[HN Gopher] Making the Monty Hall problem weirder but obvious
___________________________________________________________________
Making the Monty Hall problem weirder but obvious
Author : dyno-might
Score : 50 points
Date : 2020-10-07 11:57 UTC (11 hours ago)
(HTM) web link (dyno-might.github.io)
(TXT) w3m dump (dyno-might.github.io)
| jashmenn wrote:
| Here's what worked for me:
|
| Imagine there are 1,000 doors and you pick 1. All other doors
| except 1 are opened and you're given the offer: keep the door you
| picked, or pick this other door. What are the chances you picked
| the right door (vs. this other door)?
|
| People seem to intuitively understand that having only one door
| unopened is a massive "hint" to where the prize is.
|
| (I learned this idea from Better Explained:
| https://betterexplained.com/articles/understanding-the-monty...)
| kmm wrote:
| That explanation never worked for me, because you can turn it
| around to the situation where Monty does _not_ know where the
| car is. Say there are 1000 doors, and you pick door 429. On his
| way to open door 429, Monty stumbles, falls, and accidentally
| knocks open every door except door 128. If by some coincidence
| all opened doors happened to contain goats, you will have
| nothing to gain from switching. Very counter-intuitive, but
| just as true as the original problem.
|
| A possible intuition here is that Universes where your first
| pick was the door with the car, which initially were just 1 in
| a 1000 compared to Universes in which you picked a goat, will
| suddenly become massively overrepresented. After all, in these
| types of Universe Monty's Fall couldn't possibly have shown a
| car, whereas most of the other Universes will not survive to
| the next "round".
|
| Of course, if this happened in real life, Bayesian thinking
| would increase the likelihood of hypotheses such as, for
| example, "The door containing the car has a better lock" to
| such an extent that I would switch.
| tzs wrote:
| You can't really turn it around, because Monty knowing and
| using his knowledge of where the car is to reveal only goats
| is what makes switching advantageous.
|
| In the case of the clumsy Monty of your example, it goes like
| this:
|
| 1. There is a 1/1000 chance door 429 has the car.
|
| 2a. If it has the car, then when Monty accidentally opens 998
| doors no car will be revealed. This does not change the
| chances that 429 has that car, which remain 1/1000.
|
| 2b. If 429 does NOT have the car, then 998/999 times that
| Monty accidentally opens 998 doors, he will reveal a car,
| which presumably ends that game. There is only a 1/999 chance
| that he will not reveal the car and the game proceeds.
|
| 3. Thus, there are two cases where the game reaches the point
| of two remaining doors, with 998 revealed, the car is behind
| one of the two, and you have a chance to switch.
|
| 3a. Your door has the car, which happens 1/1000 games.
|
| 3b. Your door does not have the car, which happens 999/1000 x
| 1/999 games, or 1/1000 games.
|
| In other words, if the clumsy Monty version is played
| repeatedly, 998 out of 1000 games end without even getting
| too the point you get a chance to switch, and 2 get to where
| you get the chance. In those two, one has the car in your
| door, one not. There is no advantage to switching.
|
| In the case of the systematic Monty who knows where
| everything is and ALWAYS opens 998 goats, it goes like this:
|
| 1. There is a 1/1000 chance your door, 429, has the car.
|
| 2a. If it had the car, Monty opens 998 doors that do not have
| the car, leaving one door besides your yours.
|
| 2b. If your door did not have the car, it is one of the 999,
| and Monty systematically opens the 998 of those 999 that do
| not have the car.
|
| 3. You always reach the choice stage. You can either get
| there via 2a, which always results in the car being behind
| your door, or via 2b, which always results in the car being
| behind the other door.
|
| 3a. You get there via 2a in 1/1000 games.
|
| 3b. You get there via 2b in 999/1000 games.
|
| If you do not switch, you only if and only if you got there
| via 2a, so you only win 1/1000 games. If you always switch,
| you win if and only if you get there via 2b, so you win
| 999/1000 games.
| [deleted]
| [deleted]
| lordnacho wrote:
| For someone who doesn't get it, the problem is then whether the
| correct extension is all the doors being opened. With 3 doors
| it's the same.
|
| For me the most sensible explanation requires you to know that
| a dud door is always opened, thus the probability from the 2/3
| is the one you are switching to.
| thehappypm wrote:
| I don't think this helps me understand it.
|
| In the 1,000 doors problem, my odds of being right initially
| were something like 1/1000 and then it changes to something
| like 998/1000 or 999/1000 for switching, I can't intuitively
| grasp exactly what the odds become of winning if I switch, I
| just know it's high. Bringing it down to 3 doors doesn't help
| me much -- it's still something like 1/2 or 1/3.
| cestith wrote:
| Try thinking of it this way. It may or may not be any more
| useful. I've seen different people come to understand the
| problem from different examples. 1. Observe
| that 3/3 = 1. Pedantic, yes, but good for frame of mind here.
| 2. Pick one of three doors. (1/3 odds) 3. Gain
| information that one of the three doors is a loser.
| 4. Note your odds on choosing the original door correctly are
| still 1/3. 5. Note that if you change doors, there
| are still 2/3 doors there to choose. 5. Note you're
| not going to switch to the known loser door, so if you change
| doors you know 100% which of the other 2/3 of doors to
| choose.
|
| The intuition usually is that you're down to two doors after
| the loser door is opened, but that's not the case. There are
| still three doors. The host has just told you that if you
| trade doors, you know which door to trade for. So trade for
| it.
|
| Note there's a newer version of "Let's Make a Deal", hosted
| by Wayne Brady, but there is no option to switch after a
| losing door has been shown in that version.
| thehappypm wrote:
| Step 4, kind of a mystery. Why are my odds still the same
| even though the situation is different?
| ragnese wrote:
| You don't need to know the exact odds to understand that it's
| higher. I think that's the main takeaway of making it a 1,000
| door problem. It makes it intuitive that the correct solution
| is to switch. The exact probability doesn't matter.
| klmadfejno wrote:
| probability of winning if you switch is 1 - (1 / n) aka
| (n-1)/n.
|
| probability of winning if you don't is 1 / n.
|
| By switching, you are simply betting that your original guess
| of 1/n was wrong.
| [deleted]
| filoeleven wrote:
| I was lucky enough to get this explanation from my high school
| physics teacher, who first presented the classic Monty Hall
| problem and then illustrated the changing of the odds by
| substituting all of the lockers in our school for the three
| doors. Switching gives a clear advantage.
|
| The rest of this post is an anecdote from the same class that
| this brought to mind, and is unrelated to the topic. Maybe we
| can say it shows how good teachers engage their students or
| something, but really it's just a good yarn.
|
| We were learning about inelastic vs elastic collisions, and how
| an elastic collision has 2x the energy of an inelastic one. The
| teacher asked for a volunteer, and a bright-eyed student rose
| to the occasion. The teacher gave him some safety glasses and
| told him to lie down on the floor.
|
| The teacher took the inelastic ball and said, "Okay, I'm gonna
| drop this on your forehead now, ready?" PLONK. "Ow."
|
| "Remember that feeling! This is the elastic one, and it has the
| same mass, so it should hurt twice as much." PLONK. "Ow."
|
| The teacher asked, "So, did the second one hurt more than the
| first?" The rest of us anticipated the experimental
| confirmation of what we'd just learned about.
|
| "...I couldn't really tell the difference," said the student.
|
| "Yeah," said the teacher, "I knew you wouldn't. I just wanted
| to see if you'd let me do it."
| oh_sigh wrote:
| I think the confusion comes from the oracle-like knowledge of
| Monty Hall - he knows what is behind each door and decides which
| door to open based on that knowledge. If Monty Hall wasn't an
| oracle, and just opened a random door after the guess, people's
| intuitions would be correct.
| [deleted]
| twiceaday wrote:
| The opening of the door seems irrelevant and only serves to
| confuse. There will always be a goat door to open, who cares if
| it gets opened prior to the choice? The choice you are being
| given is "keep one door" or "choose both of the other doors."
| That is functionally the choice because Monty always opens a goat
| door. The chance of the other two doors containing a car is not
| affected.
| OpieCunningham wrote:
| This is the comment that has a car behind it.
|
| Monty's knowledge is irrelevant. The original problem can be
| modified with no change in probabilities to:
|
| Pick one of three doors. You can have whatever is behind that
| door, or you can change your pick to both of the other doors
| and win whatever is behind both of them. Should you switch?
| Obviously.
| kgwgk wrote:
| That you will never be worse by switching doesn't mean that
| you will always be better.
|
| https://news.ycombinator.com/item?id=24713352
| Tomminn wrote:
| Step 1: Take three cards face down, one of which is an ace.
|
| Step 2: Divide it into a pair of cards, and a single card.
|
| Step 3: Reveal one of the pairs of cards. If you reveal and
| ace, return to step 1 since this history is eliminated from
| the story we've been told: we know this wasn't a possible
| path to our endgame. Otherwise continue to step 4.
|
| Step 4: Which of the remaining two cards is more likely to be
| an ace?
|
| Step 5: Realize that they're just as likely as each other to
| be the ace. Step 2 didn't magically imbue the card remaining
| in the pair with extra probability juice.
|
| Long story short, you definitely need Monty to make an
| intelligent selection, so Monty's knowledge is far from
| irrelevant. It matters whether he revealed a goat by luck or
| by knowledge, because he's 2x as likely to get "lucky" in the
| case where there are 2 goats behind the doors you didn't
| choose in your original guess.
| lcuff wrote:
| I like this phraseology. How to make the answer _intuitive_ is
| the objective, and underlining that you 're choosing two doors,
| one of which is going to be wrong, really helps.
| ju-st wrote:
| Yes I finally understood it. At the beginning you are
| choosing one door vs "the others". If you could you would
| already chose "the others" because they have a 67% win
| probability but you are only allowed to chose one single
| door. Then he opens all the wrong doors of "the other doors".
| The "other doors" still have 67% probability and now only one
| door is remaining in the "other doors". Obviously that last
| remaining door now has 67%.
| basch wrote:
| That's what I dont understand about all the other explanations
| trying to "simplify" the situation. Just ask people, would you
| prefer 1 door or 2 doors. I've never understood how expanding
| it to 100 or 1000 doors is simplier than asking "which is
| better odds 1/3rds or 2/3rds."
|
| I also dont believe the original intent of the question was
| ever meant to be ambiguous with regard to whether he had
| knowledge of the goat door or whether he chose at random. The
| intent was for him to have prior knowledge or impeccable luck,
| and the wordsmithing of the question came later as, in my
| opinion, a failed rebuttal to the simplicity of the question.
| The question might have been worded to not be immediately
| obvious, but it was not intended to have different correct
| outcomes depending on interpretation.
| twiceaday wrote:
| It probably went like this:
|
| Hey lets put a car behind three doors, have people choose,
| open one of the other two doors, then ask them to switch.
|
| Sounds good Bob, but wait, won't the show end prematurely 1/3
| of the time because you will randomly open a car?
|
| Huh, good point. Um, let's sneak a peek before opening so we
| never open a car?
|
| Sounds good Bob.
| klmadfejno wrote:
| Understanding the probability tree is fairly interesting
| though.
|
| If he opens a door and tells you he knows it's a goat, you
| double your likelihood of winning by switching to the
| remaining door.
|
| If he opens a door randomly, and gets a goat, you don't
| modify your likelihood at all by switching. Saying 1 door or
| 2 doors doesn't mean you actually grasp the entirety of the
| problem.
| basch wrote:
| >If he opens a door randomly, and gets a goat, you don't
| modify your likelihood at all by switching.
|
| At best, that was intended as a red herring to make the
| correct solution less obvious. It was not intended as an
| alternate correct answer. It's not how the game ever
| worked.
|
| The initial version of the question said he opens a goat
| door. Whether he knew it would be a goat door or not is
| slightly irrelevant, because your odds of being right the
| first time were 1/3. As far as the premise of the thought
| experiment, a goat door always gets opened, either through
| peaking, premonition, or consistent luck.
|
| >Saying 1 door or 2 doors doesn't mean you actually grasp
| the entirety of the problem.
|
| I disagree, and if you don't see it as that simple, you are
| falling for the trap that makes the question fun.
| klmadfejno wrote:
| I don't have a particularly strong opinion on this, but
| if we accept that its interesting as a thought experiment
| and not a useful strat should one find themselves time
| traveled into the monte hall game show, I think it's
| worth exploring the very closely related variants of the
| thought experiment.
| basch wrote:
| If I time traveled into the show, and he opened a goat
| door, I would switch.
| twiceaday wrote:
| There is no better strategy than switching. Only under
| certain rules is there an equally good strategy in
| staying.
| majormajor wrote:
| > If he opens a door randomly, and gets a goat, you don't
| modify your likelihood at all by switching. Saying 1 door
| or 2 doors doesn't mean you actually grasp the entirety of
| the problem.
|
| What do you mean "opens a door randomly"? Is he picking
| from all three doors? Yours, and the two others?
|
| In that cases you get interesting but trivially-obvious-
| what-move-to-make scenarios like "he chose your door and
| showed you you were right" "he chose your door and showed
| you you were wrong" "he chose a different door which had
| the car"...
|
| Do you just mean the subset of "he chose randomly and
| happened to draw a goat out of one of the two you did not
| choose"? In which case switching isn't beneficial because
| you no longer are also capturing the cases that would
| otherwise be the "he chose randomly and opened the one with
| the car that you did not choose" that _are_ included in the
| original "switch or not" decision because he always goes
| to a goat?
| basch wrote:
| >Do you just mean the subset of "he chose randomly and
| happened to draw a goat out of one of the two you did not
| choose"? In which case switching isn't beneficial because
|
| Yes switching is. Switching is beneficial IF he shows you
| a goat out of the doors you didnt choose.
| kgwgk wrote:
| https://news.ycombinator.com/item?id=24713352
| buildbot wrote:
| For anyone that doesn't get it after this explanation or even the
| 1000 doors trick:
|
| Try visualizing how you'd pseudocode this game - it literally
| didn't click for me until right now, and now it seems much more
| intuitive.
| klmadfejno wrote:
| > It's important that Monty looked behind the doors before
| choosing which to open. This is where people's intuition usually
| fails. If he had chosen a door at random -- in a way that he
| risked possibly exposing a car, then the situation would be
| different. (In that case, there's no advantage or harm in
| switching.)
|
| This one took a second to rationalize. The reason it works is you
| have the same chance of winning overall (assuming you can safely
| choose the car if he opens the car by chance), it's just that the
| value of winning from switching vs. staying has been shifted into
| the probability of winning by default. The usual mental trick is
| to extend to 1,000,000 doors.
|
| If you pick one door, then are told that all of the alternatives
| except one are the correct answer, you should obviously reason
| that the door you didn't choose is the correct answer, unless you
| got the 1/1,000,000 guess. Odds of winning if you switch are
| 999,999 / 1,000,000
|
| If the host instead opens 999,998 doors randomly, you have a
| 999,998 / 1,000,000 chance of winning by default. The remaining
| two doors have equal chance of winning, giving you the same total
| odds of 999,999 / 1,000,000 no matter which you choose.
|
| Which makes sense, because both situations are, more or less,
| being given 999,999 chances to guess the lucky door.
| eithed wrote:
| If there are infinite number of doors, then does switching
| guarantee us the car?
|
| Also, Game 5: There are 2 doors. A car is randomly placed behind
| one, and goats behind the others. You pick one door. Monty looks
| behind the other doors. He chooses 0 of them with goats behind
| them, and opens them. You get two options: Option A: You get
| whatever is behind the door you picked. Option B: You get
| whatever is behind the other closed door. Should you switch?
| [deleted]
| millstone wrote:
| > But he doesn't choose the door at random. He deliberately
| chooses to show you goats. Since this is always possible, it
| tells you nothing
|
| What happens if Monty does not ever choose at random. Say that
| Monty always opens the the highest possible door.
|
| If you choose door 1 and Monty reveals door 2, then switching (to
| 3) is 100% win.
|
| If you choose door 1 and Monty reveals door 3, then switching (to
| 2) is 50% win.
|
| I think it's a crucial unstated assumption that Monty does choose
| randomly among available goat doors.
| sduff wrote:
| What works for me is scaling the number of doors to 100 (or
| more). Now the initial guess is correct only 1% of the time,
| while switching is 99%.
|
| Simulated many more variations at
| https://simonduff.net/monty_hall/
| tromp wrote:
| Most of the subtlety of the problem lies in this small sentence
| in the Side Notes: "He deliberately chooses to show you goats."
|
| This is not made as explicit in the standard formulation "the
| host, who knows what's behind the doors, opens another door, say
| No. 3, which has a goat."
|
| Which could be read as "which happens to have a goat".
|
| It's the ambiguity in Monte's door opening strategy that leads to
| different answers.
| basch wrote:
| Whether or not he knew or whether it was luck, it doesnt
| exactly matter, although traditional stats knowledge would make
| you think it does.
|
| The important thing to understand is that the premise of the
| question says he will show you a goat. If you rerun the
| experiment 10,000 times, he will show you a goat 100% of the
| time, either through peaking, premonition, or consistent luck.
|
| The problem gets trickier because people start applying domain
| knowledge of stats, and treating it as a simulation with random
| events. The goat being chosen is not random, it is an event
| that occurs 100% of the time in the premise of the thought
| experiment. Thinking about the random chance of him choosing
| the car is outside the bounds of the axiom/postulate we start
| with.
|
| tldr: it doesnt matter how he opened a goat door, all that
| matters is that he did.
| dllthomas wrote:
| But... it _does_ matter how he opened a goat door.
| Specifically, whether there was a chance that he might not
| have.
|
| I originally thought it couldn't matter. I wrote a simulation
| to demonstrate how right I was. I was wrong. I encourage
| duplicating the experience.
| tromp wrote:
| You said it well: "he will show you a goat."
|
| But the standard formulation just says Monte opened a door
| and it had a goat behind it. No explicit mention of
| intention.
|
| It matters not that he happened to do. It matters that he
| will.
| spullara wrote:
| It doesn't, as long as he always shows a goat.
| roywiggins wrote:
| What if he didn't know and just happened to open the goat
| door? Should you still switch?
| joppy wrote:
| What door Monty opens is irrelevant. He shows you a door with a
| goat, you're better off switching. He shows you a door with a
| car, you're still better off switching. What matters is that
| when you first choose a door, you have most likely chosen a
| goat.
| Tomminn wrote:
| First caveat: If he shows you a door with a car, there is
| literally no point switching. There's a goat behind both
| doors you're allowed to choose from, which are the remaining
| two.
|
| (Below I use _non-chosen_ to mean _non-chosen by the
| contestants initial choice_.)
|
| Second point: If we are in the subset of all possible
| histories where Monty picked _randomly_ revealed a goat, then
| we will have 50% of histories where both non-chosen doors
| contain 1 goat selected by our history subset, and 100% of
| histories where both non-chosen doors contain 2 goats
| selected by our history subset.
|
| Since there are twice as many possible histories where the
| non-chosen doors contains 1 goat vs 2 goats, after selection,
| we have an equal number of histories in our sample where we
| have 1 goat or 2 goats behind the non-chosen doors. Or
| equivalently, we have a 50% chance that the non-chosen doors
| contain a car.
|
| Therefore it is irrelevant whether you switch.
|
| Monty needs to make an intelligent selection to change the
| game.
| lcuff wrote:
| One of the realities we face here is that different people have
| different brains which work differently. Thus different
| phraseology is going to help. Adding variant phrases such as
| "He will never show you a car", will probably help for some.
| tromp wrote:
| It's also possible that Monty _only_ gives you the option to
| switch if you chose the car. Or only if you chose door No 1.
| All of which affect the answer. Therefore, it must be made
| fully explicit what Monty can and cannot do.
| osipov wrote:
| The analysis of the Game 3 in the article is wrong. You should
| switch.
| c3534l wrote:
| I got out three playing cards and did the experiment myself over
| and over. That was many years ago and now whenever I see
| something on the Monty Hall problem it seems so obvious as if I
| can no longer even see why people think its unintuitive. The
| reason its unintuitive is that people don't really have
| experience with anything that works this way. I'm not sure that
| the linked explanation will be that helpful to people because of
| that, although it does make it plainly obvious that Monty Hall
| knows which door has the prize and he's telegraphing to you which
| door it might be by only opening doors which don't have it in
| there. I suppose if that's the missing part of the puzzle for
| you, then that will help.
| OJFord wrote:
| I've always thought of it like the 'denominator' of probability
| for the second stage became 2/3 (vs. the usual/first stage 1).
|
| So when he reveals a goat door you know that door is certainly
| not a winner, 0 probability (of 2/3), and the switch door is
| certainly (if it were one of those) the winner, 1 probability
| (of 2/3).
|
| But I still used to manage to confuse myself thinking it's
| intuitively 'more likely' to be the original door 'now' that it
| isn't one of the others.
|
| Until I studied information theory at university and it really
| clicked - Monty's door choice has lower entropy than your
| initial pick!
|
| (But I acknowledge you can't say to the masses 'look look let's
| simplify this, if we just step back and take an information
| theoretic approach -')
| choko wrote:
| I've had issues understanding the Monty Hall problem for years
| until now. What made it click was that, at the end of the
| article, it's explained that the doors are not opened by Monty at
| random. My previous reads about the problem did not disclose
| this, so I had made the assumption that the door opening was
| random. I've also never seen the show, for what it's worth.
| [deleted]
| Tomminn wrote:
| This seems overly complicated. Here's an answer on Quora which is
| a more straight forward deweirdification.
|
| https://www.quora.com/In-the-monty-hall-problem-how-does-ope...
|
| Reprinted:
|
| Q: In the monty hall problem, how does opening the second door
| skew the probability in favor of the initially unchosen door?
|
| A: The Monty Hall problem is generally poorly described, in order
| to make the conclusion seem more surprising then it is.
|
| The actual Monty Hall game -- as imagined by the people who are
| asking the question-- is set up like this: In
| front of you are 3 doors, there is a goat behind two of
| them, and a car behind the other one. In *round 1* of
| the game, you select a *pair* of doors, from which *one*
| "goat containing door" will be *automatically eliminated
| from*, leaving only *one* door of the selected pair of
| doors in play, (and only *two* of initial *three* doors in
| play). In round 2 of the game, you guess which of the
| two remaining doors in play has the car.
|
| The choice is this: should you choose the remaining door from the
| pair selected in round 1, or should you choose the door which was
| not part of the selected pair in round 1?
|
| When phrased like this, the answer is fairly obvious: the pair of
| doors contains a car 2/3 of the time, whereas the non-paired door
| contains a car 1/3 of the time.
|
| The Monty Hall problem-- as normally described-- messes this all
| up by introducing a game show host. This is an agent who--
| seemingly by their own whim-- changes the game you thought you
| were playing, and introduces round 1 of the game once you have
| guessed the door you initially think the car is behind. The rules
| this game show host agent are following are almost never
| described to a sufficient degree to ensure the game is equivalent
| to the game laid out above. And yet, the people asking this
| problem pretend that it is exactly equivalent when they ask you
| for an answer.
|
| It's generally a poorly described problem, whose answer depends
| entirely on what kind of agent the game show host is.
|
| Don't worry if it doesn't make sense to you as it's usually
| described. If you can understand why-- in the two round game I
| describe above-- it's better to pick the door from the pair of
| doors, rather than the single door, you understand probability
| just fine.
| lqet wrote:
| > It's important that Monty looked behind the doors before
| choosing which to open.
|
| This is often not explicitly stated when the problem is given. It
| is even not a 100% clear from the statement above. Monty _always
| chooses a door with a goat_. So:
|
| 1. You choose a door.
|
| 2. Prob that there is a car behind it: 1/3
|
| 3. Prob that the car is behind the two other doors: 2/3
|
| 4. If the car was behind the two other doors (which, remember,
| has p=2/3), Monty _will choose the door without a car for you_ ,
| and the door with a car will remain closed. In this case you are
| _guaranteed_ to have the car if you switched.
|
| So with switching, the overall probability is 2/3. Without, its
| the original 1/3.
|
| If you did not understand that Monty always chooses a goat door,
| but the person giving you the problem does, or vice versa, then
| what usually happens is that both of you try to explain why your
| intuition is correct. Because most people don't talk formal
| probabilities, your explanations will be so vague that the other
| person will not realize your different understanding. You will
| discuss forever, you will both be right, and you will part ways
| with the strange feeling that maybe the other person _was_ right,
| when all along you were talking about different problems. This is
| why this problem is so notorious.
| jtsiskin wrote:
| Is it important the host looked behind the doors? If the host
| had just picked at random from the two doors, and it happened
| to show a goat, the odds would be the same.
| crooked-v wrote:
| > If the host had just picked at random
|
| The overall point is that the host is _not_ picking at
| random, and is thus affecting the outcome in a statistically
| reliable way.
| fouronnes3 wrote:
| This is gonna have anthropic principle vibes but here we
| go. Either the host looks behind the door, or they don't
| and we just don't talk about games where they accidentally
| show you the car. The probabilities are the same. We're
| already conditioned on being in the "they show you a goat"
| universe.
| samatman wrote:
| No.
|
| If you're playing blackjack, and you hit on a 19, but you
| have x-ray vision and you know a 2 is at the top of the
| deck, you're not playing the same game as you would be if
| you didn't have that knowledge.
|
| Same with the Monty Hall problem. It is a critical
| distinction whether opening the door has a 0% chance, or
| a 33% chance, of showing you a car.
| dllthomas wrote:
| Simulate it.
|
| I thought as you think. I wrote a simulation to show I
| was right. I was wrong.
|
| I have had this interchange several times. Invariably it
| goes one of two ways. They have endless reasons why they
| have to be right and they don't need to write a goddamn
| simulation, or they tell me they wrote the simulation and
| they have learned they were wrong.
| Arnavion wrote:
| Specifically, it's the difference between
| var num_stays_wins = 0; var num_switches_wins = 0; for
| (var i = 0; i < 100000; i++) { const car_door =
| Math.floor(Math.random() * 3); const my_choice =
| Math.floor(Math.random() * 3); var montys_choice; do {
| montys_choice = Math.floor(Math.random() * 3); } while
| (montys_choice === my_choice); if (montys_choice ===
| car_door) { i -= 1; continue; } if (my_choice ===
| car_door) { num_stays_wins += 1; } else {
| num_switches_wins += 1; } } console.log(num_stays_wins,
| num_switches_wins);
|
| and var num_stays_wins = 0; var
| num_switches_wins = 0; for (var i = 0; i < 100000; i++) {
| const car_door = Math.floor(Math.random() * 3); const
| my_choice = Math.floor(Math.random() * 3); var
| montys_choice; do { montys_choice =
| Math.floor(Math.random() * 3); } while (montys_choice ===
| my_choice || montys_choice === car_door); if (my_choice
| === car_door) { num_stays_wins += 1; } else {
| num_switches_wins += 1; } } console.log(num_stays_wins,
| num_switches_wins);
|
| (The difference is in the calculation of
| `montys_choice`.)
| kgwgk wrote:
| Imagine there are two contestants and each one selects a
| door. The host opens the remaining one and it happens to show
| goat. Do you think they will both increase their odds of
| winning by switching?
| mannykannot wrote:
| >> It's important that Monty looked behind the doors before
| choosing which to open.
|
| > This is often not explicitly stated when the problem is
| given, which imho is the whole reason this problem has the
| reputation of being hard to understand.
|
| If that were the only difficulty, why have so many people
| continued to have trouble accepting it even after this
| misunderstanding has been cleared up, and even after the
| correct answer has been explained to them? According to
| Wikipedia, even Paul Erdos remained unconvinced until he was
| shown a computer simulation.
|
| I recall mention of an analysis of the responses to Vos
| Savant's Parade article, concluding that a majority disputing
| the result were aware of this constraint, and I will post a
| link if I can find it again (though if a majority did not
| explain their reasoning, it may not be possible to figure out
| what assumptions they made. Nevertheless, the question in my
| first paragraph still stands.)
| brmgb wrote:
| What makes the three doors Monty Hall so counterintuitive is
| that people tend to correctly reason about the case where the
| second door is randomly opened and don't understand why it
| doesn't apply.
|
| I believe that this example makes understanding why people
| don't get it easier: you are looking for someone with one of
| your friend. You know they are in one of three rooms. Right
| before you can open the first one, your friend opens the
| second one and say: "not there". People assume the Monty Hall
| problem means that it's more likely your friend is in the
| third room and not the one you were going to open and think
| it's silly. And they are right to think that. What they don't
| get is that the case where your friend opened the correct
| door is part of the switching choice in the Monty Hall
| situation.
| adrianmonk wrote:
| It's also not always emphasized that the host never picks the
| same door you did. The host's behavior is not done in
| isolation. It is a response to your choice.
| ogre_codes wrote:
| This explanation is IMO more intuitive than the linked one.
| tasty_freeze wrote:
| It doesn't matter at all if Monty knows which doors are winners
| and losers.
|
| If Monty doesn't know, then sometimes the game will be ruined
| because he will expose the grand prize and then the game is
| moot. But if is simply lucky by showing the goat door vs he
| picked it with foreknowledge doesn't change the odds in any
| way.
| furyofantares wrote:
| In terms of the logic problem, I can't see any distinction
| between not counting moot games where he reveals a grand
| prize vs assuming he never reveals the grand prize. It's just
| two different ways to say that the only games under
| consideration are the ones where Monty reveals a goat.
|
| So I'm not getting the point you're making, unless you're
| saying that would be a more grokable way to state the
| problem?
| lqet wrote:
| It _does_ matter. If you are looking at a single instance
| where Monty got lucky and chose a goat door, then the
| probability that the door you have chosen has the car is 1
| /2, and switching doesn't change anything. This can easily be
| tested: just write a simulation that runs the experiment with
| Monty choosing a random door, and discard the instances where
| the game was "ruined" because he picked the car. In the
| remaining instances, both strategies will perform the same.
| dllthomas wrote:
| Once upon a time I was convinced of this, for reasons.
|
| I wrote a simulation to demonstrate how right I was.
|
| I was wrong and you are too.
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(page generated 2020-10-07 23:00 UTC)