Women, math, and the Monty Hall problem

[We Never Know]

It's also false. The odds remain 50/50 regardless what door you choose, or what door Monty opens.

The 3 doors are just theater because one is eliminated before you choose. And the one eliminated is always a goat. So the proposition was always 2 doors, and one with a car. This is why there is no mathematical way of proving your odds increased. Because there was no change in the odds. It was just irrelevant theater. There is only one choice, and two possible outcomes ... therefor 50/50 chance that the outcome will be the car and 50/50 that it will be the goat.

The whole idea rests on...
-first you had 1 of 3 choices
-if you switch doors after a goat is revealed you then had 2 of 3 choices.

What you have to figure out is... Is Monty trying to get you to switch from your door because its a winner or because its a loser.

 

[Debater Slayer]

You fell for the illusion that there was a 1/3 - 2/3 option. There never was.

It's not an illusion that the probability of picking a losing door is 2/3 without any knowledge in the initial scenario where the three doors are closed. I'll elaborate further in my reply to the following post:

I have no doubt that what was demonstrated in 1990 is correct. I'm just going to attempt a slightly different explanation from yours (which may not actually be different), to see what you think.

At the middle point, after Monty reveals the goat, you have two possible assumptions (as the diagram illustrates). If you assume you picked the car, you cannot infer anything from Monty's choice, but that doesn't matter, you keep your original choice as the car must be behind the door you chose originally. On the other hand, if you assume you picked a goat door, then once again your choice is determined as you know that Monty wouldn't reveal the car and revealing the goat tells you that the car has to be behind the the other door. In both cases, once the assumption is made, the second choice gives you no choice, you know where the car is. BUT, of course you don't know which assumption is correct. So, it drops back to the original choice, which is that it's more likely (2/3) that you chose a goat. So logically you choose the second assumption, decide on that basis, and switch.

Yes, I think this is also a pretty good way of explaining it. To amplify the intuitive aspect of the above, let's imagine the same problem but with 50 doors: one with a prize behind it and the rest with nothing behind them. Let's say you initially pick a door without any knowledge about any of the doors, so the probability that you pick the prize is 1/50.

Then Monty opens 48 doors, showing nothing behind them. The one door that is left besides yours is now much more likely to be the winning one in light of this new information; the probability of winning for all of the other 48 doors are now "concentrated," so to speak, in that door. You chose your door when all of the doors still had an equal probability of being the winning one, but now, after 48 of them have turned out to be duds, you're much more likely to win by switching.

 

[Debater Slayer]

I once tried to prove to a friend that switching
doubled the odds. After about a dozen tries,
the proof failed. It was unlikely, but it happened.

I think one neat thing about computer simulations of this specific problem is that they can generate so many attempts that the better odds of winning by switching become clearer as the number of attempts becomes large enough.

 

[Evangelicalhumanist]

Yes, I'm sure we can. But since I gave you the means to test the truth for yourself, and you choose to believe without bothering to try it, you've lost a learning opportunity. I myself try never to miss those.

Here is an actual simulation that you can try for yourself.

Monty Hall Problem Online. Run the monty hall game and simulation , over and over to understand the probability of this problem.

Monty Hall Problem --a free graphical game and simulation to understand this probability problem.

www.mathwarehouse.com

 

[Debater Slayer]

This problem, along with some other examples from mathematics, has long seemed to me a prime example of why intuition can be highly misleading, especially if confidence in it is not tempered by some skepticism about one's assumptions and initial reactions to or impressions about any given topic.

I think "common sense" can sometimes be as inaccurate as it is common, especially regarding subjects that don't necessarily conform to some patterns of reasoning that tend to work well in many other situations.

 

[PureX]

No. That is a misunderstanding of how the system works.

There was no “system”. Just some irrelevant theater before the actual choice.

Monty always knew which door had a goat behind it. Your claim would be true if Monty did not know and sometimes exposed the car by mistake and that did not count. But since Monty knew which door the car was behind he could always show the contestant a goat. Monty did not make his choice until after the contestant did. One's odds are two out of three if one switches because you are getting one extra chance for free. But by sticking with your original choice you would be limiting yourself to your first choice.

Here is another way of thinking of it. You are given your one out of three choices. You are then given a choice to trade your one choice for the other two doors. That is in effect the choice the Monty is given. I need to emphasize that Monty always knows so when he gives you your choice he will always show a goat first. The set up is that Monty always gives a person a choice, and since Monty never reveals a goat we know that he knew where the car was. You are always switching from what was behind one door to what was behind two doors.

 

[Evangelicalhumanist]

There was no “system”. Just some irrelevant theater before the actual choice.

Try the simulation in post #44.

 

[Koldo]

I bring psychology into it because that is what is being described in the video and hypothesis. If I have a choice of two hidden things, one being good and the other not, no amount of mathematical equations changes the fact that I have a one in two chance of choosing the good thing. It doesn't matter how many choices I started with. The only intangible factor is the psychology involved in a game show or similar situations.

Wanna check how it works? I am going to randomly choose a number between 1 and 100. You then pick any number between 1 and 100 too, trying to guess my number. I will then remove 98 incorrect numbers from the pool, leaving only two remaining (one of them will be your first pick). You will be free to either keep your number or switch to the other number left. And then I will tell you whether you guessed it correctly.

I am going to ask you to go through this with me 10 times at the same time. I have already picked 10 numbers, so you can pick your 10 numbers now. If you suspect I might cheat, we can switch roles too.

If you are right about there always being a 50% chance to guess correctly when given the choice between two alternatives, about half of the time you will guess my numbers correctly, no matter if you switch or keep your initial pick.

If however, you are wrong, and I am the one who is right, whenever you switch your pick you will nearly always guess my number correctly and whenever you stick to your initial pick you will nearly always guess my number incorrectly.

 

[Koldo]

There was no “system”. Just some irrelevant theater before the actual choice.

Wanna check how it works? I am going to randomly choose a number between 1 and 100. You then pick any number between 1 and 100 too, trying to guess my number. I will then remove 98 incorrect numbers from the pool, leaving only two remaining (one of them will be your first pick). You will be free to either keep your number or switch to the other number left. And then I will tell you whether you guessed it correctly.

I am going to ask you to go through this with me 10 times at the same time. I have already picked 10 numbers, so you can pick your 10 numbers now. If you suspect I might cheat, we can switch roles too.

If you are right about there always being a 50% chance to guess correctly when given the choice between two alternatives, about half of the time you will guess my numbers correctly, no matter if you switch or keep your initial pick.

If however, you are wrong, and I am the one who is right, whenever you switch your pick you will nearly always guess my number correctly and whenever you stick to your initial pick you will nearly always guess my number incorrectly.

 

[blü 2]

Both the math and extensive simulations demonstrate that the odds double by switching doors after one door is revealed to not contain the prize.

Saying that the odds remain 50-50 is like saying that 1 + 1 = 3; the former is just as factually and experimentally incorrect as the latter, albeit far more intuitive.

But isn't the key, as @PureX said, that the first door eliminated is ALWAYS a NO?

That is, at no stage in the game are the true odds ever 1 in 3.

 

[Evangelicalhumanist]

There was no “system”. Just some irrelevant theater before the actual choice.

The "theatre" is anything but irrelevant, but you've already made up your mind, in spite of what's been presented to you, so I guess you'll never learn this paticular truth. But as I've seen so often, truth is what you're interested in -- being right (even if only in your own mind) is all you care about.

To @Foxfyre, I'll say this about psychology. Monte Hall was certainly using psychology, because he knows that humans prefer to think that they generally make "good choices." Look how many people play lotteries by carefully selecting their own numbers, base on whatever arcane systems or preferences they have. Since the winning numbers are selected at random, none of those systems will ever improve anybody's luck.

So, Hall's assumption that people will prefer to stick with their original choice means he gives away fewer cars and more goats. Hall would be perfectly aware that very few people could break that habit and rely entirely on reason -- because if they did, Hall's method actually gave them an improved chance of winning the car.

 

[PureX]

Perhaps flexible thinking is involved here rather than being dogmatic, especially with having to explain the practical results obtained - which demonstrate the truth of changing. Would you ignore practical demonstrations for anything similar that coincided with theory?

Predictability does not determine reality. Flip a coin 100 times and you will very likely not get a 50/50 result. Yet that mathematical prediction will be the most logically accurate prediction of every single coin flip: one or a million of them.

As to the Monty thing, the theater of it creates predictions in people’s heads that do not effect the actual outcome of the choice. But a lot of people cannot recognize that what is happening in their heads is not what is actually is happening. So they think the theater is ‘real’. Humans also really like to imagine that they possess some insight into future events that they don’t actually possess. And will not let go if this fantasy easily. Casinos and TV game shows thrive because of it.

 

[Evangelicalhumanist]

But isn't the key, as @PureX said, that the first door eliminated is ALWAYS a NO?

That is, at no stage in the game are the true odds ever 1 in 3.

Not at all. The first door "eliminated" is the one chosen by the contestat at random -- and it is perfectly at random, and therefore 1 chance in 3 of having a car behind it (or more importantly, 2 chances in 3 of having a goat behind it).

Hall then eliminates another door behind which he knows for certain is a goat -- but of the two doors he can reveal, the chances of a car are 2 in 3. That does not change when he reveals one of them -- it just "concentrates" (as @Debater Slayer put it) the chances behind the door Hall does not reveal. That is to say, those 2 chances in 3 are now concentrated in the unrevealed door.

 

[Debater Slayer]

But isn't the key, as @PureX said, that the first door eliminated is ALWAYS a NO?

That is, at no stage in the game are the true odds ever 1 in 3.

The probability of winning when you choose the door is 1/3, because all of the doors are closed and you know nothing to improve your chances of picking the winning door. The act of opening the door comes after your decision, so your first decision is made without that knowledge.

If the game presented you with only two doors from the beginning, the odds would truly be 50-50, but the fact that you have to pick one of three closed doors changes the probabilities. The circumstances of the initial situation in which you pick one of the three doors are the key difference.

 

[PureX]

But isn't the key, as @PureX said, that the first door eliminated is ALWAYS a NO?

That is, at no stage in the game are the true odds ever 1 in 3.

Also, there is never a consequences of the 1 of 3 door option. So it is simply an irrelevant bit of theater.

 

[blü 2]

Not at all. The first door "eliminated" is the one chosen by the contestant at random -- and it is perfectly at random, and therefore 1 chance in 3 of having a car behind it (or more importantly, 2 chances in 3 of having a goat behind it).

Hmm.

Hmm.



Hmm.

 

[Evangelicalhumanist]

Hmm.

Hmm.



Hmm.

Okay, let's try it one post at a time. There are 3 doors, A, B and C. Behind one of them is a car, and behind the other 2 is a goat.

You have to pick a door -- so please tell us which door you've chosen, and what you think the chances are that it has a goat behind it.

(To prevent my cheating, there's a way to show how I chose which door to put the car behind, but I'm keeping it secret for now.)

 

[Foxfyre]

Wanna check how it works? I am going to randomly choose a number between 1 and 100. You then pick any number between 1 and 100 too, trying to guess my number. I will then remove 98 incorrect numbers from the pool, leaving only two remaining (one of them will be your first pick). You will be free to either keep your number or switch to the other number left. And then I will tell you whether you guessed it correctly.

I am going to ask you to go through this with me 10 times at the same time. I have already picked 10 numbers, so you can pick your 10 numbers now. If you suspect I might cheat, we can switch roles too.

If you are right about there always being a 50% chance to guess correctly when given the choice between two alternatives, about half of the time you will guess my numbers correctly, no matter if you switch or keep your initial pick.

If however, you are wrong, and I am the one who is right, whenever you switch your pick you will nearly always guess my number correctly and whenever you stick to your initial pick you will nearly always guess my number incorrectly.

I think the fallacy of your test is that with 50-50 odds you should win 50% of the time. With 50-50 odds, any person may choose rightly or wrongly 100% of the time in many guesses. But for each guess the odds remain the same. If No. 1 is a good choice and No. 2 is a bad choice I have a 50-50 or 1 in 2 chance of making a good choice. No matter how many times I get to choose between a #1 and a #2, the odds remain the same. Again the Monty Hall Paradox is based on observations that people in a given scenario will behave in a particular way. But it does not change the odds of the choice to be made.

 

[Evangelicalhumanist]

I think the fallacy of your test is that with 50-50 odds you should win 50% of the time. With 50-50 odds, any person may choose rightly or wrongly 100% of the time in many guesses. But for each guess the odds remain the same. If No. 1 is a good choice and No. 2 is a bad choice I have a 50-50 or 1 in 2 chance of making a good choice. No matter how many times I get to choose between a #1 and a #2, the odds remain the same. Again the Monty Hall Paradox is based on observations that people in a given scenario will behave in a particular way. But it does not change the odds of the choice to be made.

But the Monte Hall problem is NOT a choice between 2 things, so at no time in the game are the odds ever 50-50. That is what you are not paying attention to.

The fact that Hall eliminates one of the 3 choices part way through the game at no time ever makes this a binary choice -- it always remains trinary, but you are pretending that your first choice is past and no longer has anything to do with the game, when you are left with only 2. But those 2 are still -- like it or not -- out of 3.

 

[Watchmen]

Try the simulation in post #44.

I don’t think he will. He’s too dug in.