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Marilyn vos Savant and the Monty Hall Problem (2015)

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Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#301

Post by listenallyall »

heleninboodler wrote: post #292
HarHarVeryFunny wrote: post #239

Earlier quoted context omitted.

I think the problem statement is clear, and is independent of how the TV show actually operated. The problem posed is that you have three closed doors, behind one of which is a car, and behind the other two are goats. You get to pick one of the closed doors, and will win whatever is behind it (you want the car). One of the doors you did not pick is now opened, revealing a goat. You therefore now know the car is eithe…

It's a lot easier to see the 2/3 nature if you imagine that instead of revealing the goat and then asking if you want to switch, Monty said "you've chosen this door that has a 1/3 chance. I'm offering you to switch to choosing both of these two remaining doors, at least one of which definitely has a goat by definition." This is functionally the same offer. It's 2/3 to switch because you're effectively choosing two do…

The initial door reveal is misdirection by an informed host, such that it has no effect on the odds, which are always, players choice of door is 1/3. Imagine instead:

After the player picks a door, the host states out loud, "we started with 2 zonks, you picked one door, so there must be at least one unchosen zonk... and I'm going to show you one behind door #X (door opens with zonk)... now, would you like to switch?"

Wouldn't make for good TV but it describes what is actually happening. Knowing that the host will always show an unchosen zonk is the key to realizing that opening that door has no impact on the player's odds (1/3 originally, 1/3 after the reveal, thus switching is 2/3).

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#302

Post by rspeele »

bemusedthrow75 wrote: post #260
rspeele wrote: post #241

Earlier quoted context omitted.

> It doesn't make a difference what causes Monty to reveal a goat. Oh, but it does! See the "Monty Fall" version of the problem, in which Monty accidentally trips and opens a door, which just happens to reveal a goat. In this variant there is no advantage gained by switching, because no more information was revealed about the remaining unopened door. The information gain only happens in the original game because we k…

Nope. But I am going to leave this to someone else to explain. I'm tired out now.

Since this bugged me all day, and I suspect you are the kind of person where it bugged you all day, too, here is a better description of the "fall"/"hall" distinction.

I think we can agree that these are the six possible, equally likely, configurations of the problem starting from me having chosen door 1. G1 here is "goat 1" and G2 is "goat 2". For each possible prize behind my chosen door, there are two possible configurations of the remaining prizes.

    My Door | Door 2 | Door 3
    Car     | G1     | G2
    Car     | G2     | G1
    G1      | Car    | G2
    G1      | G2     | Car
    G2      | Car    | G1
    G2      | G1     | Car
With the "Monty Hall" problem, Monty uses his knowledge to always open a goat door. Thus we see the following revealed options and resulting 2/3s probability of switching succeeding. This is the classic version of the problem.

    My Door | Door 2 | Door 3 | Monty Reveals | Switch Result
    Car     | G1     | G2     | Either        | Lose
    Car     | G2     | G1     | Either        | Lose
    G1      | Car    | G2     | G2            | Win
    G1      | G2     | Car    | G2            | Win
    G2      | Car    | G1     | G1            | Win
    G2      | G1     | Car    | G1            | Win
With "Monty Fall", the first thing that happens is a randomly chosen door, that isn't our own, reveals a goat. This is interesting. In the classic problem we were always going to see a goat next, because those are the rules Monty plays by. But in this case, the fact that we randomly found one wasn't guaranteed.

Essentially, you are blindfolded and throw a dart at the 2x6 grid of cells under the headers "door 2" and "door 3", and I tell you that the cell you've hit is a goat. What do you know about the row you hit being a switch-or-stay row? Well, half the possible goats you might've hit are in the first 2 scenarios where you should stay, and half the possible goats are in the last 4 scenarios where you should switch. So you're at 50/50. You don't have any new information to switch on.

    My Door | Door 2 | Door 3
    Car     | G1(a)  | G2(b)
    Car     | G2(c)  | G1(d)
    G1      | Car    | G2(e)
    G1      | G2(f)  | Car
    G2      | Car    | G1(g)
    G2      | G1(h)  | Car
You are just as likely to be looking at (a), (b), (c), or (d) (so you should stay) as you are to be looking at (e), (f), (g), or (h) (so you should switch). It is 50/50 in this version of the problem.[footnote]

This may make it confusing going back to the original. I seem to have shown that both ways make sense but still, how is it different? Imagine it like Monty is doing a random dice roll for which door to open, and he simply juices the outcome by correcting it to the goat door when a car door is selected, since he can't reveal a car and spoil the game. Now we have these equally possible scenarios (a) through (l) for his fair dice roll...

    My Door | Door 2 | Door 3
    Car     | G1(a)  | G2(b)
    Car     | G2(c)  | G1(d)
    G1      | Car(e) | G2(f)
    G1      | G2(g)  | Car(h)
    G2      | Car(i) | G1(j)
    G2      | G1(k)  | Car(l)
Which he corrects, avoiding cars, to:

    My Door | Door 2 | Door 3
    Car     | G1(a)  | G2(b)
    Car     | G2(c)  | G1(d)
    G1      | Car    | G2(f,e)
    G1      | G2(g,h)| Car
    G2      | Car    | G1(i,j)
    G2      | G1(k,l)| Car
Now we are back to the original game scenario where we see a goat no matter what. And we can see that 8 of the possible ways we might have arrived at seeing this goat come from "switch" rows while 4 come from "stay" rows.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#303

Post by rspeele »

rspeele wrote: post #302
bemusedthrow75 wrote: post #260

Earlier quoted context omitted.

Nope. But I am going to leave this to someone else to explain. I'm tired out now.

Since this bugged me all day, and I suspect you are the kind of person where it bugged you all day, too, here is a better description of the "fall"/"hall" distinction. I think we can agree that these are the six possible, equally likely, configurations of the problem starting from me having chosen door 1. G1 here is "goat 1" and G2 is "goat 2". For each possible prize behind my chosen door, there are two possible con…

[footnote] If this part of the explanation is bugging you, consider these related problems:

Problem 1. There are two opaque, externally identical bags, each containing 2 marbles. One bag contains 2 black marbles. The other bag contains 1 black marble and 1 white marble.

You choose a bag and draw a marble from it, without looking inside. The marble is black. What should you conclude are the odds that the remaining marble in the chosen bag is black?

Answer: .elbram kcalb dnoces a dnif ot ylekil sdriht-owt era ew oS .gab etihw-dna-kcalb eht morf si eno ylno dna ,gab kcalb-lla eht morf era elbram kcalb a werd ew hcihw ni soiranecs elbissop eht fo owT

Problem 2. There are three opaque, externally identical bags. One bag contains 2 black marbles. The other two bags each contain 1 black marble and 1 white marble.

Again you choose a bag and draw a marble from it, without looking inside. The marble is black. What should you conclude are the odds that the remaining marble in your chosen bag is black?

Answer: .tnecrep ytfif era elbram kcalb dnoces a gniward fo sddo ruO .gab kcalb-lla eht nesohc gnivah fo sddo ytfif-ytfif ta won era ew oS .sgab etihw-dna-kcalb tnereffid owt eht morf era owt dna ,gab kcalb-lla eht morf era elbram kcalb a werd ew hcihw ni soiranecs elbissop eht fo owT

You can connect Problem 2 to our random door opening and a goat being revealed, in the Monty Fall (with an F) problem.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#304

Post by bazoom42 »

apetresc wrote: post #192
bazoom42 wrote: post #186

Earlier quoted context omitted.

> it just wouldn't make sense in the context of the game show for the presenter (who knows where the goats are) to ever open a door to reveal the car and give you the option to switch. None of it makes sense because it is not a real game show, it is a thought experiment.

Apropos the topic of being confidently incorrect, it was indeed a real show: https://en.wikipedia.org/wiki/Let%27s_Make_a_Deal Hosted by a real Monty Hall: https://en.wikipedia.org/wiki/Monty_Hall It even had the gag prizes like goats (or, as in the photo from the wiki page, a llama)

The name of the problem is obviously taken from the real game show, but the real show did not work like in the thought experiment. So you cant just apply game show logic.

The “monty hall problem” (in the form stated in the article) is really a trick question, since it hinges on some assumptions which is never stated.

Even when stated correctly, the problem is a fun and counter-intuitive problem. But when leaving out critical information, you create a very differet problem of guessing what the unstated assumptions are.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#305

Post by hgomersall »

kragen wrote: post #270
hgomersall wrote: post #236

Earlier quoted context omitted.

But if the car door is picked, there's no further game to play. Surely the only interesting thing to ask is, conditioned on seeing a goat, what's the probability the third door contains a car. The cases of observing a car are irrelevant since that's not the scenario. I'm still not convinced it's any different whether Monty Hall knows or not so long as the goat door is opened. Edit: having written this, thinking about…

your first paragraph is what i thought before running the simulation, but it turned out to be wrong. if my simulation isn't convincing to you, try writing one that is

I'm convinced i think, just not initially by the simulator! I needed to reason my way separately but I think I'm happy with the sim now.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#306

Post by hgomersall »

tsimionescu wrote: post #278
hgomersall wrote: post #236

Earlier quoted context omitted.

But if the car door is picked, there's no further game to play. Surely the only interesting thing to ask is, conditioned on seeing a goat, what's the probability the third door contains a car. The cases of observing a car are irrelevant since that's not the scenario. I'm still not convinced it's any different whether Monty Hall knows or not so long as the goat door is opened. Edit: having written this, thinking about…

> the fact they all contain goats is pretty suggestive that I have the car, or at least 50/50 That's exactly the point. If he doesn't know, then it's exactly 50/50 and there is no reason to switch. If he does know, then it's 1/NUM_DOORS versus NUM_DOORS-1/NUM_DOORS, so you'd be crazy not to switch. The point is, if he picks at random, in the 100 door case, the vast majority of the time he will open the car door while…

This is why one can't ignore the car picked cases, so the simulator is correct.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#307

Post by mcv »

orblivion wrote: post #85
saalweachter wrote: post #81

Earlier quoted context omitted.

_Let's Make A Deal_ was on the air for twenty (thirty?) years at the time of this controversy. There was sufficient evidence for Savant's Monty -- the car was revealed 0% of the time, not 33 or 66% of the time.

I've never seen the show; I'm curious, did he always open one of the doors?

And if there's actual show where this experiment was done dozens of times, surely there's tons of data to confirm the correct approach here. Did anyone ever document all of that?

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#308

Post by lynguist »

HarHarVeryFunny wrote: post #281
petesergeant wrote: post #273

Earlier quoted context omitted.

> The simplest way to explain why switching is the correct strategy In my experience the most _intuitive_ explanation is to simply ramp it up to 100 doors, with Monty opening 98 of them, to make it clear that switching offers you the benefit of all unopened doors.

Maybe different explanations resonate with different people. Another explanation: After initial selection there is 1/3 chance car is behind your door, and 2/3 chance it is behind one of the other two doors. When one of the other doors is opened revealing a goat, we NOW know (new information to be taken advantage of!) the 2/3 chance represented by those two other doors lies with the unopened one. So, your choice is st…

“People” seem to think whenever you have a number of options, each option is equally likely.

I get into the car today - either I die in a crash or I don’t.

Hmm, I didn’t die, I survived the 50:50.

Their model doesn’t allow accounting for varying probabilities for different outcomes.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#309

Post by mayd »

grumbel wrote: post #61
Jimmc414 wrote: post #35

I never understood this one. Not choosing a new door seems like an implied 50/50 choice to keep your original selection. edit: I know it works out on paper, but if I put myself in the situation, I know that if I decided not to choose again that is a choice and I am re-choosing the same door.

The remaining doors are not random. The one you picked still has a 1/3 chance of winning, that doesn't change with opening another door. However switching doors is essentially the same as picking all the other doors at once, since the door with a goat is not opened, thus the 2/3. Imagine the thing with 100 doors and imagine the host isn't opening any doors, but just telling you what's behind them and it becomes prett…

Case 1: Suppose there are n doors and Monty does not know what is behind any of the doors. You choose one door at random. Then Monty opens (n-2) of the remaining doors AT RANDOM until there is only one other door left. By chance none of the doors he opened had the car behind it. Then he asks you if you want to switch. Should you switch or not?

Case 2: Suppose there are n doors and Monty knows what is behind every one of the doors. You choose one door at random. Monty deliberately opens (n-2) of the remaining doors from the left to right, skipping the door with the car. Then he asks you if you want to switch. Should you switch or not?

Whether n=3 or n=100, it seems to me that it does not matter whether Monty Hall has complete knowledge or zero knowledge of the location of the car. You are required to make a choice under the condition where there is only one other unopened door and all the other doors did not reveal a car. The player's original choice was correct with probability 1/n and the probability of the complementary event must be (n-1)/n. The player's strategy of switching will result in a win with probability of (n-1)/n.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#310

Post by mathgeek »

listenallyall wrote: post #298
HarHarVeryFunny wrote: post #239

Earlier quoted context omitted.

I think the problem statement is clear, and is independent of how the TV show actually operated. The problem posed is that you have three closed doors, behind one of which is a car, and behind the other two are goats. You get to pick one of the closed doors, and will win whatever is behind it (you want the car). One of the doors you did not pick is now opened, revealing a goat. You therefore now know the car is eithe…

It absolutely is not "independent of how the show operated." The host knowing in advance where the car is, (thereby ensuring the revealed door is always a zonk) is essential. If the host just picked a random door every day, and sometimes did reveal the car prematurely, then it would in fact be a 50-50 decision to switch.

The math here, for those wondering, is that “informed host has a 0 chance to pick the car” becomes “ignorant host has a 1/3 chance to pick the car”. Contestant’s original pick is still 1/3 and the unpicked door goes from 2/3 (informed) to 1/3 (ignorant).
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