The Monty Hall problem (video) | Probability | Khan Academy (2024)

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  • kradler97

    12 years agoPosted 12 years ago. Direct link to kradler97's post “Isn't it always going to ...”

    Isn't it always going to be a 50/50 chance? If #3 is eliminated you are just choosing between #1 and #2. Switch or no switch it is still a 1/2 chance. I don't understand how your odds increase by always switching. Thanks for anyone's help

    (207 votes)

    • jonat3

      12 years agoPosted 12 years ago. Direct link to jonat3's post “That's the initial trick....”

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      That's the initial trick. You are tricked into believing here that the 3rd curtain/door doesn't matter, since it's seemingly "eliminated" from the equation. This is an illusion and is meant to fool you into thinking the 3rd curtain doesn't matter. You HAVE to consider the 3rd curtain as well. Why? Because you made your pick BEFORE the 3rd curtain got eliminated. It would only be a true 50/50 chance if this 3rd curtain was eliminated BEFORE you made your pick and not after.

      To see this more easily, you just have to exaggerate the numbers to the extreme. Suppose you had 1000 doors and only one door had the prize. The chances of picking the right door would be extremely small, right? Ok, you pick a door and afterwards, the host eliminates 998 doors, leaving the door you picked and a second door. Now,you tell me if you think it likely that your initial pick would just HAPPEN to be the right door chosen out of 1000 doors. It sure as hell wouldn't be 50%.

      (624 votes)

  • Carley Fletcher

    12 years agoPosted 12 years ago. Direct link to Carley Fletcher's post “Could this question also ...”

    Could this question also be understood by multiplying them as dependent variables? Given that you will always switch. P(winning if you first choose wrong)= 2/3 * 1/2 = 1/3. P(winning if you first choose right)= 1/3 * 1/2 = 1/6. The 1/2 probability comes from the two remaining doors, either you have the correct one or the wrong one. It does make sense to switch because you are more likely to choose the incorrect door at the beginning.

    (39 votes)

    • leo.e.walsh

      12 years agoPosted 12 years ago. Direct link to leo.e.walsh's post “What makes the Monte Hal ...”

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      What makes the Monte Hal Problem interesting in that the host knows, and will always chose the goat. So I tried to break it down by probability of winning the car behind doors A, B or C...

      A. Initial Conditions:
      1) P(A) = P(B) = P(C) = 1/3
      2) P(A)+P(B)+P(C) = 1 (or 100%). This is just a way of stating that the prize can be anywhere.

      B. You Choose 'A' -- can be any door, but we can call any door you choose 'A'
      ...This leads to the following probabilities:
      1) P(A) = 1/3
      2) P(not A) = 2/3.
      * Also note that P(not A) = P(B)+P(C)

      C Monte Opens 'B' -- can be any door, but we can call any door he chooses 'B'
      ...Monte will never open the door with the prize, or he will be fired. So we have the following certain probability:
      1) P(B) = 0/3 = 0

      D. Monte says, "Do you want to switch?"
      ...So, let's look at our certain probabilities:
      1) P(A)=1/3 (From section A)
      2) P(not A)=2/3 (From section B)
      3) P(B) = 0 (From section C)

      E. Using the last 2 certainties, we end up with the following
      1) P(not A) = P(B)+P(C). So, substituting, we end up with
      2) P(B)+P(C) = 2/3
      3) Since P(B) = 0, Then we can substitute '0' for P(B), which gives us
      ... 0+P(C) = 2/3.
      ...This simplifies to
      *4) P(C) = 2/3

      F. So it makes sense to switch...

      Notice that Monte's new information tells you nothing about A. Nor does it change the initial odds. It only gives you information about * P(not A)*.

      I really hope this helps. The notation really took work to put together. It is more mathematically rigorous than Sal's demonstration, but I was a mathematics major at university. But I love his explanation for the simplicity...

  • Brian D

    12 years agoPosted 12 years ago. Direct link to Brian D's post “Also why in "deal or no d...”

    Also why in "deal or no deal," with 30 cases, if you could get it down to your case and only one other case, you should switch. You get a 96.67% chance of getting your million dollar case. Of course it would take quite a bit of luck to get that far, but nevertheless the theory behind the situation is identical.

    (13 votes)

    • narcolepticned

      12 years agoPosted 12 years ago. Direct link to narcolepticned's post “The difference is that th...”

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      The Monty Hall problem (video) | Probability | Khan Academy (17)

      The difference is that the 28 other cases are not all eliminated for the player by the host. I think that the host of "Deal or No Deal" doesn't even know which cases have the good prizes. Ultimately the situation is different.

      (26 votes)

  • Syomantak Chaudhuri

    12 years agoPosted 12 years ago. Direct link to Syomantak Chaudhuri's post “Why is the probability 2/...”

    Why is the probability 2/3?Lets take it this way,the hosts shows you the door that doesn't have the prize and then you are left with two possible outcomes.One of them has the prize.Therefore,P(the outcome would be req. prize)=1/2

    (4 votes)

    • Matthew John Ting

      11 years agoPosted 11 years ago. Direct link to Matthew John Ting's post “No, Syomantak, you're cha...”

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      The Monty Hall problem (video) | Probability | Khan Academy (22)

      No, Syomantak, you're changing the situation in an essential way. In the case you've given, you are only choosing once. In the actual Monty Hall problem, you choose twice, and the probability of the second choice is affected by the first. Dependent probability.
      Think of it this way:
      if you choose the correct door the first time, switching will always make you lose.
      If you choose one of the two wrong doors the first time, switching will always make you win.
      What's the probability of choosing the wrong door the first time?
      Yes, that's right, 2/3.
      Another expansion of the thought that several people have brought up:
      instead of 3 doors, think of 10. Monty will do a similar thing once you choose for the first time, but, instead of revealing 1 goat, he will reveal 8.
      you have the same scenario:
      If you choose the right door the first time, switching will make you lose.
      If you choose any of the 9 wrong doors the first time, switching will always make you win.
      What's the probability of picking the wrong door the first time? 9/10
      So, what's the probability of winning if you always switch? 9/10
      And, what's the probability of winning if you never switch? 1/10.

      (44 votes)

  • Ian

    12 years agoPosted 12 years ago. Direct link to Ian's post “The 2/3 probability when ...”

    The 2/3 probability when always switching does not make practical sense. Knowing ahead of time that a wrong answer will be identified for you, that just eliminates one pick from the possibilities, so the probability is always 50% that you will pick correctly, whether you stick to your initial pick or switch. It appears to me that they get the 2/3 probability by mixing the initial probability of picking the first time correctly with the altered probability once you know which one is incorrect.

    • ahenzinger

      12 years agoPosted 12 years ago. Direct link to ahenzinger's post “That would be correct if ...”

      The Monty Hall problem (video) | Probability | Khan Academy (26)

      That would be correct if it was independant probability. But, this is dependant probability, because if you should switch or not depends on if you picked the car or one of the goats in the initial pick. This means that the second pick depends on the first pick.
      There is a higher chance of picking a goat in the initial pick, so their is a higher probability of winning, if you swith in the second pick.
      Thats why there is not a 1/2 probability, but a 2/3 one.
      Hope this helps you understand :)

      (17 votes)

  • jacob.c.wan

    11 years agoPosted 11 years ago. Direct link to jacob.c.wan's post “i'm confused."if you pi...”

    i'm confused.

    "if you pick wrong the first time, you should switch." / "if you picked wrong, and switch, you will always win."

    it seems like the condition for these probabilities is "if you picked wrong." but the contestant will NEVER KNOW if they picked wrong - we are under the assumption that the contestant DOESN'T know they picked wrong the first time.

    he doesn't know what he picked at all.

    and so, with that condition out the window, doesn't this fall apart? if someone could explain, that'd be great =/

    (9 votes)

    • ArDeeJ

      11 years agoPosted 11 years ago. Direct link to ArDeeJ's post “The win isn't guaranteed ...”

      The Monty Hall problem (video) | Probability | Khan Academy (30)

      The win isn't guaranteed even if you switch, but more likely.

      (12 votes)

  • Ralph co*ckburn

    7 years agoPosted 7 years ago. Direct link to Ralph co*ckburn's post “I remember having this ar...”

    I remember having this argument with one of my high school math teachers. I finally convinced him by scaling this problem up to 1 million doors. If you start with 1 million doors, and are asked to pick one, then 999,998 are opened, only leaving your door and one other surely the other door is far more likely to be the one.

    When it comes to most math puzzles my initial instinct is to scale it up and see if there's an obvious result, it really helps sometimes.

    (17 votes)

  • Danilo Souza Morães

    8 years agoPosted 8 years ago. Direct link to Danilo Souza Morães's post “i think its easier if you...”

    i think its easier if you think there are 100 doors and only one of them has the prize. Then you pick one door and Monty Hall opens all the other 98 doors leaving only your door and another one closed. What is the probability that you picked the right door out of 100 doors? It is very unlikely so you will want to switch. The monty hall problem has 3 doors instead of 100. It is still more likely that you pick a goat.

    (14 votes)

    • Zahara Sherif

      7 years agoPosted 7 years ago. Direct link to Zahara Sherif's post “The probability of winnin...”

      The probability of winning is 1/3 because there are 3 doors and 2 doors are wrong and 1 door is right so the chance of losing is higher than the chance of winning. You said if a person picks door 2 the Monty Hall will close door 1 and 3. If a person picks door 1 which is wrong the Monty Hall will close door 3 and give you chance to switch to the right answer, so it means they want always people win the prize. What is the point of playing the game and there isn't losing?

      (0 votes)

  • Paul Mackilligin

    10 years agoPosted 10 years ago. Direct link to Paul Mackilligin's post “An old puzzle and a good ...”

    An old puzzle and a good one. I enjoy asking maths teachers and maths graduates this one because they almost always get it wrong! It is deeply counter-intuitive that what happened in the past - your past decision - can change the chances of there being a car behind either of the two doors in front of you in the present. I mean, whatever happened in the past, right now you've got two doors, and if the car is not behind one then it is behind the other. Only two possibilities so a 50/50 chance, right?

    You can do the maths and STILL not believe it because it's so counter-intuitive.

    The way I got over the counter-intuitive aspect of it was the same as the way Peter Collingridge does below (or above?) which is to do a thought experiment and imagine that there is a huge number of doors, and all but two are opened by the game show host, leaving the same situation as before, namely that there are two doors, with a goat behind one and a car behind the other. In the case of a hundred doors or a thousand doors it is intuitively obvious that, for all practical purposes, you may as well assume that you intially chose a goat, and that you are therefore almost always going to win a car if you switch. Once you see that, you can mentally reduce the number of doors until there are just three, but you will still retain the intuition that you are better off switching.

    In Philosophy this method of reasoning is called 'reductio ad absurdum' (reduction to the absurd) and it's a really useful tool in all kinds of situations.

    Great video.

    (8 votes)

    • Edwin Rios

      8 years agoPosted 8 years ago. Direct link to Edwin Rios's post “The Monty Hall problem wa...”

      The Monty Hall problem was shown on the movie 21, where door no. 3 was first revealed as a goat, instead of a new car, but the contestant is given a second opportunity to change his initial answer, as part of an exercise related to variable change.
      I had to think hard about this short video to reach onto my own understanding, since it was not explained clearly on the movie. The premise of winning the car after switching from door 1 to door 2 doesn't seem intuitive, or even logical, being that you only have a 50% chance after door 3 was revealed as a goat. However, when door 3 was reveled as a goat, you already won 1 selection. You didn't get the car, but you did win a selection of, lets say, eliminating a door that had a goat. Now you are offered a second selection. Since you already eliminated 1 door with a goat, does that mean that over time you will hold a 100% record for eliminating all doors with goats? Maybe not, and since you were able to eliminate the first one already, there is a chance that the second option may not allow you to eliminate the second goat with the same choice. So you have to think that this could also be defined as a game of eliminating goats. This does not guarantee that the car will be behind door 2, but given 2 opportunities to eliminate goats, chances are you can't eliminate them by maintaining the same choice always. So you are actually working on the premise of being 50% right and 50% wrong with the same choice. That is where the logic is, far from the emotion of believing that given a second opportunity, you will always be right with your first choice.

      (5 votes)

  • NachoMan008

    a year agoPosted a year ago. Direct link to NachoMan008's post “Hi Sal - Aren't the odds...”

    Hi Sal - Aren't the odds of winning and losing precisely the same whether you stay put or change? The odds of winning initially are 1/3, losing 2/3. Obviously, only once will the show tip its hand, so, in the second pass, you're shown one of the losing choices and the odds of picking a winner from the remaining two choices becomes 1/2. The odds of picking a loser from the remaining two choices is also 1/2.

    (4 votes)

    • Ananta

      a year agoPosted a year ago. Direct link to Ananta's post “No, the odds of picking a...”

      No, the odds of picking a winner from the remaining two choices becomes 2/3 because you have a 1/3 chance of picking the right door the first time, making the chance of picking the right door of the other two doors 2/3, but then the host reveals one of the doors that doesn't have a fantastic prize in it making the chance of picking the right door concentrated on the last door.

      (3 votes)

Video transcript

Let's now tackle aclassic thought experiment in probability, calledthe Monte Hall problem. And it's called theMonty Hall problem because Monty Hall was the gameshow host in Let's Make a Deal, where they would set upa situation very similar to the Monte Hall problemthat we're about to say. So let's say thaton the show, you're presented with three curtains. So you're the contestant, thislittle chef-looking character right over there. You're presented withthree curtains-- curtain number one, curtain numbertwo, and curtain number three. And you're told that behindone of these three curtains, there's a fabulous prize,something that you really want, a car, or a vacation, or sometype of large amount of cash. And then behind theother two, and we don't know which theyare, there is something that you do not want. A new pet goat or anostrich or something like that, or abeach ball, something that is not as goodas the cash prize. And so your goal is to tryto find the cash prize. And they say guesswhich one, or which one would you like to select? And so let's say that youselect door number one, or curtain number one. Then the MonteHall and Let's Make a Deal crew will make it alittle bit more interesting. They want to show youwhether or not you won. They'll then show you oneof the other two doors, or one of theother two curtains. And they'll show you oneof the other two curtains that does not have the prize. And no matterwhich one you pick, there'll always be atleast one other curtain that does not have the prize. There might be two,if you picked right. But there will always beat least one other curtain that does not have the prize. And then they willshow it to you. And so let's say that theyshow you curtain number three. And so curtain numberthree has the goat. And then they'llask you, do you want to switch to curtain number two? And the question here is,does it make a difference? Are you better off holdingfast, sticking to your guns, staying with the original guess? Are you better off switchingto whatever curtain is left? Or does it not matter? It's just randomprobability, and it's not going to make a differencewhether you switch or not. So that is the brain teaser. Pause the video now. I encourage youto think about it. In the next video, we willstart to analyze the solution a little bit deeper, whetherit makes any difference at all. So now I've assumedthat you've unpaused it. You've thought deeply about it. Perhaps you havean opinion on it. Now let's work itthrough a little bit. And at any point,I encourage you to pause it and kind ofextrapolate beyond what I've already talked about. So let's thinkabout the game show from the show's point of view. So the show knowswhere there's the goat and where there isn't the goat. So let's door numberone, door number two, and door number three. So let's say that ourprize is right over here. So our prize is the car. Our prize is the car, andthat we have a goat over here, and over here wealso have-- maybe we have two goats inthis situation. So what are we goingto do as the game show? Remember, the contestantsdon't know this. We know this. So the contestant picks doornumber one right over here. Then we can't liftdoor number two because there'sa car back there. We're going to liftdoor number three, and we're going toexpose this goat. In which case, itprobably would be good for the person to switch. If the person picks door numbertwo, then we as the game show can show either door numberone or door number three, and then it actuallydoes not make sense for the person to switch. If they pickeddoor number three, then we have toshow door number one because we can'tpick door number two. And in that case, itactually makes a lot of sense for theperson to switch. Now, with that out ofthe way, let's think about the probabilitiesgiven the two strategies. So if you don'tswitch, or another way to think about this strategy isyou always stick to your guns. You always stick toyour first guess. Well, in that situation, whatis your probability of winning? Well, there's three doors. The prize is equally likelyto be behind any one of them. So there's three possibilities. One has the outcomethat you desire. The probability of winning willbe 1/3 if you don't switch. Likewise, yourprobability of losing, well, there's two waysthat you can lose out of three possibilities. It's going to be 2/3. And these are theonly possibilities, and these add up toone right over here. So don't switch, 1/3chance of winning. Now let's think aboutthe switching situation. So let's say always--when you always switch. Let's think about howthis might play out. What is yourprobability of winning? And before we eventhink about that, think about how you wouldwin if you always switch. So if you pick wrongthe first time, they're going to show you this. And so you should always switch. So if you pick doornumber one, they're going to show youdoor number three. You should switch. If you picked wrongdoor number three, they're going to showyou door number one. You should switch. So if you picked wrong andswitch, you will always win. Let me write this down. And this insight actually camefrom one of the middle school students in the summer campthat Khan Academy was running. It's actually a fabulousway to think about this. So if you pickwrong, so step one, so initial pick is wrong,so you pick one of the two wrong doors, and then steptwo, you always switch, you will land on the car. Because if you pickedone of the wrong doors, they're going to have toshow the other wrong door. And so if youswitch, you're going to end up on the right answer. So what is the probability ofwinning if you always switch? Well, it's going to be theprobability that you initially picked wrong. Well, what's the probabilitythat you initially picked wrong? Well, there's twoout of the three ways to initially pick wrong. So you actually have a2/3 chance of winning. There's a 2/3 chance you'regoing to pick wrong and then switch into the right one. Likewise, what'syour probability of losing, given that you'realways going to switch? Well, the way that you wouldlose is you pick right, you picked correctly. In step two, they'regoing to show one of the two emptyor non-prize doors. And then stepthree, you're going to switch into theother empty door. But either way, you'redefinitely going to switch. So the only way to lose, ifyou're always going to switch, is to pick the rightthe first time. Well, what's theprobability of you picking right the first time? Well, that is only 1/3. So you see here, it'ssometimes counter-intuitive, but hopefully this makessense as to why it isn't. You have a 1/3 chance of winningif you stick to your guns, and a 2/3 chance of winningif you always switch. Another way tothink about it is, when you first makeyour initial pick, there's a 1/3 chancethat it's there, and there's a 2/3chance that it's in one of the other two doors. And they're going toempty out one of them. So when you switch,you essentially are capturing that2/3 probability. And we see that right there. So hopefully you enjoyed that.

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