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Fair coins tend to land on the same side they started

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Re: Fair coins tend to land on the same side they started

#241

Post by Sohcahtoa82 »

evouga wrote: post #216
tacitusarc wrote: post #186

Earlier quoted context omitted.

For sufficiently analytical folks that works, but for lay people it tends to still be confusing. The best way I’ve heard it explained to help people get it through intuition is by changing the number of doors and goats. Say there are 100 doors, and they all have goats except one, which has a car. You pick door 1. Monty then proceeds to open doors 2 through 48, skips door 49, and then opens the remaining doors. After…

The situation is now counterintuitive in the other direction: if Monty Hall had opened those 48 doors at random and they just happened to not contain the car, then there is no advantage to switching, though many people would insist otherwise.

> if Monty Hall had opened those 48 doors at random

The fact that Monty Hall opens the doors deterministically (not randomly) is KEY.

In the original problem, Monty ALWAYS opens a door with a goat. In using 50 doors, Monty would ALWAYS open doors containing goats, and not the car. It's not random.

Knowing it's not random, it should be very intuitive.

Re: Fair coins tend to land on the same side they started

#242

Post by s1artibartfast »

Natsu wrote: post #238
s1artibartfast wrote: post #231

Earlier quoted context omitted.

I always feel like there is something fundamental missing from the examination of Monty Hall problems. I think it has to do with the difference between "probable outcome in reality" and "probably outcome based on personally known information". Lets say when you get down to doors #1 and #49, Monty brings in someone new, with no information and says pick a door. For that new person, standing right next to you, doors #1…

There is something missing: regular stats don't differentiate between doing things and observing things and these two are not at all the same. If I have a digital thermometer and I observe it to show a high temperature, then I will note an association between that and feeling warm. But if I merely set the thermometer gauge to a high value artificially, it's not going to make me feel any warmer. This ambiguity is reso…

I think it is more fundamental than that, and not even mathematical. I think the issue is that people conflate or blur the difference between reality and their models of reality.

Your personal, information limited calculation of the chance a car is behind door #1 has no impact on if there is a car behind door #1. Reality is binary and constant. There was always a car there, or there always wasn't.

Most people correctly intuit that of course the real probability that the car is behind door #1 cant change with reveled information. It isn't a quantum car. They just get caught up on the fact that predictive chance is a attribute of the model, not the real door.

Re: Fair coins tend to land on the same side they started

#243

Post by bigstrat2003 »

cortesoft wrote: post #195
dfxm12 wrote: post #172

Earlier quoted context omitted.

The Monty Hall problem is especially unintuitive if you've ever watched Let's Make a Deal, since the problem set up is oh so close to, but not exactly, the set up of the Big Deal in the show. It's too easy to conflate the rules of the show with the math problem, which will lead to confusion. I think seeing the results of a simulation also elucidates the set up of the math problem vs reading a proof.

Yeah, I feel like the Monty Hall confusion goes away if you are explicit about the rules: "Hall will always open one of the two non-chosen doors and will never reveal the prize" I think most people who don't understand the problem miss that critical detail.

No, I don't think that detail makes it any easier. I know that but I still really can't accept the correctness of the Monty Hall strategy (I have to basically just take it on faith and stop trying to understand it). I was trying to put my finger on why, and I think it's this.

After Monty eliminates one of the three doors, then the prize is behind one of the two. If someone were to come in this point, with no prior knowledge whatsoever, their chance of picking the correct door at random is 1/2. And that is still true even if they pick the door which our contestant is being asked whether or not to switch from! This is a real mind fuck to try to accept, that the same state of what's behind each door leads to different odds of making a correct random choice, depending on when you make the choice.

I honestly don't think I'll ever be able to "get" the Monty Hall strategy. I think I get why it works (choosing to switch means you're going from a 1/3 probability to 1/2), but it makes no sense at all. It seems like even if you choose to stay on the same door, your probability is 1/2 (the same as if Joe came in off the street and chose the same door as you). Like I said, I just have to take it on faith.

Re: Fair coins tend to land on the same side they started

#244

Post by LoganDark »

4star3star wrote: post #179
cantrevealname wrote: post #128

Earlier quoted context omitted.

> Why would you test it? I recall conversations on Usenet decades ago about the Monty Hall problem[1] in which people gave elementary proofs that probabilities don't change by opening a door. Even from mathematicians and statisticians. People were very insistent that the analytical solution was simple and obvious and that switching doors didn't change anything. The only thing that changed some people's minds was a pr…

The way the problem makes sense to me is this. Doors: Goat Goat Car You pick a door. Monty shows you a Goat. You switch or stay. Monty will never show you the Car before offering a switch. He always shows you a Goat. It doesn't matter which Goat he shows you - it's just "not the Car". If your first choice is a Goat, switching will win you the Car. If your first choice is a Car, switching will win you a Goat. You have…

This is the best explanation I've seen for the problem so far. Thank you

Re: Fair coins tend to land on the same side they started

#245

Post by alexdowad »

s1artibartfast wrote: post #217
naniwaduni wrote: post #171

Earlier quoted context omitted.

Fundamentally, the trouble with the Monty Hall problem isn't that analysis comes to the wrong answer, it's that people often come to the wrong model when reasoning about it informally. It's not any harder to do the "correct" analysis than to write up a simulation. It's mostly just easier to convince yourself that the simulation matches the problem description when it reaches the unintuitive result.

Fundamentally, I think the real trouble with the Monty Hall problem is that the assumptions of the game are not clearly stated. Because of this, people come up with different models.

That's absolutely right; further, if you explicitly model the behavior of the game show host, you can exhibit models under which "it's better to switch" and models under which "it doesn't matter if you switch or not".

Re: Fair coins tend to land on the same side they started

#246

Post by Natsu »

s1artibartfast wrote: post #242
Natsu wrote: post #238

Earlier quoted context omitted.

There is something missing: regular stats don't differentiate between doing things and observing things and these two are not at all the same. If I have a digital thermometer and I observe it to show a high temperature, then I will note an association between that and feeling warm. But if I merely set the thermometer gauge to a high value artificially, it's not going to make me feel any warmer. This ambiguity is reso…

I think it is more fundamental than that, and not even mathematical. I think the issue is that people conflate or blur the difference between reality and their models of reality. Your personal, information limited calculation of the chance a car is behind door #1 has no impact on if there is a car behind door #1. Reality is binary and constant. There was always a car there, or there always wasn't. Most people correct…

I agree that the map is not the territory, but there is a better model here that captures the difference.

The car isn't moving, as you say, but that intervention by the host lets us trade one door for both of the other doors.

Re: Fair coins tend to land on the same side they started

#247

Post by benchaney »

bigstrat2003 wrote: post #243
cortesoft wrote: post #195

Earlier quoted context omitted.

Yeah, I feel like the Monty Hall confusion goes away if you are explicit about the rules: "Hall will always open one of the two non-chosen doors and will never reveal the prize" I think most people who don't understand the problem miss that critical detail.

No, I don't think that detail makes it any easier. I know that but I still really can't accept the correctness of the Monty Hall strategy (I have to basically just take it on faith and stop trying to understand it). I was trying to put my finger on why, and I think it's this. After Monty eliminates one of the three doors, then the prize is behind one of the two. If someone were to come in this point, with no prior kn…

The best probability estimate you can make is constrained by the information you have available. The new person showing up has less information than the existing constant, so it makes sense that their best estimate would be less precise. Similarly, if someone with x-ray vision walked up in the middle of the game, they could pick the car 100% of the time, because they have access to more information than either of the existing contestants.

Your last paragraph isn't correct though, By switching you go from a 1/3 probability to a 2/3 probability. Based on the information the original contestant has, switching gets the car 2/3 of the time.

Re: Fair coins tend to land on the same side they started

#248

Post by mewpmewp2 »

asimpletune wrote: post #159
mewpmewp2 wrote: post #30

Earlier quoted context omitted.

That's amazing, but I guess it won't help when the person can choose the bias? Because according to the study the person can choose the bias by choosing which side start up. So if the person wants tails based on what you've said, they should always 1. Do the first throw starting tails up. 2. If the first one is tails, then they now want to start second one heads up. 3. If the first one is heads, they will want to try…

But wasn't the bias in the paper something like 50.5% vs 49.5%?

Yes, but I used more extreme numbers for ease of calculation and to clearly indicate the direction of a probability.

Re: Fair coins tend to land on the same side they started

#249

Post by bigstrat2003 »

benchaney wrote: post #247
bigstrat2003 wrote: post #243

Earlier quoted context omitted.

No, I don't think that detail makes it any easier. I know that but I still really can't accept the correctness of the Monty Hall strategy (I have to basically just take it on faith and stop trying to understand it). I was trying to put my finger on why, and I think it's this. After Monty eliminates one of the three doors, then the prize is behind one of the two. If someone were to come in this point, with no prior kn…

The best probability estimate you can make is constrained by the information you have available. The new person showing up has less information than the existing constant, so it makes sense that their best estimate would be less precise. Similarly, if someone with x-ray vision walked up in the middle of the game, they could pick the car 100% of the time, because they have access to more information than either of the…

I don't see how a new contestant has less information, though? They know that one of the two doors contains the prize, which is all the previous contestant knows either.

Re: Fair coins tend to land on the same side they started

#250

Post by cortesoft »

bigstrat2003 wrote: post #243
cortesoft wrote: post #195

Earlier quoted context omitted.

Yeah, I feel like the Monty Hall confusion goes away if you are explicit about the rules: "Hall will always open one of the two non-chosen doors and will never reveal the prize" I think most people who don't understand the problem miss that critical detail.

No, I don't think that detail makes it any easier. I know that but I still really can't accept the correctness of the Monty Hall strategy (I have to basically just take it on faith and stop trying to understand it). I was trying to put my finger on why, and I think it's this. After Monty eliminates one of the three doors, then the prize is behind one of the two. If someone were to come in this point, with no prior kn…

I have fun trying to explain this problem. Let me see if I can give an explanation that will help you.

So lets say you have just picked a door in the beginning. You know you have a 1/3 chance of being right.

If I then tell you, "I will give you two options... you can either bet you are right, or bet that you are wrong"

You would obviously choose to bet you are wrong, correct? Because you know you only have a 1/3 chance of being right with your guess, which means you have a 2/3 chance of being wrong. The smart bet is that your original guess was wrong.

This is actually what is happening in the game if you think about it. You pick a door and it has 1/3 chance of being the right one; since we know Monty is only going to ever reveal a goat and never the prize, we don't even NEED Monty to reveal the door at this point - we know he is going to reveal a goat, no matter what. We don't even have to wait to see which door he reveals, since that isn't going to give us more information (it is going to be a goat, no matter what). So when he asks you if you want to switch doors, he isn't asking you to switch to ONE of the other two doors, he is asking if you want to switch to having BOTH other doors as your choice. Whether he reveals the goat before or after you choose to switch doesn't matter, because you know it will always be a goat.

If that is still not clear, lets just write out all the options:

There are three doors, A B C. One has a prize, the other two have goats. Let see what happens with your two options (switch or dont switch).

In our first example, you pick door A and you are going to switch.

1/3 of the time the prize is behind door A. If the prize is behind door A, and you switch, you lose. This is 1/3 of the time, and you lose for switching.

1/3 of the time the prize is behind door B. You picked door A, so Monty reveals door C. You switch to the remaining door (B) and you win.

1/3 of the time the prize is behind door C. You picked door A, so Monty reveals door B. You switch to the remaining door (C) and you win.

Add up all those choices, and 2 out of the 3 times you win.

Now lets imagine that we DON'T switch.

1/3 of the time the prize is behind door A. Monty reveals one of the other doors, but you don't switch. You win.

1/3 of the time the prize is behind door B. Monty reveals door C, but you don't switch from A. You lose.

1/3 of the time the prize is behind door C. Monty reveals door B, but you don't switch. You lose.

So in this not switching world, you win 1/3 of the time.

In summary, switching wins 2/3rds, not switching wins 1/3.

Does that help at all?

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