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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)

#311

Post by psd1 »

XCabbage wrote: post #194
NoboruWataya wrote: post #161

Wow, I knew about the problem and was vaguely aware that it had generated some controversy amongst statisticians but I had no idea about the insanely arrogant and obnoxious (not to mention wrong) abuse Marilyn had received from the "intellectual elite". I have little sympathy for the "ambiguous question" defence. Not only is Marilyn's interpretation grammatically valid, it just wouldn't make sense in the context of t…

> 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 No, but it would make perfect sense for him to open a door at random and - if doing so reveals the car - tell you sadly that you've lost as a disappointed klaxon plays. Indeed this is almost the only possible way that the initial door o…

> there's no tension

There certainly is. It's a high-stakes scenario and the door-opening presents new information. It makes the conclusion feel less certain.

Revealing the car would conclude the game, of course. Unless the contestant were holding out for the better goat.

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

#312

Post by swayvil »

WarOnPrivacy wrote: post #57
kevinmchugh wrote: post #24

Earlier quoted context omitted.

The "smartest person in the world" tag is a marketing push. It works with her name. It wouldn't work as well if she was named Marilyn Smith.

> The "smartest person in the world" tag is a marketing push. It reflects some broad understandings of the 1980s and what we valued.

She wears a suit with shoulder pads. Dates Michael Douglas.

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

#313

Post by HarHarVeryFunny »

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.

That doesn't change anything.

Obviously if the host reveals the car then you've lost, but we're not talking about such a case. Regardless of whether that is happening some of the time, there is always a 2/3 chance your initial pick was wrong and you'd be better off switching.

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

#314

Post by btilly »

zachm0 wrote: post #280
jncfhnb wrote: post #267

Earlier quoted context omitted.

It’s still wrong. He needs to declare (or at least he needs to consistently) open an empty box; not any random box or a box of his choosing at his whim. If he opens a random box; you have not learned anything about the keys GIVEN he opened an empty box This paper adds that in as an assumption after the prompt, which I’m pretty sure is not the original prompt

If I understand correctly, you’re saying Monte’s intention (randomly picking an empty box vs purposely picking an empty box) is effecting the odds that the box in hand has keys? Also, do you have any evidence that this isn’t the original?

It's how probability works under Bayes' theorem.

The probability of A given B is the probability of A and B divided by the probability of observing B. And the probability of observing B depends on counterfactuals of various sorts. "What would happen if...?" And that's where intention comes in.

In this case B is "Monty opens an empty box". The probability of the event B depends on Monty's knowledge and intent. If Monty knows where the prize is, and always avoids it, then Monty always opens an empty box. Probability 1. If Monty is clueless, then Monty opens an empty box with probability 2/3. And if Monty is knowledgeable and malicious, then Monty opens an empty box with probability 1/3.

Event A is that you have found the prize and Monty found an empty box. We're assuming that this is the probability that you initially found the prize, and so has probability 1/3. And so we get that Savant's Monty leaves you with odds (1/3)/1 = 1/3 of having the prize, ignorant Monty leaves you with odds (1/3)/(2/3) = 1/2 of having the prize, and malicious Monty leaves you with odds (1/3)/(1/3) = 1 of having the prize.

I find it absurd that I've never looked at it this way and recognized the fourth possibility. HELPFUL Monty knows the answer, and is giving you every chance. So if you had the prize, helpful Monty would show you that you're a winner, otherwise helpful Monty will give you another chance. What helpful Monty changes is the probability of A and B. If you had the prize, you would have been shown it. Therefore the probability of A and B is 0, and you really, really want to take Monty's hint and switch.

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

#315

Post by listenallyall »

HarHarVeryFunny wrote: post #313
listenallyall wrote: post #298

Earlier quoted context omitted.

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.

That doesn't change anything. Obviously if the host reveals the car then you've lost, but we're not talking about such a case. Regardless of whether that is happening some of the time, there is always a 2/3 chance your initial pick was wrong and you'd be better off switching.

If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. Thus switching, at this point, is a 50-50 proposition. Or 1/3 to 1/3 if that makes more sense. Switching is the advantageous move only when the host always reveals a zonk first, and that's why it does in fact matter "how the TV show operates."

Another show, Deal or No Deal, often came down to 2 suitcases (from an initial 26), one of which was worth a six-figure amount and the other a tiny amount. Something like $200,000 vs $500. The logic of the Monty Hall problem would suggest that since the player had 1/26 chance of initially choosing $200,000, they should switch suitcases at the end. But because all 24 previous reveals were random, there is actually no advantage to doing so.

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

#316

Post by HarHarVeryFunny »

listenallyall wrote: post #315
HarHarVeryFunny wrote: post #313

Earlier quoted context omitted.

That doesn't change anything. Obviously if the host reveals the car then you've lost, but we're not talking about such a case. Regardless of whether that is happening some of the time, there is always a 2/3 chance your initial pick was wrong and you'd be better off switching.

If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. Thus switching, at this point, is a 50-50 proposition. Or 1/3 to 1/3 if that makes more sense. Switching is the advantageous move only when the host always reveals a zonk first, and that's why it does in fact matter "how the TV show operates." Another show, Deal or No Deal, often cam…

> If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car.

No... Remember one door has been opened and the car was not there, so we now know that it is guaranteed the car is behind one of the other two doors. If there was only 1/3 chance of it being behind either door, then that's the same as saying there's only a 2/3 (1/3 + 1/3) chance of there being a car anywhere, when we know for sure (3/3 chance) that it is somewhere!

The reality is that the contestant's original pick has a 1/3 chance of being right, and the other other two doors have a combined 2/3 chance of the car being behind one of them. Once one of those doors has been opened and the car shown not to be there, then this now means this combined 2/3 chance lies with the other door (which you should therefore switch to). Note that the probabilities now add up as expected: 1/3 chance original pick + 2/3 remaining choice = 3/3.

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

#317

Post by default-kramer »

HarHarVeryFunny wrote: post #316
listenallyall wrote: post #315

Earlier quoted context omitted.

If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. Thus switching, at this point, is a 50-50 proposition. Or 1/3 to 1/3 if that makes more sense. Switching is the advantageous move only when the host always reveals a zonk first, and that's why it does in fact matter "how the TV show operates." Another show, Deal or No Deal, often cam…

> If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. No... Remember one door has been opened and the car was not there, so we now know that it is guaranteed the car is behind one of the other two doors. If there was only 1/3 chance of it being behind either door, then that's the same as saying there's only a 2/3 (1/3 + 1/3) chance of…

What do you think is incorrect about the code here which produces 1/2 if Monty randomly chooses? https://news.ycombinator.com/item?id=39515082

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

#318

Post by [deleted] »

NoboruWataya wrote: post #161

Wow, I knew about the problem and was vaguely aware that it had generated some controversy amongst statisticians but I had no idea about the insanely arrogant and obnoxious (not to mention wrong) abuse Marilyn had received from the "intellectual elite". I have little sympathy for the "ambiguous question" defence. Not only is Marilyn's interpretation grammatically valid, it just wouldn't make sense in the context of t…

[deleted]

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

#319

Post by HarHarVeryFunny »

default-kramer wrote: post #317
HarHarVeryFunny wrote: post #316

Earlier quoted context omitted.

> If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. No... Remember one door has been opened and the car was not there, so we now know that it is guaranteed the car is behind one of the other two doors. If there was only 1/3 chance of it being behind either door, then that's the same as saying there's only a 2/3 (1/3 + 1/3) chance of…

What do you think is incorrect about the code here which produces 1/2 if Monty randomly chooses? https://news.ycombinator.com/item?id=39515082

Nothing .. I was wrong.

Switching only wins when Monty avoids opening the car door. If Monty opens a random door, then switching makes no difference.

The flaw in my logic above for the random case is that it's 1/3 chance orig pick was right, 1/3 chance unopened door, plus 1/3 chance that Monty opened the door with the car behind meaning you never even got a choice to switch.

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

#320

Post by kurikuri »

HarHarVeryFunny wrote: post #316
listenallyall wrote: post #315

Earlier quoted context omitted.

If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. Thus switching, at this point, is a 50-50 proposition. Or 1/3 to 1/3 if that makes more sense. Switching is the advantageous move only when the host always reveals a zonk first, and that's why it does in fact matter "how the TV show operates." Another show, Deal or No Deal, often cam…

> If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. No... Remember one door has been opened and the car was not there, so we now know that it is guaranteed the car is behind one of the other two doors. If there was only 1/3 chance of it being behind either door, then that's the same as saying there's only a 2/3 (1/3 + 1/3) chance of…

The host’s knowledge does contribute to the 2/3, because that knowledge prevents an outcome from occurring. Let us consider the switching strategy and instead say selecting the correct door awards 6 points. Our question: what is the expected value of this strategy when the host knows, vs when the host doesn’t know.

Consider the scenario where the player’s first choice is the winner. This happens 1/3 of the time, and the host’s knowledge doesn’t change their behavior. With the switching strategy, the player always loses in this case, so the contribution of this scenario towards the expected value for either case(host knowing vs not) is 0.

Consider the other scenario where the player’s first selection does not contain the car, and note that this situation happens 2/3 of the time. For convenience, label the player’s selected door 1, and the car containing door 2.

If we assume the host has knowledge, then the host will select door 3. If the player decides to stay, they lose. In this scenario, the switch strategy always wins, so its contribution towards the expected value when the host knows is (2/3 * 6) = 4.

If we assume the host doesn’t have knowledge, he will pick either door 2 or 3 with equal probability. In the event he selects door 2, the winning door, so half the time, the player loses because they can’t switch. This occurs half of the time within this scenario, so this sub -scenario contributes 0 points to our expected value. If the host selects door three, then the switch strategy wins, meaning this sub-scenario contributes (2/3 * 1/2 * 6) = 3 points to our overall expected value.

When the host has knowledge, the scenario where the player loses because of the host selection will not occur, leading to the 2/3-probability. Without the host having this knowledge, our event space changes and the player switch strategy only has a 1/2 chance of winning.

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