[HN Gopher] Tricking Monty Hall
___________________________________________________________________
Tricking Monty Hall
Author : zdw
Score : 45 points
Date : 2023-07-15 12:06 UTC (10 hours ago)
(HTM) web link (ignorethecode.net)
(TXT) w3m dump (ignorethecode.net)
| rain1 wrote:
| This is incorrect, the goats and car are behind doors. They are
| not inside cardboard boxes.
| tzs wrote:
| Here's a way to think of it that some people fine helpful. Assume
| N boxes, the host knows where the prize is, always reveals N-2
| goats after the content picks leaving just the contestant's box
| and one other box unopened, and always gives the contestant a
| chance to switch after revealing the N-2 goats.
|
| A person who does not switch wins if and only if their initial
| pick was correct. The probability of that is 1/N where N is the
| number of boxes.
|
| A person who switches wins if and only if their initial pick was
| _not_ correct. The probability of that is 1-1 /N.
|
| Switching is (1-1/N)/(1/N) = N-1 times as likely to win.
| voyager1 wrote:
| Shameless plug, I am fascinated by Monty Hall and I had a hard
| time proving it to my friends so I made this:
| https://kr1stjans.github.io/monty-hall/ :D feedback appriciated
| :)
| arithma wrote:
| I haven't done the analysis, but one major assumption that the
| whole result is based on is whether the rules of the game are
| pre-established or just revealed after the first pick. With an
| adversarial game host who has the option to reveal or not, maybe
| the result changes, but more importantly, it explains the
| intuitionistic refusal for some to buy the argument.
| jncfhnb wrote:
| In an adversarial setup you basically get 1/3 wins like you
| started with because the optimal host move is to provide no
| information whether they reveal or not
| mrob wrote:
| In an adversarial setup, the optimal host move is to permit
| you to switch only if you picked the car. You still get 1/3
| wins with optimal play (never switching), but it's better for
| the host because it allows people who think it's the
| traditional game to get zero wins.
|
| If the host strategy isn't explicitly specified, it's
| reasonable to assume an adversarial setup, because seeing the
| player lose after switching would be more entertaining for
| the audience.
| listenallyall wrote:
| > seeing the player lose after switching would be more
| entertaining for the audience
|
| Have you ever watched game shows? Because it is far more
| satisfying and fun for the audience when players win, not
| lose. If you are out there rooting for people to lose,
| you're a miserable bastard.
| mrob wrote:
| Now that I think about it, I have watched a few episodes
| of game shows, and I think you're right. They are always
| set up so the audience identifies with the players and
| the hosts are the adversaries, so you want the players to
| win.
| jncfhnb wrote:
| I don't think the premise is that the host is allowed to
| change your ability to switch at all or that the player is
| misinformed about the rules...
|
| Obviously a game can be made unwinnable if you simply lie
| about the rules
| aaron695 wrote:
| [dead]
| mikrl wrote:
| I did a presentation on this for a discrete math course in
| undergrad.
|
| IIRC I did the N box case and took the limit to infinity and,
| unsurprisingly, got the result that you should always switch.
| fsckboy wrote:
| if you play this game repeatedly, 2/3 of the time your first
| guess will be a goat, and Monty will then always show you the
| other goat, and if you switch you will see the car behind the
| remaining door.
|
| if you play the game once, you should follow the same strategy.
|
| don't explain more than that, the person just needs to grok that.
| a restatement of the main idea might help:
|
| 2/3 of the time your first choice is a goat therefore 2/3 of the
| time Monty is showing you the other goat therefore 2/3 of the
| time if you switch you will be switching to the car.
| rahimnathwani wrote:
| I like this explanation. Another equivalent one that I wrote
| about a while back[0]:
|
| You can choose to have:
|
| - The most valuable prize that's behind door A, or
|
| - The most valuable prize that's behind doors B or C
|
| When you look at it this way, it's obvious that you would rather
| have the most valuable prize from the 'other' doors, and the way
| to do that is to switch.
|
| [0] https://www.encona.com/posts/monty-hall-problem
| andy800 wrote:
| To the best of my knowledge, this "question" was first presented
| to the general public around 1990 via a popular syndicated
| newspaper column, and quickly "went viral" the way things did
| back then, by being repeatedly discussed in newspaper and
| magazine columns all across America.
|
| It's fascinating that 33 years later, there is still no simple
| explanation of the optimal strategy, such that most people would
| say "a-ha" when they hear it. In fact, most people actually
| express skepticism or disbelief when presented with the "correct"
| strategy.
| xg15 wrote:
| The reasoning that made me get it was basically:
|
| The key is that the host is quite constrained in which box he can
| show you: He must _always_ show you a box with a goat and he can
| _never_ show you the box you initially picked. So the choice the
| host makes is less random than your (initial) choice and can give
| you new information.
|
| In particular, only two different things can happen:
|
| a) You initially picked the car. In that case, the host is free
| to pick any of the two remaining boxes since both have goats in
| it and neither were picked by you. In this case, it would be
| obviously unwise to switch.
|
| b) You initially picked a goat. In that case, the host has no
| choice at all: They _must_ pick the one box with the second goat,
| which leaves the last remaining box as the one with the car. So
| you should absolutely switch here.
|
| If you knew whether you're in situation a) or situation b), you'd
| already have won the game: You could switch as needed and would
| always get the car.
|
| Obviously though, you don't know. However you know that a)
| happens when you initially picked the car and b) when you didn't.
| You also know, you initially picked the car with 1/3 probability.
| So you know that with 1/3 probability you're in situation a) and
| with 2/3 probability you're in situation b).
|
| Therefore, if you're just always pretend you're in situation b)
| and switch, you'll win the car with 2/3 probability.
| nkrisc wrote:
| You can distill it all to this:
|
| If your initial guess is correct, then switching always loses.
|
| If your initial guess is wrong, then switching always wins.
|
| Everyone who understands the game can agree on those facts
| without controversy. If they don't then there's no point
| proceeding because they do not understand the rules of the
| game.
|
| Therefore, the odds of winning when you switch are the inverse
| of the odds of winning from your initial guess.
|
| If your initial guess wins 1/3 of the time, that means always
| switching loses 1/3 of the time and conversely always switching
| wins 2/3 of the time.
| diputsmonro wrote:
| A slightly different reframing of this idea makes it even
| more clear to me:
|
| When you make a choice, you divide the doors into two sets:
| the set containing your choice, and the set containing the
| other doors. Obviously your set has a 1/3 chance of
| containing the prize, and the other set has a 2/3 chance.
|
| When the host reveals a door in the second set, it has no
| effect on those initial probabilities. The set as a whole
| still has a 2/3 chance of containing the door, and now you
| have the choice of selecting the only item in that set which
| you know isn't wrong.
|
| This reasoning is just as obvious when you scale up the doors
| too. With 100 doors, your set has a 1% chance of containing
| the prize, and the set of unchosen doors a 99% chance. All
| but one door in that set are revealed to be wrong, then you
| get a "50/50" choice of swapping to that last remaining
| item... of course that's what you want to do!
| ComplexSystems wrote:
| The one which made it clear for me was: imagine you have 4
| doors instead. You choose one at random. The host then chooses
| _two_ of the remaining doors he knows has goats. Is it in your
| best interest to switch to the one he didn 't open?
|
| Suppose you have 100 doors. You choose one at random. The host
| then chooses _98_ of the remaining doors he knows has goats. Is
| it in your best interest to switch?
|
| In all of these situations, the basic question is: what is more
| likely - that you chose the one with the car, or that you
| didn't, and the door remaining is only remaining because the
| host _knew_ it was the one with the car?
| TulliusCicero wrote:
| Yeah, simply imagining a much larger number of doors makes
| the logic obvious.
| ksaj wrote:
| I used an entire deck of cards (52 of them) to demonstrate
| this to a friend, and he still wouldn't believe it. I gave up
| explaining it after that.
|
| It's really easy to get people arguing about this one, even
| though the correct answer is so well documented.
| moffkalast wrote:
| Any explanation that doesn't involve scaling up the number of
| doors to a arbitrarily large amount where the probability
| becomes obvious is really lacking if you ask me. The only
| reason this is so confusing in the first place is because it's
| only 3 doors.
|
| With an infinite amount of doors, the chance of picking the
| right one at first is zero. When the host opens all doors
| except one other, it's almost certain that it's the door with
| the car.
| travisjungroth wrote:
| It's good for the aha intuition moment for some people. It's
| the thing that made me get it as a kid. There's something
| unsatisfying about it though. You can have 100 doors, and
| open 98 to leave the initial guess and 1 more. But this is
| scaling up the rule as "all but one". I don't see an obvious
| reason that the rule shouldn't scale up as "open one more
| door". Then you have 99 closed and 1 open and the result
| isn't nearly as obvious, even though switching is the better
| move.
| moffkalast wrote:
| Yeah I thought about that, both +1 door and infinity-1 door
| options are both equally unfaithful to the original I
| suppose. Keeping the 50% door opening ratio would make more
| direct sense if you wanted to scale it up for other
| reasons, but isn't nearly as clear in intuitively showing
| the concept.
| diputsmonro wrote:
| I agree that it feels weird at first, but thinking about it
| this way fixed that for me:
|
| The whole reason for the apparent paradox is having the
| "50/50" choice at the end. You either keep the door you
| picked, or swap with the other door, so it feels like no
| advantage.
|
| If they only revealed a single door, then you would have
| multiple doors to swap to, so you wouldn't have the "50/50"
| choice. The only way to keep the "50/50" choice at the end
| is to reveal all but one door and give you the option to
| switch to it.
| FartyMcFarter wrote:
| Yeah, this is exactly the easiest way to think about the
| problem. In summary, by switching you win when you initially
| picked a goat, which happens two thirds of the time.
| [deleted]
| hurril wrote:
| This is the first thing to get, the second is that Monty
| won't pick the car. If you do not understand these two, then
| you will say 1/2.
|
| I've heard the analogy with 1 000 doors and opening 998 of
| them before but that doesn't explain anything at all. If you
| think that makes you understand then you're mistaken; that
| made you understand something else.
|
| The key is that you have 1/3 of being right, 2/3 of being
| wrong. So in 2/3 of the cases, you will be offered to pick
| the car. So 2/3 switching gives you the car and 1/3 loses you
| the car. So always switch, it has better odds.
| silisili wrote:
| I wonder: has anyone went through the old tapes and confirmed
| that the people who stayed won about a third of the time, and the
| switchers 2/3 of the time?
|
| I agree with the math, but wonder how unscrupulous the show was.
| stavros wrote:
| What helped me visualize is this:
|
| Say the game has a million boxes. You pick one at random, the
| host opens the other 999,998 boxes (which all contain goats), and
| asks if you should switch.
|
| At that point, it's very obvious you _really should_ switch, as
| the original box has a one in a million chance of being correct,
| but the second box that is now unopened is _wink wink nudge
| nudge_ maybe the one you want.
| mrob wrote:
| The million box analogy isn't very clear IMO, because it
| assumes the host always opens all boxes except one. In the
| original scenario, "all except one" is identical to "exactly
| one", so "exactly one" might be the correct host behavior.
| stavros wrote:
| There is no "correct" host behavior. One behavior helps
| explain why the probability is 66%, the other doesn't.
| oefnak wrote:
| No, the host will only open 1 box, not 999,999. Otherwise it's
| not the same problem at all.
| stavros wrote:
| How is it different?
| IanCal wrote:
| I see this posted a lot but it doesn't click with me. It only
| feels obvious once I know how the probabilities work - meaning
| I have to get the base case first.
| chefandy wrote:
| Picture, in your mind or maybe draw on paper, 20 brightly lit
| doors in a row where 19 are labeled L and one is labeled W.
| Out of those 20 doors, one door gets randomly selected. It's
| almost certainly going to be an L. After choosing it, the
| lights go out above 18 of the other L doors so only the W
| door, and the L door you probably chose, remained lit.
|
| If you stayed with your existing selection-- one of two doors
| that remains lit-- you've still almost certainly got an L
| door. If you switch to the one other door, with all of the
| other L doors eliminated, the only way it's not the W door is
| if you chose the W to begin with.
| IanCal wrote:
| Two things about this
|
| 1. It requires that I already _get_ the point about the
| probabilities.
|
| 2. You don't get a benefit in Monty hall if the host picks
| randomly - the description requires that they deliberately
| leave the winning door available. Any situation that gives
| you the same intuition but you just picture as happening
| without intent has given you the wrong feeling.
|
| I don't find that increasing the number of doors makes it
| feel any different.
| latexr wrote:
| > You don't get a benefit in Monty hall if the host picks
| randomly
|
| But the host _doesn't_ pick randomly, they _always_ pick
| a goat. That is an integral part of the Monty Hall
| problem.
|
| Picking randomly wouldn't make sense. You'd pick a goat,
| then the host could randomly reveal the car and the game
| would end anticlimactically.
| chefandy wrote:
| Exactly-- the entire point is that the host always
| eliminates every losing answer that you didn't choose
| unless you chose the right answer initially. If you keep
| your pick, you have a 1 in 3 chance of winning, and if
| you switch, you have a 1 in 3 chance of losing.
| xg15 wrote:
| That's really a good way to visualise it. The host opens all
| the other boxes except your pick _and one specific other box
| somewhere in the middle_. It 's clear that something special
| must be going on with that box.
| patrickthebold wrote:
| I like the variant: You pick a door. An earthquake hits and one
| door opens that is a goat. The earthquake of course had know
| knowledge of that it's just what happened.
|
| Should you switch?
|
| I really thought I understood the problem until I got that
| variant wrong.
|
| There's a good video with a rant about the problem I will link to
| it when I find it.
| lisper wrote:
| Yep, this is the key. The host knows where the prize is, and
| the rules of the game force him to use and reveal some of that
| knowledge when he chooses the door to open.
| slavik81 wrote:
| That rule is fundamentally what changes the odds, but it's
| worth noting that Monty only revealing goats was not one of
| the rules of the actual Monty Hall gameshow, nor was it
| explicitly stated in the original framing of this problem
| (though arguably it was implied). The somewhat ambiguous
| description of the problem in the newspaper column that
| popularized it may have contributed to the confusion.
| lisper wrote:
| Yes, definitely. However, it's a pretty reasonable
| inference, because Monty would not open a door without
| being sure that it would not contain the prize because that
| would just destroy the dramatic tension. "Oops, sorry,
| guess you lost. Didn't see that coming. Better luck next
| time."
| kgwgk wrote:
| > Should you switch?
|
| Why not? There is no harm in doing so if the probabilities are
| the same anyway...
| necovek wrote:
| One thing to note is that 33% chance is still pretty good odds.
| Stats like that only really work with large numbers, and when
| you've got a single chance, neither of the outcomes is really
| surprising.
|
| IOW, if every participant switched, the show would be handing out
| prizes 2/3rds of the time, but that 1/3rd of participants would
| still be going home empty handed. Do you want to be that one? If
| participants randomly chose whether to switch or stick, how often
| would the show hand prizes out?
|
| Now we've got multiple conditional probabilities.
|
| And that choice is what makes casinos, lotteries and games like
| these work: we want to succeed despite the odds.
| jncfhnb wrote:
| If participation randomly chose whether or not to switch they
| would walk away with a prize 50% of the time. Assuming randomly
| == equally likely
| evandale wrote:
| My favourite casino game is craps. I find it amazing they
| created a game that has two basic strategies you can play and
| the one that has better odds is the "dark side" and is frowned
| upon to play that strategy.
|
| Sometimes I'll lose a ton of money and a superstitious pass
| line bettor will make a quip about how I'm playing the dark
| side and deserve it for betting against the table. My usual
| reply is along the lines of: we're all playing against the
| casino and the dark side beats the casino more often than the
| pass line ;)
| ryandrake wrote:
| Craps is a social game, probably more so than any other table
| game in a casino. I find the minuscule odds advantage of
| playing Don't Pass Line bets to be not worth "being the bad
| guy" at a table full of superstitious gamblers out having
| fun, but that's just me. I'm sure the casinos love this
| artificial stigma.
|
| I always wondered: how juicy would they have to make Dont
| Pass to overcome the social stigma at the table? Would it be
| enough to pay instead of push on 12? How much house edge
| would cause everyone to switch to the dark side? I have no
| doubt casinos have spent millions of man-hours of research to
| understand this.
|
| EDIT: In addition, most craps games offer "odds" bets once a
| point is established which have zero house edge, so by taking
| the maximum odds, you reduce the house edge on Pass and Don't
| Pass to the point where it's not really probabilistically
| significant which side you play.
| ksaj wrote:
| It reminds me of S. Korean "fan death."
|
| The common belief that sleeping with a fan on can kill you
| by suffocation (among other "explanations") was and is a
| way to convince people to conserve electricity, and nothing
| more.
|
| The artificial stigma is a way to convince people to not
| use the more successful strategy against the house.
| latexr wrote:
| > is a way to convince people to conserve electricity
|
| According to Slate, the belief is older than that:
|
| > Internet conspiracy lore sometimes blames the legend on
| a 1970s-era government campaign to conserve electricity,
| but in fact these warnings are generations older, dating
| almost back to the introduction of electric fans to
| Korea.
|
| https://slate.com/human-interest/2013/01/fan-death-
| korean-mo...
|
| Apparently we don't know how the myth began.
|
| https://en.wikipedia.org/wiki/Fan_death#Origins_of_the_be
| lie...
| andy800 wrote:
| > casinos have spent millions of man-hours of research
|
| You'd be surprised, casinos, in fact, spend almost zero
| time actually working through the math of their games. All
| the established table games are taken for granted, and for
| anything new -- entirely new games or side/extra bets --
| the responsibility lies on the game inventor to have a
| trusted 3rd party validate the math (this is usually
| enforced by the gaming commission) and the casino simply
| refers to the validated results when considering
| implementing the new game/wager.
|
| Any consideration of "social stigma" about a wager is nil.
| As an example, baccarat is fundamentally a boring game with
| zero decisions, however the actual gameplay, where players
| have all kinds of superstitions and are allowed to touch,
| tear, fold the cards, can be extremely social and often
| quite exciting. Casinos could deal blackjack in a similar,
| social fashion where all players share a single hand, but
| none do. And certainly, no casino would shift the odds in
| the players' favor (pay on 12 on Don't) for some kind of
| social impact, as any positive benefit would be 10x, 100x
| subsumed by professional bettors exploiting the positive
| EV.
| deeg wrote:
| Here's my attempt to explain the Monty Hall with code:
| https://github.com/DeegC/monty_hall_paradox
| johnfn wrote:
| The reasoning I've always used to understand this problem is to
| imagine that you and the host both choose at exactly the same
| time. Then 2/3 of the time you've both simultaneously selected
| both boxes with goats, and therefore 2/3 of the time the
| unselected box is the car.
| jncfhnb wrote:
| I think this approach shown in the article is not great. It adds
| additional layers of thinking where none are needed. You don't
| lie. It also doesn't feel like it would help explain variants
| like 5 doors, and only one goat is revealed.
| LukasMathis wrote:
| I would argue that if you start out knowing that you will
| switch, you essentially _do_ lie, because you initially pretend
| to pick a box (or door) that you know you will not end up
| opening.
|
| It is true that this approach adds additional layers of
| thinking, but the problem is that without those layers, the
| solution to the problem is simply not intuitive for the vast
| majority of people. Adding these additional layers helps at
| least some people gain a more intuitive understanding of why
| switching helps.
| jncfhnb wrote:
| Ehhh. Idk. I think it is not great.
|
| A much easier framing on a similar level is simply saying you
| pick a door, and then the host says you can either open the
| chosen door, or both other doors one at a time.
|
| It's the same thing. But if you don't get why the state is
| unchanged even after opening one of the doors I don't think
| you get it with this whole lying thing
| evandale wrote:
| This is a good way of putting it and I think would resonate with
| some people!
|
| I've not been able to get my dad to understand why you should
| switch with 1,000,000 boxes because he refuses to believe the 3
| box and 1,000,000 box case is the same. I can't even convince him
| a 3 and 5 box problem are the same.
|
| I actually think I could convince him switching is correct using
| this strategy because I suspect if he thinks he's tricking or
| cheating at the game he'll be more open to the explanation.
| arithma wrote:
| Are we siblings
| ksaj wrote:
| I think the only way to demonstrate it to lay people is to
| rephrase it as the odds of picking the wrong one right off the
| bat. If it's 2/3 odds that you chose wrong, and I remove one
| card, did that change the odds that you already chose wrong?
|
| The trick is focusing on the 1/3 odds of winning, and not
| mentioning there is an alternate (and correct) way of looking
| at the question - the initial odds of losing.
| Swizec wrote:
| My problem with Monty Hall is that it's not a repeated game.
| Average probabilities don't help _me_.
|
| This is the point Taleb makes in a lot of his books. You are
| not the average probability. You only get one play through
| [life] and the only thing that matters is how the dice fall for
| you. Even a 90% average chance of getting $prize, is still a
| 10% chance of walking away with nothing and you better be ready
| to get that one because you very well might.
|
| The strategy to increase your average probabilities works in
| poker. Because you play multiple hands and even multiple games.
| In Monty Hall, you're just making a guess and no matter how
| clever you are about it, it's still just a guess. There is no
| strategy because the game doesn't last long enough for strategy
| to matter.
|
| Now if you had an iterated monty hall, then yes, bring out the
| maths.
| LukasMathis wrote:
| If I tell you that I'm rolling a die, and ask you whether I
| rolled a six or not a six, offering you 100$ if you guess
| right, you very obviously should say "not a six", even if we
| only play the game once.
| kgwgk wrote:
| No, reading Taleb I realized that if it's not repeated it's
| just a guess and winning or losing is a 50/50 thing. There
| is no strategy because the game doesn't last long enough
| for strategy to matter. [/s]
| HWR_14 wrote:
| It depends on if I look at the die before I decide on the
| "N or not N" options.
| Ste_Tokyo wrote:
| [flagged]
| jdechko wrote:
| I made the same argument the last(?) time a Monty Hall
| article made the front page.
| atq2119 wrote:
| This does not work as an argument against the Monty Hall
| problem.
|
| It does work against simplistic arguments about which of two
| or more probabilistic games one should prefer, because it's
| basically an argument that linearity if expectations doesn't
| necessarily apply in non-repeated games.
| Null-Set wrote:
| It sounds like you embrace the frequentist interpretation of
| probability, but the bayesian aka subjectivist interpretation
| of probability says you can make predictions on the
| likelihood of single events based on your knowledge.
|
| https://en.wikipedia.org/wiki/Probability_interpretations
| jncfhnb wrote:
| No. Wrong. That is not what frequentists would say.
| IanCal wrote:
| > There is no strategy because the game doesn't last long
| enough for strategy to matter.
|
| It makes it much more likely that you win a car. You don't
| see any value in doubling your chances of a big win?
| Swizec wrote:
| > You don't see any value in doubling your chances of a big
| win?
|
| You're improving your _average_ chances. But you're not
| improving _this game_ 's chances. If you play 10 games,
| your average chances matter. If you're playing 1 game,
| reality was already set before you started the game. The
| car doesn't move because you improved your average
| probability.
| IanCal wrote:
| Unless you are arguing that you don't really have a
| choice because your actions are predetermined you are
| absolutely increasing your chances of winning this game.
| kgwgk wrote:
| > You're improving your average chances. But you're not
| improving this game's chances.
|
| If the strategy won't improve your chances in any single
| game in a series of games - how can it improve the
| average chances?
| thrdbndndn wrote:
| I don't get what you're trying to say.
|
| The car doesn't move, but your pick moves, even in this
| single game.
| jncfhnb wrote:
| What do you think "this game's " chance of winning is by
| following the strategy vs not following the strategy?
| andy800 wrote:
| Yes, the whole point is that the car doesn't move. When
| you initially chose, you had a 1-in-3 chance of correctly
| choosing the door with the car. The host is now offering
| you to switch over to, essentially, "all the doors you
| didn't choose", which means 1-(1/3). As you state, you've
| only got one opportunity to play, and the car itself
| hasn't moved, but you can switch your choice. Stick with
| the initial 1-in-3 chance, or go to the other side, i.e.
| 2-in-3 chance?
| JackFr wrote:
| > But you're not improving this game's chances.
|
| You're quite wrong. That is precisely what you are doing.
| jncfhnb wrote:
| Sounds like taleb has undermined your understanding of math
| then.
|
| A 90% chance is worse than a 99% chance. The possibility that
| you get nothing is not an argument to ignore the context.
|
| Consider Russian roulette. Are you seriously going to argue
| that the number of bullets in the gun doesn't matter if you
| only play one time?
| Swizec wrote:
| > The possibility that you get nothing is not an argument
| to ignore the context.
|
| Ah but I'm not ignoring the context. I'm expanding the
| context beyond "fun math puzzle" to "wow this is a game not
| worth playing because there's no edge".
|
| Again, only applies to the non-iterated version. If you
| find an analog of monty hall that gives you many at-bats,
| play away.
| jncfhnb wrote:
| No it doesn't! This is an astoundingly stupid argument.
| Doubling your odds of winning is an edge.
| rahimnathwani wrote:
| "wow this is a game not worth playing because there's no
| edge"
|
| The cost of entry is $0.
|
| The prize is a brand new car.
|
| The optimal strategy gives you a 2/3 chance of winning.
| activitypea wrote:
| I feel like most explanations are overcomplicating it.
|
| By the rules of the game, the option of switching is asking you
| the question "Do you think you chose the correct door out of the
| three?". If you think you did, you don't switch. If you think you
| didn't, by rules of the game, switching will get you to the right
| door. Since you had a 33.3% chance to choose correctly and 66.6%
| to choose incorrectly, betting against yourself and switching is
| probably the correct choice.
| c7b wrote:
| Like a lot of things in math, I think Monty Hall becomes easier
| to see when you look at some extreme cases. 3 doors is the lowest
| amount of doors where the paradox appears, ie no way to go
| smaller. So try a much bigger problem. I like this version:
|
| After Sheherazade has told the calif 1001 stories, she tells him
| that (only) one of them is true, not fiction. She has him guess
| which one, and he picks one (say story #500). Then Sheherazade
| tells him that the true story is either the one he picked, or
| another one, say #312.
|
| I think in this formulation, it is much easier to see that the
| calif would be well advised to switch his guess to #312, as his
| initial guess only had a 1/1001 chance of being correct. Monty
| Hall is the same problem with 3 instead of 1001, but the same
| principle holds.
| jamjamjamjamjam wrote:
| I honestly wouldn't mind a goat.
| Ekaros wrote:
| What this problem always fails for me is that the show host
| doesn't pick randomly. What if there was option of opening the
| car?
|
| Also, does the goat come butchered or do you need to do it on
| stage?
| cman1444 wrote:
| This is an extremely important part of the problem that is
| always glossed over in explanations of the Monty hall problem,
| and it really bothers me that it's ignored because it is
| crucial to determine what is the "correct" decision.
|
| If the game show host is opening a door at random (i.e. it's
| possible that he opens the door with the car behind it OR the
| door you already picked) then the outcome is 50/50 whether you
| switch or not. But if the host knows what's behind the doors,
| and purposefully opens one of the remaining two with the goat,
| then you should switch doors to increase your odds to 66
| percent.
|
| On the above article this is only briefly addressed and not
| explained. In many tellings it is not addressed at all.
| jjnoakes wrote:
| > If the game show host is opening a door at random (i.e.
| it's possible that he opens the door with the car behind it
| OR the door you already picked) then the outcome is 50/50
| whether you switch or not.
|
| I don't think this is true. Assuming you mean that if the
| host opens my box, I switch to one if the other two at random
| with equal probability (even if he opened my box and it was
| the car), and assuming if the host doesn't open my box that I
| stay or switch to the other unopened box with equal
| probability (even if the host opened the box with the car), I
| think the odds are 33% that I win the car, and 66% that I
| don't.
| dylan604 wrote:
| Not "rigging" the decision, would be pointless in the game.
| Yes, it is failed to be mentioned, but if the host revealed
| the car, then what's the point?
| travisjungroth wrote:
| It's also helpful (but not essential) to realize that the
| choice of which door to open has two parts, a rule and a
| decision. If the participant is doing a switch or a stay
| strategy, the decision aspect has no impact on winning.
|
| The rule part is Monty can't reveal the car. The decision
| comes up if the participant chooses the car first. Monty will
| have two doors to choose from to open. But if you have a stay
| strategy you'll win for either choice, and if you have a
| switch strategy you'll lose no matter what. So his decision
| doesn't matter.
|
| This makes it very easy to simulate. It also might lead to
| people to understanding the solution.
|
| If you pick wrong first, you'll see staying always loses and
| switching always wins. The probability of picking right first
| is 1/3, of picking wrong first 2/3. Since switch always works
| when you pick wrong and you pick wrong 2/3 times, go with
| that.
| tromp wrote:
| In this article it's addresses with this sentence
|
| > The show's host then reveals the contents of one of the two
| remaining boxes, but, importantly, always a "goat box."
|
| but it could be made slightly more explicit as:
|
| The show's host, who knows the location of the car, is then
| compelled to always reveal one of the two unselected boxes as
| containing a goat.
| andy800 wrote:
| As far as solving the problem in an academic sense, your
| explanation is clearer. However, one of the things that
| makes the whole thing fascinating is that the actual
| players are not told this specific information, that the
| host knows. If you watch the show religiously it might dawn
| on you, hey, he must not be picking the preview door
| randomly because he has never accidentally revealed the car
| -- but that's beyond most people's observation.
|
| In practice, the revelation of one door is likely to make
| the player even more committed to the original choice --
| the odds (seemingly, not actually) just went from 33% to
| 50%, momentum is good, I'm feeling lucky, no way am I going
| to switch now!
| drexlspivey wrote:
| If the host picked randomly then there would be a 1/6 chance
| that the show would end rather anticlimactically and you would
| be indifferent in switching. Now this 1/6 becomes your edge
| turning your 1/3 to 1/2 after the goat is revealed.
|
| EDIT: 1/3 not 1/6
| IanCal wrote:
| 1/3 not 1/6. There are six outcomes of you picking a door and
| the host opening one of the remaining doors. In two of them
| you pick a goat the first time and then the host reveals the
| car.
| [deleted]
| jdechko wrote:
| I wonder if a decision tree would help more people since
| it's possible to visualize every possible outcome of every
| possible game.
| IanCal wrote:
| That's the thing that makes it click for me, and why
| there's a difference if the host is random or not.
| IanCal wrote:
| If they pick randomly, there's no benefit to switching. No harm
| either.
|
| If it's random, 1/3 of the time you're right on your first
| guess and switching doesn't help. 2/3rds of the time you're
| wrong - if the host knows and deliberately doesn't open the
| door showing the car switching means you win. If however he
| opens randomly, half of those cases show the car and you can't
| win - so the breakdown is 1/3 switching wins, 1/3 switching
| fails and 1/3 the car is shown and you lose regardless.
| atum47 wrote:
| I've made this "intuition" for the Monty Hall into a game [1] so
| I can test it for myself.
|
| I remember being intrigued by the notion that you're better off
| switching.
|
| 1 - https://victorribeiro.com/montyhall/
|
| Keep in my that this was made a long time ago, I wasn't even a
| programmer back then, I was more a webmaster.
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