Instigator / Pro
14
1500
rating
12
debates
50.0%
won
Topic
#5230
The solution to the Monty Hall problem is to switch, it is not a 50% chance
Status
Finished
The debate is finished. The distribution of the voting points and the winner are presented below.
Winner & statistics
After 2 votes and with 10 points ahead, the winner is...
Tickbeat
Parameters
- Publication date
- Last updated date
- Type
- Standard
- Number of rounds
- 5
- Time for argument
- Three days
- Max argument characters
- 10,000
- Voting period
- One week
- Point system
- Multiple criterions
- Voting system
- Open
Contender / Con
4
1264
rating
363
debates
39.81%
won
Description
There are three doors. Behind two of them is a goat. But behind one of them is $1,000,000 of cash. But you don't know what is behind which doors, so you pick one at random.
Then the judge says, "Now, I won't show you what's behind your door, but I will show the other door that has a goat behind it that you didn't pick." He opens a door that you didn't pick, and there is, in fact, a goat behind it.
Now you are presented a choice: do you want to stick with your choice of door, or switch your choice to the other door that the judge didn't open?
This is known as the Monty Hall problem, and it was a real scenario taking the form of a game show. And most people would assume that it doesn't matter, because you have a mere 50/50 chance. But I am here to show you how that's wrong. If you think that you do in fact have a 50% chance of picking the money after Monty Hall eliminates the other goat door that you didn't pick, consider joining this debate to tell me why you think that, because you're wrong. Good luck!
Round 1
Let me introduce you to this diagram:
Door orientations: Goat, Goat, Money.
Choice 1:
Door 1
Host reveals goat behind door 2
Outcome 1:
You stick with your choice.
GOAT
Outcome 2:
You switch your choice.
MONEY
Choice 2:
Door 2
Host reveals goat behind door 1
Outcome 1:
You stick with your choice.
GOAT
Outcome 2:
You switch your choice.
MONEY
Choice 3:
Door 3
Host reveals goat behind door 1
Outcome 1:
You stick with your choice.
MONEY
Outcome 2:
You switch your choice.
GOAT
Here's what this diagram means:
This diagram lists the three possible choices you could have made, what the host reveals based on your choice, and then the two possible outcomes based on that, that being keep, or switch.
Now let's tally up all the outcomes of keep, and all the outcomes of switch:
Keep:
GOAT
GOAT
MONEY
Switch:
MONEY
MONEY
GOAT
As we can see, a higher percentage of the outcomes that resulted from switching lead to you ultimately choosing money than the keep. I had debated my dad about this for quite a while, and he was convinced that there is a mere 50% chance, until I showed him this diagram. So, let me explain why this diagram is true.
Initially, when you pick a door at random, there is a 2/3 chance that you picked a door with a goat behind it. This is because there are two goat doors, and one cash door. So, at this moment, there is a higher likelihood that you have chosen a goat door.
Then, the host reveals the door that you didn't pick which had a goat behind it. Now what? Well, what happens when you switch? Well, because the host has revealed the other goat door, this means that if you were to switch, whatever is behind your choice of door is always going to flip, no matter what. If you had happened to pick a goat before, and you switched, then that automatically means that you will switch to money, and vise versa. This means that in every possible switching scenario, every time you picked a goat door, you switched to a money door. And, every time you picked a money door, you switched to a goat door.
So, when you switch, your odds are inverted. Initially, there is a 2/3 chance that you picked a goat door, and a 1/3 chance that you picked a money door. So, in two of those scenarios, you accidentally picked goat. Therefore, in two of those scenarios, you switched to money.
Contrast that with when you don't switch. Initially, there is a 2/3 chance that you picked a goat door, and a 1/3 chance that you picked a money door. So, in two of those scenarios, you accidentally picked goat. But since you didn't switch, this means the outcome is exactly the same as your first odds, meaning that in two of those scenarios, your outcome is goat, and in only one of them is your outcome money.
When you switch, then when you pick goat, you will always switch to money. And when you pick money, you will always switch to goat. So, in all of the switch scenarios, all of the scenarios in which you initially picked goat, you ended in money, and in all of the scenarios in which you initially picked money, you ended in goat. And since there are more scenarios where you initially picked goat, because there are two goat doors and only one money door, this means that there are more scenarios in which you end up with money if you switch.
Thanks for starting the debate.
Your argument is wrong because you only presented 3 options, when there are 4 options possible.
Since the correct doors do not change regardless of what we choose,
We can easily solve the puzzle by this simple example done by deduction of options.
Door 1 are correct doors. There are 3 doors. Doors 2 and 3 are incorrect doors.
Option first:
Door 1 picked, Door 2 open, Door 3 remains
Switch = bad
Option second:
Door 1 picked, Door 3 open, Door 2 remains
Switch = bad
Option third:
Door 2 picked, Door 3 open, Door 1 remains
Swich = good
Option fourth:
Door 3 picked, Door 2 open, Door 1 remains
Swich = good
We see that by considering all possible options, the number of times swich is good is same as number of times swich is bad.
This mathematical statistic method, also known as probability by number of all equally possible options, demonstrates equal probability.
Assuming equal number of selections of each option, and equal number of each option happening, which must be when correctly calculating probability of selection,
We see that number of times switch is bad is same as number of times switch is good.
Any deviation from this math can only be due to:
1. Correct doors changing, which is not a possibility in this case.
We know that correct doors stay correct regardless of our choice.
2. Statistical deviation, which happens even in cases where chances are 50 50.
For example, you can toss a coin 10 times and get heads 9 times out of 10, despite that chances are 50 50.
Thus, no simulation cannot account probability, nor can examples.
3. What my opponent did:
Failure to account for the fact that choosing correct door the first time has 2 possible results, not just one, while choosing one of the incorrect doors at start has only one possible result, apparently the other incorrect door opening.
My opponent might assume that you have 66% chances to choose incorrect door, and that, therefore switch is 66% likely to be correct.
However, this is just an assumption that probability is something which remains the same when number of options reduce.
It is like this.
There are 3 doors. I tell you that door 1 are wrong.
So either door 2 or door 3 are correct.
When number of options reduce, so does probability.
While the probability of choosing right doors is 1/3 when there are 3 possible doors, when the number of options is reduced to 2, probability changes with that.
When choosing between two options, both equally possible, possibility cannot be 66% for one.
And the door number 2 or 3 never carry 66% probability of any kind, so to assign any of them 66% probability is a mathematical error.
It is true that when there are 3 doors, 2 of which are wrong, you have 66% chance to pick one of the wrong doors.
But the second time you are choosing, you are choosing between 2 doors, not 3, so same probability cannot apply.
In fact, during second choice, there are only 2 doors, one correct and one wrong.
There is no reason to believe that with equal number of selections for both doors would yield something different than 50 50.
It is like this.
You realized door 1 are incorrect.
You have door 2 and door 3 left.
One of these is correct, the other is wrong.
If you selected door 2 fifty times, and door 3 also fifty times, you would be right 50% of the time.
Thus, with this simple logic, we see that chances are indeed equal.
Round 2
You have made an error that may or may not have been intentional in your math.
Look at option 1 and 2 in his diagram. Notice the difference between them? No, you don't, because they are exactly the same. He listed the same choice twice to make it look like there are four choices.
No, there are not four choices, there are only three, because there are only three doors.
And, your example of Monty Hall showing you one goat door and then you choosing a door is also flawed. You see, this is fundamentally different from the actual Monty Hall problem, as your choice was made before the host showed you the goat door, not after.
And, I have already explained why my diagram works, and as you can clearly see, it goes through the three possible choices you could have made, and the two possible outcomes for each of those choices, giving a total of six possible outcomes, and more outcomes ended in money when you switched versus when you kept.
How can you explain this? This is a direct proof, so the only way he can continue with his argument is by trying to discredit my proof, because otherwise, it is still proof.
So what was his reason?
"Failure to account for the fact that choosing correct door the first time has 2 possible results, not just one, while choosing one of the incorrect doors at start has only one possible result, apparently the other incorrect door opening."
Sorry, but this is straight up a lie. I did in fact account for the two possible outcomes per choice. When you look at the diagram, you can see that each choice has sections labeled "Outcome 1" and "Outcome 2." These are the two possible outcomes that you can have based on your choice, being keep, or switch. Outcome 1 is always the keep outcome, and outcome 2 is always the switch outcome.
So sorry, but I did account for the fact that choosing the correct door the first time has two possible results. Anyone with eyes can clearly look at choice 3 in the diagram, and see that it does in fact branch off into two outcomes, and only in choice 3 does outcome 2 (switch) end in goat. But for the rest of the choices, outcome 2 ends in money, which is why it's more favorable.
I have now directly called out both of your arguments against my proof, therefore, if you want to validate your claim, you must first prove that my proof is wrong, and I will tell you why it is in fact a proof.
No, there are not four choices, because there are only three doors. And in your attempt to list the supposed four choices, you listed the same choice twice.
My opponent misunderstood the choices.
When you select correct doors, which are doors 1, there are 2 possible following results: either door 2 open either door 3 open.
My opponent presented this as just 1 result, when in fact its 2 different results which must be taken into account, as door 2 are not same as door 3.
If they were treated as same, they wouldnt be 2 doors but one.
My opponent says "No, there are not four choices, there are only three, because there are only three doors.".
Number of options isnt equal to number of doors, or its components.
This is because when choosing correct doors, correct doors carry 2 options, not one.
My opponent wants to say that if we choose correct door 1, it is same option if fake doors 2 or fake doors 3 open.
However, this is not the same, as door 2 and door 3 are different doors, which my opponent knows because he treats them as 66% of all doors.
If 2 and 3 were same thing, then 2 and 3 would be one and we would just have 1 incorrect door during first choice.
But because they are separate doors, each option must be treated separately.
We see by simple logic that choosing correct doors, doors 1, has 2 possible results: door 2 opening or door 3 opening.
Where choosing incorrect doors, door 2, has only one possible result, door 3 opening.
And choosing incorrect door 3 also has 1 possible result, door 2 opening.
So total amount of statistical possibilities before second choice is 4, not 3.
My opponent also says that choice is made only before goat door are shown, but this ignores his own description, in which you have ability to choose again after the number of doors was reduced to two.
"Choosing to switch" is just a different name for choice, where you choose between two doors.
My opponent again claims that his diagram works, but his diagram doesnt include 2 possibilities of door opening when the first choice results in correct doors, yet includes 2 possibilities when incorrect doors are first choice.
So his statistical error treats two fake doors as one in one case, but in other case it treats them as two doors again, which is why his diagram is statistically incomplete, as when correct doors are selected, the judge has 2 options, to open fake door 2 or fake door 3, and lacks those options when incorrect doors are selected.
His diagram also assumes that probability of 1 incorrect door becomes 66% when other incorrect doors are excluded, but this is clearly false as same amount of selection of both remaining doors always yields 50% chance, something my opponent conceded.
My opponent says that he accounted for two possibilities of correct doors being open, but from his diagram we see that one option is clearly lacking.
"Choice 3:
Door 3
Host reveals goat behind door 1
Outcome 1:
You stick with your choice.
MONEY
Outcome 2:
You switch your choice.
GOAT"
Possibility which my opponent didnt include:
"Choice 3:
Door 3
Host reveals goat behind door 2
Outcome 1:
You stick with your choice.
MONEY
Outcome 2:
You switch your choice.
GOAT"
This possibility is unaccounted for in opponent's diagram, it is not even mentioned.
He ignored the fact that choosing correct doors has 2 possible outcomes for judge(host), to open door 1 or door 2.
My opponent accounted for 3 possibilities, and excluded 1 possibility, therefore creating a mathematical error of excluded possibility which ignores that judge(host) makes a choice as well, and his 2 possible choices add to the statistical result.
If being presented with 3 doors creates 1/3 chance, then judge being presented with 2 doors creates additional 1/2 chance which affects statistical possibilities.
Further, my opponent cannot account for how in his version of math, after one door is excluded, the other doors somehow inherit the 66% probability.
In all statistics, when number of equally correct options reduce from 3 to 2, the two options remaining still remain equally correct.
So if 3 options each have 1/3 chance, excluding one does not yield 2/3 chance to the other, but 1/2 chance to the remaining.
My opponent's case is built on an assumption that probability of each option remains same when number of equally likely options reduce.
We can see with simple example that this is false.
If I have 3 boxes which all have 1/3 chances of being correct, and only 1 is correct,
Removing 1 incorrect box leaves only 2 boxes.
When choosing between those two boxes, each has 1/2 chance of being right, because when third box was removed, the probability of each box changed.
Further, in my opponent's diagram, it is clear that as soon as you make the first choice, number of choices reduces to two doors.
Out of those two doors, one correct and one incorrect, the chances are obviously 50%.
My opponent defends that the door you chose holds 66% chance of being incorrect.
That is only true when there are 3 doors, since his first probability calculation includes 3 doors.
When number of doors reduce, same probability cannot apply.
Since my opponent's diagram involves a reduction of options from 3 to 2,
we see that no matter which incorrect door we choose, the other incorrect doors get removed from equation, leaving only 1 correct and 1 incorrect door.
My opponent cannot account for the fact that 2 separate incorrect doors are treated as one in his diagram in post results and in pre results.
In fact, after one incorrect door is excluded, my opponent treats the remaining incorrect door as 66% probability for no reason.
My opponent's case rests on these assumptions:
1. Reduction of options doesnt affect probability (absurd).
2. Two possibilities when correct doors are chosen, my opponent treats as one possibility, despite that there are clearly two possibilities of judge(host) to open either first incorrect door or the second incorrect door, each having 50% probability on their own (Statistical error).
3. That when one incorrect doors are excluded, the other incorrect doors somehow have same probability as when there were 3 doors.
These assumptions are clearly false, as when choosing 1 out of 3 doors, there is only 1/3 possibility per door, and removing one doors does not give the other 2/3 possibility.
Thats because when observed step by step, removing one incorrect doors means that now there is 50% chance that you chose the right door during first selection.
Round 3
Here is a diagram taking into consideration your new argument:
Goat, Goat, Money
Choice 1:
Door 1
Host reveals goat behind door 2
Outcome 1:
You stick with your choice.
GOAT
Outcome 2:
You switch your choice.
MONEY
Choice 2:
Door 2
Host reveals goat behind door 1
Outcome 1:
You stick with your choice.
GOAT
Outcome 2:
You switch your choice.
MONEY
Choice 3:
Door 3
Host reveals goat behind door 1
Outcome 1:
You stick with your choice.
MONEY
Outcome 2:
You switch your choice.
GOAT
Host reveals goat behind door 2
Outcome 3:
You stick with your choice.
MONEY
Outcome 4:
You switch your choice.
GOAT
Now let's tally it up:
Keep:
GOAT
GOAT
MONEY
MONEY
GOAT
MONEY
MONEY
Switch:
MONEY
MONEY
GOAT
GOAT
MONEY
GOAT
GOAT
It appears as though they are equal. So, case closed, right? Wrong. Here is why it is invalid to use this as proof to say it is 50/50, and it's going to be difficult to put into words, but I'll try my best:
Because there are only three doors, there are only three choices you can make. But in the scenario in which you pick money, the host can reveal one of two doors this time. This means that there are now four possible outcomes.
The reason this doesn't work is because this doesn't increase your likelihood of making the right decision. Your odds are based on two things alone: your initial choice, and your final choice. Which door the judge decides to open does not at all change what your outcome is going to be. And if something is gauranteed to not change your outcome, then it won't affect your odds. This means that your likelihood of having chosen correctly are still the exact same as before. This is why I didn't include this in my previous diagram, as it was not relevant.
Here is once again an explanation of why it's better to switch:
You are more statistically likely to pick goat than to pick money. And when you switch, you always flip your choice, no matter what. So, every time you pick goat, you switch to money. And every time you pick money, you switch to goat.
There are more scenarios in which you pick goat and switch to money than pick money and switch to goat.
In a switching scenario, if you pick goat, you end in money.
You are more likely to pick goat.
Therefore, you are more likely to end in money.
Your argument suffers from its own premise.
How do you know that the probability of choosing incorrect doors is 66%?
Because
Premise 1: there are 3 doors, 2 of which are incorrect.
However, when one of the incorrect doors are excluded, premise 1 no longer applies as there are no 3 doors anymore, but two.
So you cannot prove that chances dont change when number of doors change, and it is obvious that 2 doors, one correct and one incorrect, produce 1/2 chances.
Further, in your counter argument, you assume equal number of selection of each door.
You assume if 10 people choose correct door 1, 10 people would choose incorrect door 2, and 10 people would choose incorrect door 3.
However, thats an assumption.
With just slight deviation,
For example, 15 people choosing correct door the first time and 15 other choosing one of two incorrect doors, it brings probability to 50%.
Your case cannot account for deviation nor predict when deviation happens.
Your argument, as demonstrated before, suffers from whats in probabilities is known as "conjunction fallacy".
Apparently, there are only two possible options:
1. Real pattern follows theoretical probability
Or
2. Real pattern doesnt follow theoretical probability
Since your entire argument is in conjunction with 1 and depends on 1 being true in order for your side to be true, 1 must be true for your side to be true.
It follows that probability of your case is probability of 1, and cannot be more than probability of 1, because conjunction laws say that two things in conjunction must both be true for conjunction to be realized, and the probability of conjunction cannot be higher than probability of any one part of conjunction.
If conjunction says "A and B", if A has 10% chance, conjunction cannot have more than 10% chance.
What is the probability of 1 in your conjunction?
Virtually impossible when odds are too similar.
For example, if you tossed a coin 100 times, there is only a small chance you would get 50 heads and 50 tails.
Despite that theoretical probability says that its 50 50, real pattern rarely matches theoretical probability.
Many gamblers lose a fortune because they think that 66% probability means that such pattern will be achieved in real life in their case.
In reality, being 66% likely to succeed theoretically can easily translate into 0%, 30%, 50%...
Think of it this way.
If you toss a coin 100 times, it could happen that you get 80 heads and 20 tails.
Theoretical probability cannot account for real life circumstances, as complete randomness is virtually impossible, and even complete randomness often doesnt agree with theoretical predictions in limited number of experiments.
Since you cannot prove that your real life case follows theoretical probability of 66% and that it is its solution, using theoretical probability is not proven to be correct in real life case.
It would be the same if you made the topic:
"Tossing coin 100 times will yield 50 heads and 50 tails".
In overwhelming majority of cases, you would be wrong, despite that theoretical probability says that chances are 50%.
So how is it possible that probability doesnt account for real life cases?
Because person's choice and circumstances all affect probability, and they cannot be accounted for.
To put it simply, just because only 1 out of 3 doors are correct, you cannot prove that person will select incorrect doors 2 out of 3 times.
Theoretically, it should happen, but real examples almost never follow theoretical probability.
You also assume that person's choice follows random pattern, but it doesnt.
Some people are more likely to pick correct doors out of 3, so solution for them is obviously not to switch.
Some people are likely to be a part of pattern which favors correct doors in first choice, making switch undesirable.
In conclusion, you cannot prove that theoretical probability follows real life pattern in a way which would favor your case, and since there are obviously patterns which oppose your case, it would be nonsense to say that in those patterns the solution is to switch.
Your case rests upon assumption that each doors will be selected equal amount of times, because only equal amount of selection yields your theoretical results.
But since you only get 1 chance at game, not 3 chances, there is no way for you to know that choosing to switch will benefit you, as the odds of your real pattern following theoretical pattern are low.
To believe that your one chance will follow theoretical probability is an assumption.
It obviously wont be true in 34% of cases in theory, and in practice it might fail even in 50% of cases sometimes.
So you cannot prove that theoretical chances are what will happen to you for sure.
"The solution to the Monty Hall problem is to switch" depends upon real patterns following theoretical probability, something which your case cannot really predict since the difference between odds is too small for there to be a consistent real pattern, and many patterns will deviate from your theoretical predictions and some will even have 50% pattern, proving that chances are indeed 50% in that pattern.
Round 4
I am not saying that switching is a guarantee that you are going to win the money, and real life results can vary, precisely because it is random.
What I'm saying is that when you switch, you have a higher probability of winning than if you didn't switch.
And, I didn't think I'd have to explain this again, but the reason is because you when you switch, your choice is reversed. If you picked goat, you will always switch to money. And if you pick money, you will always switch to goat. This means that in every scenario in which you pick goat, you switch to money. And since you have a higher probability to pick goat, this means you have a higher probability to end up with money, if you switch.
When you use the argument stating that having a choice between two options is 50/50, you have a fundamental misunderstanding of how this works. Your odds are first defined at the BEGINNING, and there, you have a 2/3 chance of picking goat. Therefore you have a 2/3 chance of switching to money instead of goat.
The funny thing is, I predicted that you were going to create scenarios in which you have only two choices from the very beginning, and compare them to the Monty Hall problem, before we even got started. So if you keep using those kinds of arguments, I'm going to keep responding to them the exact same way, but try to explain it slightly better so I can get it through to you.
Since your side of the topic is a mathematical truism, I feel like I am supposed to just concede.
So yeah, I concede.
You win! Your side of the topic is correct.
Round 5
Since the debate is basically over now, I am posting this so that the end can pass by more quickly, instead of having to wait three days for me to forfeit.
Good debate.
Good debate 😁
GG.
I cant always debate on the side I agree with.
Sometimes one must change sides to make it more interesting.
Well yeah, but I thought the point was that if you agree with the premise of the debate, then don't join it.
Its a debate.
I am supposed to disagree, even if I dont believe in what I am saying.
You cant really have a debate unless we take opposite sides in topic and unless we disagree.
So the point is to disagree, so to say.
Of course, I do plan to concede in next round, as your side of the topic is mathematical truism.
It is so hard to explain this to you 💀
Eh, arguing against truisms is not easy.
I forgot to put this in my debate argument, but in addressing your response to me pointing out the flaws in your example:
In a scenario in which the host reveals the goat door and THEN you choose your door, the difference between that and the Monty Hall problem is that you never had the opportunity to choose before the host revealed the goat behind one of them. When he reveals the goat behind one of them before you choose, he has now eliminated one of the doors you could have chosen.
In the actual Monty Hall problem, all three doors are available for you to choose, because the host didn't reveal the goat door until after you chose.
That's the difference between the Monty Hall problem, and the scenario you invented, in which there are now only two initial choices, instead of three initial choices like in the actual Monty Hall problem.
I am certain that switching doesnt make more sense.
I was just stating the opinion I found on google.
On google, anyone can write any assumption.
But after doing the mathematical calculation myself, I realized that guy on google is wrong.
Oh, I didn't even realize you were my opponent. You seemed to agree with the fact that switching makes more sense. Why did you join this debate?
Given that I am your opponent, revealing it in comments is not really "revealing" of any kind to your opponent.
Well don't reveal it all in the comments...
Yeah, apparently,
With 2 fake doors and 1 proper door,
You have 66% chance to pick fake doors during first choice.
So when given second choice, the doors you first picked are 66% likely to be fake doors, so switching makes more sense.
To be honest, I don't know if anybody is even going to accept this debate, considering I have already devised a proof in Microsoft paint, and it doesn't even involve any math equations.
Apparently this used to be hotly debated even after the mathematical proofs were devised.
There are of course kritiks, but pro is seeking someone who simply disagrees with the soundness of the math when applied in the real world.
I think this is basically a truism if judge has an interest for you to open wrong doors.