Monty Hall problem explained visually

http://blog.vctr.me/monty-hall/

by vicapow • 13 years ago

55 77 13 years ago

blog.vctr.me

jbellis• 13 years ago

The best explanation I've seen of the Monty Hall problem is still Oscar Bonilla's: http://oscarbonilla.com/2009/05/the-monty-hall-problem/

sanoli• 13 years ago

Best explanation I've seen was using statistics: simulate the problem 1000 times. Look at results when you switch and when you don't switch doors. Or do it in your head: If I switch doors, I will win with 2 out of 3 doors. Otherwise, I'll win only with one door.

bsirkia• 13 years ago

Very cool way of illustrating it. It looks like you just used iframes for the animations so they were in separate pages with distinct stylesheets, JS resources, etc., why did you do that rather than having the Jquery animations happen in the main document? Was it just to keep things nicely modular?

Scriptor2• 13 years ago

One approach that really helped me out was to extend the problem from three doors to a million doors.

You're allowed to pick one door, out of a million. Then, the host opens 999,998 of the other doors which he knows have a goat. This leaves the closed door you picked and one remaining closed door.

So, there's now two possibilities. Out of one million doors you somehow just happened to have guessed right and picked the right one. Or, you're wrong. If you're wrong then the correct door would be the other closed one.

jiggy20114• 13 years ago

Does it matter whether or not the host knows? I don't see how it changes the odds in any way, unless of course he opens the door with the car (very likely with 1 million doors).

pbreit2• 13 years ago

The host knows where the car is and so leaves it hidden. If some one shows you this with even just 10 doors, it will be immediately and patently obvious that you must switch.

This is, indeed, the most effective way to explain it, imo.

ghaff• 13 years ago

It does. If the host doesn't know, there's a 1/3 probability that the game will end (you won't get a chance to switch) if the host picks the car. If this doesn't happen, it's 50:50 whether you keep your door or switch to the remaining unopened door.

Rudism• 13 years ago

If the host knows, you should switch. If he doesn't know, it doesn't matter--you have a 50/50 chance of winning either way.

ds9• 13 years ago

The fact that you asked this question shows what's wrong with the supposed "problem". In the usual presentation, it's a "trick question" because the person presenting it omits the crucial information that the offeror is intending to act adversely to the decsion making subject.

Without that information, the interlocutor naturally assumes that the offeror opens doors randomly - in which case the fact that he knows what's behind each door is an irrelevant "red herring" and the correct conclusion is different.

In the random-opening case the intuitive conclusion is right - that the agent (offeree)'s decision does not affect the odds. But if the question-poser revealed that the offeror is trying to maneuver the offeree into the result that's less valuable for the offeree, then the supposed mystery of the whole thing evaporates and the correct conclsion is obvious.

coldtea• 13 years ago

Does it matter whether or not the host knows? I don't see how it changes the odds in any way

Of course it matters.

The host uses his knowledge to AVOID opening the door with the car (and thus ending the game prematurely).

So he essentially gives out this information: "the door with the car is one that I, the host, haven't opened".

scrabble• 13 years ago

Yes, it matters. The host opens every door other than the one with the car - or one with the goat if you picked the car.

The host is not opening just one door.

bluedevil2k• 13 years ago

I do the same thing by expanding it out to more doors, but when talking about it or explaining it to others (great cocktail party discussion btw), I use 100 doors, to make the percentages easy to follow.

Example - there are 100 doors, you pick 1 that has the grand prize. There's a 1% chance you picked the prize on your selection. Monty Hall eliminates 98 of the doors, meaning there are 2 left. There's STILL a 1% chance you picked the door with the prize, but that means there's a 99% chance the only remaining door has the prize. Obviously, you should switch.

Scriptor2• 13 years ago

Ah, I used one million because it really hammers the scale of it in, but I like that with 100 you can easily track the percentages.

Jtsummers7• 13 years ago

Another way is to take a step back and forget about revealing the goats. You have your initial selection (using your numbers) with a 1% chance of picking the prize. Now you're presented the option to leave it shut and open all other doors keeping the prize if it's in any one of them. So if the prize has equal odds of being in any given door, then opening 99 of them gives you a 99% chance of finding the prize versus the 1% chance of only opening one door. This has helped some people I've talked to realize that the odds the prize is in that remaining door (in the proper Monty Hall problem) is actually the sum of the odds it's in all the doors you didn't select.

EDIT: I've also demo'd this using 3 cards (an ace and two others), but I like the idea of demonstrating with a larger scope. Call the ace of spades the prize, let the player select one card from the deck at random. Now they can keep that card or scour the remaining 51 cards for the ace of spades. A good tactile demonstration that they have a 1/52 chance (~2%) of having selected correctly the first time versus ~98% that they were wrong.

jwr• 13 years ago

What I find hard to undestand is why I should switch instead of randomly choosing again?

JamesArgo• 13 years ago

This is the most intuitive description I've found:http://lesswrong.com/lw/2b0/bayes_theorem_illustrated_my_way...

dopamean• 13 years ago

This is a great explanation I hadn't thought of before. Thanks.

frankc• 13 years ago

The way I explain is to expand it but also change the frame of reference to so its clear that the host is an adversary. Imagine we play a game called "who has the Ace of Diamonds"? I deal you one card face down and I deal me 51 cards. I look at my cards and then choose 50 of them to show you, none of which are the Ace of Diamonds. Do you want to keep your card or take the one I have not turned over?

basch2• 13 years ago

I will never understand how one million doors is easier to understand than 3. Switching doors means you get two doors out of three doors. 66% > 33%. 66/33 illustrates the benefit much better than 1 vs 2 out of a million.

I think what confuses people is that they think a random door is revealed, whereas the problem says a goat is specifically chosen before the offer to switch.

pbreit2• 13 years ago

For this to really work, it must be clarified up front that Monty is going to give you the opportunity to switch every time the game is played.

bernardom2• 13 years ago

Like everyone who encountered this in their intro stats class, I love this problem. And it bothers everyone the first time.

Are there other problems like the Monty Hall problem, perhaps in other disciplines?

DanielStraight• 13 years ago

This could be the most bothersome problem of all:

http://www.airplaneonatreadmill.com/

ksmiley• 13 years ago

I know of one other mathematical fact that matches the Monty Hall problem in the loudness of its opponents: 0.999(...) exactly equals one. Despite a number of proofs existing, you can still find ten large pages of arguments on Wikipedia's talk page.

http://en.wikipedia.org/wiki/Talk:0.999.../Arguments

tomgagnier• 13 years ago

This problem is known as "Restricted Choice" in the card game bridge. It occurs when deciding whether to play for the drop of a missing honor card or finesse. See http://en.wikipedia.org/wiki/Principle_of_restricted_choice for more.

drblast2• 13 years ago

The two envelopes paradox is a good one:

http://consc.net/papers/envelope.html

And another that I've figured out the "right" answer to but think there might be more to it is this:

Suppose I set up a two-player game as follows:

  1. One player is the "odd" player, the other is
  the "even" player.

  2. Each player may choose a single number per turn,
  either one or two.

  3. The result of each turn is the sum of the two players'
  choices.  For example, if the even player picks 1 and the
  odd player picks 2, the result of that turn is 3.

  4. If the result is odd, the even player pays the odd
  player that amount, and vice-versa.  In our example 
  result of 3, the even player would pay the odd player 3
  dollars.

Question: Is this game fair? If not, is it possible to give either player a per-turn stipend to make the game fair? If so, how much is that amount?

tomca32• 13 years ago

I'll bite.

I guess that the odd player can always choose 1. In that case he can either win 3 or lose 2. Making that the favorable choice for the odd player.

Even player could always choose 2 in which case he can either win 4 or lose 3. Making this the favorable choice for the even player. However, if both players play their favorable choice, odd player always wins: 2 + 1 = 3.

I'm not sure if I got this right, but that's my thinking so far...

jcampbell12• 13 years ago

I like this question:

Flip a coin. If it is heads, flip again and answer Yes for heads and No for tails. If the first coin is tails, answer the question: "Have you ever cheated on your spouse?"

Now, what is the probability the person is a cheater given they answered yes?

bernardom2• 13 years ago

Wait, is there an interesting answer to this?

I'll label the four outcomes 1Y, 1N, 2Y, 2N. Let X = probability person cheated on spouse.

If I flip heads first, I get yes or no equally. So the first two outcomes are: 1Y(50%) or 1N (50%).

If I flip tails, I answer 2Y (X) or 2N (1-X).

So the distinct pdf is: 1Y (25%) 1N (25%) 2Y (0.5X) 2N (1-0.5X)

And combined, Y = 0.25+0.5X

What am I missing? (Wait, oh, is this a stupid joke about whether they lied about cheating on their spouse? Dammit.)

mildavw2• 13 years ago

Did I read somewhere that this technique is actually used to gather survey data where the subject may have reason to lie? They can answer yes but it remains plausible that it was because of two coin flips coming up heads so they are not individually implicated. When you aggregate the data over many subjects, however, you have a better idea of how many actual cheaters there were than if you asked directly.

Anyway, here is the code below: http://jsfiddle.net/nQ3Gu/

  var runs = 1000,
    honest_admissions = 0,
    automatic_admissions = 0;

  function flip() {
    return Math.random() < 0.5 ? 'h' : 't';
  }

  for (var i = 0; i < runs; i++) {
    if (flip() == 't') {
        honest_admissions++;
    } else if (flip() == 'h') {
        automatic_admissions++;
    }
  }
  document.write('The "yes" is honest ' + honest_admissions / (honest_admissions + automatic_admissions) + ' of the time.')

2/3 of yeses are honest and 1/3 didn't have to answer the question!

basch2• 13 years ago

It bothers everyone because the teacher never _stresses_ the fact that the host doesn't open a door at random. It's purposely deceptive.

ryanthejuggler• 13 years ago

Heh, Monte Carlo simulation of Monty Hall problem. Mildly amusing... if only we could work a Monty Python reference in there somewhere!

Watching this play out really kills any remaining gut response that would say "it's the same either way!"

gknoy• 13 years ago

Just make it show alive/dead parrots, or spam, instead of goats. ;)

"Our table has spam, spam, spam, spam, eggs, spam, spam, and spam. You've chosen your dish, now let's see what else was on the table...."

;)

mathattack• 13 years ago

I like the simulation at the bottom. Sometimes it's best to just say, "The math gets confusing, let's run this a couple times and see what happens."

Kliment• 13 years ago

The best explanation I've seen was in series 3, episode 3 of Jame May's Man Lab, where they repeatedly iterated the problem and showed that when repeated enough times, the frequency of winning/losing tends to a different value with a change of door than without (by having the presenters open booby-trapped beer cans and either get splashed with beer or not)

mactitan• 13 years ago

some background: http://en.wikipedia.org/wiki/Monty_Hall_problem Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong

(per above post At least this made it to stats intro class)

intellegacy2• 13 years ago

The rules of the problem aren't explained very well. most of the contention probably is based on unclear boundaries of the game.

zcarter• 13 years ago

I always explain this problem with stars: I'm thinking of a star in the night's sky... guess which one.

Alright, good guess, but would you consider changing? You can either keep your original guess, or switch to this one other star I'm pointing to now.

Everyone switches.

brainburn2• 13 years ago

That doesn't explain anything except that people will be influenced because you pointed to another star.

CWuestefeld• 13 years ago

I liked the simulation, and it reminds me of a tangentially related probability issue from early in my career.

While still in college, I had a part-time job testing a spreadsheet product (back in the days when there were other commercial choices). One of the things we testers did to keep ourselves entertained was that we each picked a casino game, and implemented that game in spreadsheet macros. My game was blackjack, but another guy did craps.

Just for kicks, the guy who built the craps game re-wired it as an autoplay simulation, and left it playing against itself overnight. When we came back in in the morning, we found that the virtual player was rich, which seemed odd. So the next night we left it running again, and again the following morning, the player had made a bundle.

It turns out that my coworker had accidentally discovered a bug in the random number generator: the distribution of the results was a bit non-uniform.

error54• 13 years ago

Nice explanation. Small critique: I think this page would be better if the animations could be played manually rather than on a continuous loop which is quite distracting. Also, maybe add in a bit where the user can play the game themselves to see the logic work out.

Something like this was useful when I was first learning about the Monty Hall problem.

http://www.grand-illusions.com/simulator/montysim.htm

jbindel• 13 years ago

There's no need for the animations at all. Besides not letting the reader work through the problem at her own pace, the animations hide the previous steps.

LanceH• 13 years ago

I've usually found the problem poorly worded (probably intentionally). An assumption that Monty knows which door is which is required but often not stated. If Monty doesn't know which door is which, the odds are 50-50 like you'd imagine.

cma• 13 years ago

He just has to not open a door with the car behind, it doesn't matter whether intentional or not. If he forgot his intentionality but takes the exact same actions by chance, the stats don't change for the player.

ubercow132• 13 years ago

This isn't right though, it makes all the difference whether he acts knowingly or happens to pick a goat; it's a 50/50 chance on changing if he chose a goat by luck (see other posts ITT)

taternuts• 13 years ago

That's a good looking demo, quite smooth. Implementing this was actually a recent DailyProgrammer challenge (http://www.reddit.com/r/dailyprogrammer/comments/1qdw40/1111...)

officemonkey• 13 years ago

In another 50 years, no one is going to know who "Monty Hall" is.

Except for this "problem" which people will still be arguing about.

damontal• 13 years ago

got into a huge fight with a family member over this. i even programmed a simulation and showed it run over 1000 tries and he still refused to budge from the 50% answer.

stefan_kendall• 13 years ago

Play a cash game with cards for a while. Dollar bets, dollar return.

Play 50 games every time you see him until he gets it.

x0n• 13 years ago

I have a much easier explanation than this: By offering you the switch, the host has given you two doors to open - it's irrelevant that he opened one for you - so, it comes down to picking one door (the one you're on) or two doors (of which he has opened one for you, again, irrelevant - the effect is the same) ergo, it's 1/3 or 2/3.

allthatglitters• 13 years ago

Given two choices (equal probability; redundant?) - which is the case once you have two doors - switching cannot change the odds. Have someone flip a coin. Call it. But before the flipper looks at the coin - (whether he KNOWS the result or not) he asks you to consider switching. Explain how the option of switching changes the probability of a correct or incorrect call. Inference?

alco2• 13 years ago

I can't see how your generalization of the problem to more doors is correct.

In the classical problem with 3 doors, if you pick a door with a car, the host will only open 1 other door.

In your demo, if I use 40 doors and pick one, the host opens all other doors but one. Why?

If the one of those 40 I pick has a car, will the host open all remaining doors? That doesn't seem to be the case for the 3-door variant.

Jtsummers7• 13 years ago

In the Monty Hall problem the door with the car is never revealed until the end. In the 3 door version if you've selected the prize door, the goat revealed is selected at random. If you've selected a goat, the other goat is revealed and the car remains hidden behind the remaining (at this point unselected) door. So with 40 doors, after your selection 38 goats would be revealed. It may be that you selected the car by chance (1/40 chance), in which case 38 of the 39 remaining doors would be selected at random (as it doesn't matter). If you selected a goat only the other 38 goat doors would be opened (but the contestant can't tell the difference between these circumstances).

alco2• 13 years ago

Still, why? Why open 38 doors? I don't think this emphasizes the difference in probability, it just creates artificial bias.

Let's analyze the classical problem. After you pick a door, one of the remaining two doors is opened.

With more doors (say, 11), it wouldn't make for an interesting show if you opened 9 more doors after the first one. A more reasonable approach would be to open one additional door and let the player choose to switch or stay with their choice each time. But that would only complicate the explanation, of course.

Let's dissect the 11-door case. We will do the following: the player picks a door, then the host a) opens one other door; b) opens 50% of the remaining doors; c) leaves one door closed.

IMO, the (a) case is the closest generalization of the original problem. You have a 1/11 chance to pick the car door on your first try. Then, one other door is opened, and if you switch to another random door, the best case probability of you getting the car is 10/11 * 1/9 = 10/99. This number is slightly bigger than 1/11.

In the (b) case you get a better result when switching, but to me that just looks like adding artificial bias. Obviously, the (c) case looks best because you've removed so much choice by leaving only one other door closed. Adding more doors does one thing: improves the probability of switching. That's good if this is what you're aiming for. But this turns it into a completely different problem.

So, you can see that the result depends on what you want to get. If you try to stay faithful to the original problem statement, adding more doors actually makes the benefit of switching less obvious.

In summary, I'm convinced that 3 is not an accidental number in the original problem statement. Whoever came up with it, they knew how to remove space for speculation by providing the only reasonable thing to do: open 1 door (or 50% of the doors) and leave 1 other door closed.

brainburn2• 13 years ago

To get people thinking I generally have them agree there are 9 possible games to play, so the final win-chance cannot be 50%

jcampbell12• 13 years ago

No one thinks the final win chance is 50%. They mistakenly think it is 1/3 with either strategy.

intellegacy2• 13 years ago

I suspect that much of the contention around this famous problem is based on the original boundaries / rules of the problem being unclear.

The problem as stated sometimes leaves out crucial information that affects the outcome.

Example:

Game show, 3 doors. Contestant chooses one door. The host opens another door, which has a goat. Do you switch?

In this situation, it is not explained whether the host randomly picked one of the two remaining doors to open, without knowledge that one of them would contain a goat. This is the key missing information. If the host ALWAYS reveals the goat, then it is beneficial to switch as 2 out of 3 times the contestant wins by switching.

However, if the host is simply opening another door at random, and the door happens to contain a goat, then there is no advantage to switching, as the odds are 50/50.

edit: it appears that from the original version of the problem stated in Parade magazine, the implication is that the host knowingly reveals a goat.

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"

toddmorey2• 13 years ago

Here's how it helps me to look at it. As others have said, you can play with the numbers, but the mechanics are basically this:

First step: you are choosing one option from all available choices. Second step: you are now betting whether your original choice was the correct one.

By consciously narrowing down all the remaining options to one, the host is essentially asking you: "Do you think you guessed right?" As long as there are more than two available options at the start, the odds are greater than you guessed wrong.

As others have said, the important dynamic here is that the host _knows_ where the prize is and proceeds to open all the other doors, leaving only yours and one remaining closed. That remaining door then symbolizes an aggregate of all the doors you didn't select.

dllthomas• 13 years ago

Consider the version where Monty does not know what door the car is behind. He randomly opens one of the other two doors, and (per the rules) if he reveals a car, game over too bad. This time, he happened to reveal a goat. Why is the question not "did you guess right originally?"?

I'm quite convinced that it is different if the host knows versus if he does not, but I've never quite been able to put my finger on why in an intuitive way.

Jtsummers7• 13 years ago

Whether we arrive there by chance or by deliberate action, once we know that the open door has a goat, the odds for winning by switching remain the same. The odds only change for the overall game (where there's now a 1/3 chance we lose/win when the host reveals the car). If the goat is revealed, we're still in the original game: 2/3 for switching, 1/3 for staying.

EDIT: Leaving my wrong answer, but bvk is correct.

toddmorey2• 13 years ago

That's a great question! I think the person who can articulate that intuitively should win the car!

My stab at it: If he doesn't know, then he's essentially another contestant, and his odds are the same 1 in 3 as yours. You've made selections one after the other, but since your selection has yet to be revealed as either right or wrong, you are essentially just picking different options from the same scenario. There are exactly equal chances that you won the car, he flubs and reveals the car, or he reveals a goat.

bvk• 13 years ago

Imagine the host doesn't know. If you guess right, he opens a door with a goat behind it and you find yourself in the classic situation.

But if you guess wrong, there's a 50% chance that the host reveals a car by accident.

So the probablities look like:

33%: you guess right, the host reveals a goat 33%: you guess wrong, the host reveals a goat 33%: you guess wrong, the host reveals a car

So, of the situations where the host reveals a goat, the car will be behind your door 50% of the time.

If the host knows and avoids the goat, the probabilities look like this:

33% you were right; the host reveals a goat 33% you were wrong; the host reveals a goat behind door B 33% you were wrong; the host reveals a goat behind door C

So in the situations where the host knows about and avoids the goat, 66% of them have the car behind the other door.

In the case where the host doesn't know about the car, picking the car on your first choice makes it more likely for the host to reveal a goat. That's the difference.

yconst• 13 years ago

Looking this up on Wikipedia reveals it's relationship with the Three Prisoners Problem[1] as well. The solution presented in that case helped me in reasoning about it.

[1] http://en.wikipedia.org/wiki/Three_Prisoners_problem

mercer• 13 years ago

I've tried to think of the best way to make this obvious. This is what I came up with, and I'd like to hear if it's accurate:

Contestant picks a door out of three. 1/3rd chance he's got the car. Then he is given the choice to either stick with his door, or do a coin-toss where you have to pick heads.

In this case, it seems obvious that option for the coin-toss is the better option, because it gives you a 50% chance of winning.

The confusing part of the problem is that you feel like you're doing one thing, where really it's two independent games where one clearly offers a bigger chance of winning.

That said, some people here have mentioned that it's not actually a 50% chance if you switch doors. So maybe my explanation is incorrect? And if it's incorrect, is it still correct enough to show that switching doors (i.e. opting for the coin-toss) is the better option?

Monty Hall problem explained visually

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