch to 1, CAR.
Look what happens when you aren't a retard
So you are really saying you pick one door out of a million, see 999,998 options deliberately removed as wrong, leaving one right and one wrong answer remaining between the door you picked and another door they selected to remain, and think, "Eh, it's just 50/50 then, no real difference if I switch or not."
I'll give you the benefit of the doubt here and assume you're trolling. Even if you weren't, this problem couldn't be laid out any simpler in this thread than it already has.
Hahaha really? The Door 1 initial pick is only one of three, it's irrelevant whether he opens 2 or 3, it's the same thing.
Notice how you don't address the staying option. You already figured it out, you're just being silly now.
The change in probabilities is illusory. You can see this if you change the facts only slightly.
>There are 1000000 doors
>behind each is a pile of money, ranging from $1 to $1000000
>The host tells you to pick the door with $1000000 behind it
>The host then removes every door, containing $2 to $999999
>What are the odds that you picked the door with $1, and what are the odds that you picked the door with $1000000?
The odds of you having initially picked any door are 1/1000000. Doesn't matter if you picked $1, $549,844, or $999,999. If 999,998 doors are removed, leaving $1,000,000 and one other dollar amount as the prize, of course you would switch, because the chance of you picking $1,000,000 was 1/1000000.
Just because choosing the winning door and switching to either losing door has the same loss result doesn't mean one of those doors can be excluded from a stat table. The image contains every possible game set, and there are 12. Only in 6/12 games does switching get a win.
You're mischaracterizing and overcomplicating the argument. It doesn't matter at all what door Monty shows a goat behind. The two goat doors are functionally identical.
You could even consider it a different way. Imagine the host doesn't even reveal a goat. You choose a door, then he asks you if you want to stick with your door, or get what's behind both incident doors. Obviously you take the two doors. It's the exact same thing.
His table is purposely misleading you.
This also shows every possibility and the chances of them happening. It shows the it's better to switch 2/3 of the time.
But that table says its a half chance that a goat is revealed when door 1 is the car.
The truth is it is 100% guaranteed a goat is revealed even if you chose the car. Both other doors are guaranteed to have goats if you chose the car first. 1/2 + 1/2 is 1, it doesn't magically depreciate to 1/6 + 1/6.
Funny thing, in high school I had such difficulty getting my classmates to accept the correct answer (it's always better to switch -- it's an increase from a 1/3 to 2/3 chance) that I wrote a fucking computer simulation that ran through the scenario 20,000 times, switching for the first ten thousand and staying for the second. The little simulated dude won the car 2/3 of the time in the first half, 1/3 of the time in the second half.
>"But anon, that only happened because it was a simulation! The equations might show a 66% chance, but in the real world it would be 50%."
"No, dude, there's no fucking equations in the program. Look here! It literally just plays out the scenario twenty thousand times and stores the results."
>"NO. YOU MUST HAVE DONE SOMETHING WRONG."
I've literally never seen so many smart people reject something when there was literal proof in front of them.
I tried this earlier. The simulation was coded incorrectly and is flawed. The placement of the goats and the car never changes, so playing manually allows you to always win cause the car is always in 1 spot and never moved after multiple tries. The auto-play feature therefore only randomizes the choice, of which 1/3rd of the time it will win because theres only 1 car. It doesnt randomize the location of the goats/car however, which is an important factor to how that particular simulation is flawed. Both location and choice must be randomized to get a proper result.
Also, before somebody points it out, I'm aware I way overused the word 'literally' in this post, and yes I'm a little embarrassed. Still doesn't make it a 50% chance, though, sorry.
No, it's showing that when you pick the door with the car in it, which goat door is revealed is randomly selected. So if the car is always behind door one and you choose it, half the time door 2 is revealed, and half the time door 3 is revealed. That's what op's table is getting wrong.
Is this one coded better? It honestly doesn't matter which simulation you do, the results will be the same.
Which goat is revealed is useless information though. If you stuck with your car choice and a goat is revealed, your final choice is between a car and one goat. If you stuck with your car choice and the other goat is revealed instead, your final choice is between a car and the other goat. That is what op's image shows, that there are two unique sets being played between [car and one goat] and [car and other goat], both sets lose if you switch.
It is in contrast to the table here >>16661830
Which assumes there is only 1 set if your first choice was the car. It removes information from the problem because it simplies 2 events down to 1 event without changing the weight of new 1 event.
Say i choose door 1. It has a car. I dont know that though. Two individual games can be played from here.
1) Goat1 is revealed, and if I switch I get Goat2, a loss.
2) Goat2 is revealed, and if I switch I get Goat1, a loss.
Say I chose Goat1 first though. Only 1 set can be played
1) Goat2 is revealed, and if I switch I get a car, a win
If i chose Goat2 first
1)Goat1 is revealed, if I switch i get a car, a win
There is only 1 chance of winning the car if i chose either goat first and then switch, and the chance of selecting a goat is 2/3.
1×(2/3) = 2/3
There are 2 chances of winning the car if I chose the car first and then stayed, and the chance of selecting the car is 1/3.
2×(1/3) = 2/3
Since two opposing values of 2/3 cancel each other out, the end result is 50/50 that staying or switching will guarantee a win. Both choices have the same chance of being correct.
>Say i choose door 1. It has a car. I dont know that though. Two individual games can be played
Those two individual scenarios will only happen 1/6 of the time. Op's table acts as if they happen the same amount, which is incorrect.
Granted, although they can get so close that for something like this it really is a technicality. You obviously know that, I'm just writing for the crowd.
Also worth noting that based on what little I've read on the subject, actual human choices would almost certainly be vastly less random than the computer's.
Dude, you're just wrong, and people have explained how and why you're wrong several times. There's several simulations available on the web, and if you're at all technologically-minded, it doesn't take that much know-how to code up one yourself.
Run any simulation. Switching doors will net the good result 2/3rds of the time. If you were to test this in reality -- it'd take a lot of patience since you'd have to run the scenario hundreds or thousands of times, but in theory you could have your roommate place candy & rocks in bags or something -- switching bags would net the good result 2/3rds of the time. I fucking promise you.
The way you think statistics work is not how they work in reality, however unintuitive it might seem.
>Say i choose door 1. It has a car. I dont know that though. Two individual games can be played from here.
>1) Goat1 is revealed, and if I switch I get Goat2, a loss.
>2) Goat2 is revealed, and if I switch I get Goat1, a loss.
As anon stated, these scenarios only happen when you originally pick the car. You have 1/3 chance of originally picking the car. Scenario 1 and 2 happen half the time when you pick the car. So (1/3)/2=1/6.
Also, which one of you fuckers keeps editing the Wikipedia page to read "... vos Savant's incorrect response ..." (instead of 'correct response')? Because that's incredibly immature, but also kind of hysterical. I know it's one of you.
I'm not a physicist, and I'm leery of making any authoritative statements outside my field, but I do know that true randomness doesn't actually exist anywhere outside of quantum mechanics. My understanding is that it very much does exist in quantum mechanics, e.g. in radioactive decay, although I'm especially leery of making any pronouncements here, since anything I might say might be met with "haha, that's what you read in the pop-sci press xD but here's the truth: [ten pages of completely random and occasionally paradoxical gibberish]" You'd have to go to a physicist for more, and they'd probably scoff and direct you to a philosopher for a discussion of whether 'does true randomness exist?' is even a meaningful question.
This simulation was coded incorrectly. It uses the function FLOOR which is defined as
>Floor function returns the greatest integer not greater than x
between the values of 0 and 3.
if x is 0.0, it returns 0.
if x is 0.3, it returns 0.
if x is 0.9, it returns 0.
if x is 1.0, it returns 1.
if x is 1.3, it returns 1.
if x is 1.9, it returns 1.
if x is 2.0, it returns 2.
if x is 2.3, it returns 2.
if x is 2.9, it returns 2.
if x is 3, it returns 3
and then nothing cause it doesn't go above 3.0
it's not solving for 3 doors. It's solving for the chance that a number value falls between 0 and 3. Any number greater than or equal to 0 but less than 1 returns 0. Any number greater than or equal to 1 but less than 2 returns 1. Any number greater than or equal to 2 but less than 3 returns 2. A number that is 3 returns 3. Four values: 0, 1, 2, 3
This is important later in the function which calls for a comparison between that value to something the program is intentionally looking for.
Yes. You always switch. The first time you choose there's a 1/3 chance. After you know where one goat is then when you switch your answer there's a 1/2 chance to select the correct door. This is the only correct answer, get on my level.
I'm learning to code right now and am coding it correctly yes. I have experience with coding but not this language, so i'm reading the code and changing the parts to remove floor and properly accomodate for value ranges tied to integers. It's taking some time probably another 30-45 minutes. This board is slow so it should be good.
Math major here.
Everyone calm your horses. Without understanding what a probability actually means you can interpret the answer to be 1:1 or 1:2 depending on how you word it. They are the answers to 2 separate questions and the confusion comes from not recognizing the distinction between the probability of you guess the right door, and the probability of a particular door being the right door.
If you picked at random, the probability of you piking the correct door is 50/50 which is a correct answer. However that is not quite the same as the probability of which door is the correct door. Its all about how much information you have at that point.
In my right hand I have a prize, in my left hand I have nothing. The probability of you picking with no further information that just pick a hand is 50/50. However if you know that the prize is in my right hand (as in I told you) then you now have access to more information. Therefore even though if you were to pick a hand at random it would be 1:1, if you knew the answer the probability is 1:0.
This is what the monty hall problem gets at.
There is a 1/3 probability that the answer is the door you picked. If you would like for me to explain why 1/3 is the answer then I can. I just answered why 1/2 is wrong.
car is behind door 1 for the sake of simplicity
pick door 1, monty opens 2, switch, goat
pick door 1, monty opens 3, switch, goat
pick door 2, monty opens 3, switch, prize
pick door 3, monty opens 2, switch, prize
pick door 1, monty opens 2, stay, prize
pick door 1, monty opens 3, stay, prize
pick door 2, monty opens 3, stay, goat
pick door 3, monty opens 2, stay, goat
am I wrong?
cmon guys everyone knows the monty hall problem. how about you fags leave this hellish inferno for a second and go to wikipedia.
fuck, even OP's picture is straight outta wikipedia.
Monty Hall flips a coin 999,999 times. It lands on heads 333,333 times. It lands on tails 666,666 times.
Monty Hall flips the coin for the 1,000,000th time. What are the odds that it will land on heads?
Yes you are wrong. Youre not taking into consideration indistinguishable cases.
>pick door 1, monty opens 2, switch, goat
>pick door 1, monty opens 3, switch, goat
The door monty opens doesnt effect whether or not you have a better chance of winning if you switch therefore you would only abstract 1 case for the scenario. Same with always stay.
Is the coin flip fair. Then its 50/50. The trend will be towards a 50/50 distribution as it approaches infinity though however 1000000 isnt close to infinity to bet on that.
If you flipped a coin an infinite amount of times and say so far a 33/66 distribution however then you must assume the coin flip is in some way not random and fair and you should with tails.
I've never read such pseudo intellectualism. The Monty Hall problem makes perfect sense.
You pick a door. It's a 1/3 chance of being the car. So it's not likely you picked the car. The host eliminates a goat. Since it wasn't likely you picked the car, you switch.
It's that simple. I'm not gonna explain the numbers to you fucktards cause you won't get it. Also to the guy giving the 1-10000000$ comparison. The probability is still the same and your an idiot. Its not likely at all you choose the max value. So you switch when the host reduces.
That's all the problem is saying.
It's counted as two situations, but they only happen 1/6 of the time.
Opening a random goat door only happens when you originally choose the car. You originally choose the car 1/3 of the time. Opening the first or second goat door both happen 1/2 the time of that 1/3 chance.
>It's counted as two situations
For indistinguishable probabilities, you can count them as separate things and will lead to the same result, however it makes the math a bit worse. Its better to not that its an indistinguishable system in play and simplify.
I'm bad at whatever language this is and have hit a roadblock in rewriting it. Still working on it, i've managed to implement the given example's graphical bar% but i just can't seem to be getting them to actually take the results from the code yet.