# MAth question from the movie 21 **URL:** \ **Category:** Factual Questions\ **Created:** [July 11, 2010, 2:01am UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268 "2010-07-11T02:01:56Z")\ **Posts on this page:** 18\ **Page:** 2
**Author:** ![cosmosdan](https://avatars.discourse-cdn.com/v4/letter/c/a6a055/32.png) [@cosmosdan](https://boards.straightdope.com/u/cosmosdan)\ **Post date:** [July 11, 2010, 1:02pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/21 "2010-07-11T13:02:39Z")
> [@x-ray\_vision](#): > > Also, in the Monty Hall problem, if Monty randomly showed you what is behind one of the two remaining doors and it just happened to be a goat and then offered you the chance to switch, it doesn’t matter if you do or not- your probability of winning is 1/2 either way. That’s how it looks on the surface but evidently according to statistics, that’s not true. ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [July 11, 2010, 1:14pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/22 "2010-07-11T13:14:25Z")
> [@cosmosdan](#): > > That’s how it looks on the surface but evidently according to statistics, that’s not true. It is true. The fact that Monty has prior knowledge of what is behind every door and always reveals a goat is crucial. ---
**Author:** ![x-ray\_vision](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/x-ray_vision/32/351_2.png) [@x-ray\_vision](https://boards.straightdope.com/u/x-ray_vision)\ **Post date:** [July 11, 2010, 1:40pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/23 "2010-07-11T13:40:34Z")
> [@cosmosdan](#): > > That’s how it looks on the surface but evidently according to statistics, that’s not true. What **Pedro** said. Let’s look at Cecil’s example you quoted from (it’s the same as the 99 goat example). The reason you switch is because Cecil mentions that the dealer looks at the remaining 51 cards and purposefully turns over 50 that are not the ace of spades. This gives you no new information about the card you selected, so your chances of having it be the ace of spades is still 1/52. However, since the dealer made sure that if the ace of spades is in the remaining 51 cards he would not turn it over, if you switch to the remaining card your probability of having ace of spades becomes 51/52. But what if the dealer did not look at the remaining 51 cards and randomly flipped over 50 of them and it just so happened that none of them are the ace of spades? It doesn’t matter if you switch. The probability is 1/2 that your original pick is the ace of spades and the probability that the card that the dealer didn’t flip over is also 1/2. ---
**Author:** ![cosmosdan](https://avatars.discourse-cdn.com/v4/letter/c/a6a055/32.png) [@cosmosdan](https://boards.straightdope.com/u/cosmosdan)\ **Post date:** [July 11, 2010, 2:07pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/24 "2010-07-11T14:07:26Z")
> [@Pedro](#): > > It is true. The fact that Monty has prior knowledge of what is behind every door and always reveals a goat is crucial. Okay. I failed to consider the random element of Monty’s open door. I think the problem assumes hen does know where the car is. If he didn’t he might reveal the car and ruin the segment. ---
**Author:** ![Telemark](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/telemark/32/372_2.png) [@Telemark](https://boards.straightdope.com/u/Telemark)\ **Post date:** [July 11, 2010, 5:30pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/25 "2010-07-11T17:30:15Z")
> [@cosmosdan](#): > > I think the problem assumes hen does know where the car is. If he didn’t he might reveal the car and ruin the segment. Yes, the classic version of the problem has Monty knowing where the car is and always opening a goat door. The problem is different if either of those two assumptions are removed. ---
**Author:** ![The\_Flying\_Dutchman](https://avatars.discourse-cdn.com/v4/letter/t/9fc348/32.png) [@The\_Flying\_Dutchman](https://boards.straightdope.com/u/The_Flying_Dutchman)\ **Post date:** [July 11, 2010, 6:17pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/26 "2010-07-11T18:17:08Z")
> [@Pedro](#): > > It is true. The fact that Monty has prior knowledge of what is behind every door and always reveals a goat is crucial. True, but if you read the problem, you’ll note that there is no reason for the contestant to assume that Monty has prior knowledge. Monty could simply have started opening any door and decide to give the contestant a second chance before revealing the result by opening the second door. Considering that the contestant has one extra choice in the matter, decide if Mony knows or doesn’t know, I feel compelled to challenge the 2/3 probability for switching. Intuitively I expect it is halfway between 1/2 and 2/3 and should be 7/12. ---
**Author:** ![x-ray\_vision](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/x-ray_vision/32/351_2.png) [@x-ray\_vision](https://boards.straightdope.com/u/x-ray_vision)\ **Post date:** [July 11, 2010, 6:21pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/27 "2010-07-11T18:21:32Z")
> [@The\_Flying\_Dutchman](#): > > True, but if you read the problem From where? The problem usually includes the caveat that Monty knows what’s behind the doors and opens a door that will reveal a goat. ---
**Author:** ![cosmosdan](https://avatars.discourse-cdn.com/v4/letter/c/a6a055/32.png) [@cosmosdan](https://boards.straightdope.com/u/cosmosdan)\ **Post date:** [July 11, 2010, 6:25pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/28 "2010-07-11T18:25:04Z")
> [@The\_Flying\_Dutchman](#): > > True, but if you read the problem, you’ll note that there is no reason for the contestant to assume that Monty has prior knowledge. Monty could simply have started opening any door and decide to give the contestant a second chance before revealing the result by opening the second door. > > Considering that the contestant has one extra choice in the matter, decide if Mony knows or doesn’t know, I feel compelled to challenge the 2/3 probability for switching. Intuitively I expect it is halfway between 1/2 and 2/3 and should be 7/12. good for you. It’s accepted that the problem changes depending on what the host knows or doesn’t know. Cecil comments in one link that if everyone knows that Monty always opens a goat door after the first guess is made , then they can obcviously play the stats and switch. If he only occasionally offers them a 2nd chance then it’s harder to figure the stats. ---
**Author:** ![x-ray\_vision](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/x-ray_vision/32/351_2.png) [@x-ray\_vision](https://boards.straightdope.com/u/x-ray_vision)\ **Post date:** [July 11, 2010, 6:32pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/29 "2010-07-11T18:32:05Z")
> [@The\_Flying\_Dutchman](#): > > Considering that the contestant has one extra choice in the matter, decide if Mony knows or doesn’t know, I feel compelled to challenge the 2/3 probability for switching. Intuitively I expect it is halfway between 1/2 and 2/3 and should be 7/12. But there isn’t one extra possibility; there’s an endless amount. Maybe Monty doesn’t know. Maybe he knows and only will reveal a goat. Maybe he knows and will only allow contestants the choice to switch when they have already chosen the car, guaranteeing they lose if they switch. Maybe he only offers contestants a chance to switch so on average keeping your first choice will result in a win 50% of the time, etc. ---
**Author:** ![The\_Flying\_Dutchman](https://avatars.discourse-cdn.com/v4/letter/t/9fc348/32.png) [@The\_Flying\_Dutchman](https://boards.straightdope.com/u/The_Flying_Dutchman)\ **Post date:** [July 11, 2010, 7:14pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/30 "2010-07-11T19:14:01Z")
> [@x-ray\_vision](#): > > But there isn’t one extra possibility; there’s an endless amount. Maybe Monty doesn’t know. Maybe he knows and only will reveal a goat. Maybe he knows and will only allow contestants the choice to switch when they have already chosen the car, guaranteeing they lose if they switch. Maybe he only offers contestants a chance to switch so on average keeping your first choice will result in a win 50% of the time, etc. All your extra possibilities are subject to only one of the two choices I mentioned , namely that Monty knows, and thus irrelevant. ---
**Author:** ![x-ray\_vision](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/x-ray_vision/32/351_2.png) [@x-ray\_vision](https://boards.straightdope.com/u/x-ray_vision)\ **Post date:** [July 11, 2010, 7:21pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/31 "2010-07-11T19:21:23Z")
> [@The\_Flying\_Dutchman](#): > > All your extra possibilities are subject to only one of the two choices I mentioned , namely that Monty knows, and thus irrelevant. It’s not irrelevant. That “Monty knows” does not affect probability; it’s what he does or doesn’t do that affects probability. ---
**Author:** ![The\_Flying\_Dutchman](https://avatars.discourse-cdn.com/v4/letter/t/9fc348/32.png) [@The\_Flying\_Dutchman](https://boards.straightdope.com/u/The_Flying_Dutchman)\ **Post date:** [July 11, 2010, 7:53pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/32 "2010-07-11T19:53:10Z")
> [@x-ray\_vision](#): > > It’s not irrelevant. That “Monty knows” does not affect probability… Yes it does. That is why it is specified in the “official” problem. ---
**Author:** ![x-ray\_vision](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/x-ray_vision/32/351_2.png) [@x-ray\_vision](https://boards.straightdope.com/u/x-ray_vision)\ **Post date:** [July 11, 2010, 8:07pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/33 "2010-07-11T20:07:02Z")
> [@The\_Flying\_Dutchman](#): > > Yes it does. That is why it is specified in the “official” problem. It doesn’t. In the “official problem” (wherever that is written), Monty knows where the car is and gives contestants the choice to switch doors after revealing a goat behind one of the remaining doors. In post #26 you stated _"Considering that the contestant has one extra choice in the matter, decide if Mony knows or doesn’t know, I feel compelled to challenge the 2/3 probability for switching. Intuitively I expect it is halfway between 1/2 and 2/3 and should be 7/12. "_ Where are you getting the numbers 1/2 and 1/3 to average out? ---
**Author:** ![The\_Flying\_Dutchman](https://avatars.discourse-cdn.com/v4/letter/t/9fc348/32.png) [@The\_Flying\_Dutchman](https://boards.straightdope.com/u/The_Flying_Dutchman)\ **Post date:** [July 11, 2010, 8:24pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/34 "2010-07-11T20:24:11Z")
> [@x-ray\_vision](#): > > It doesn’t. In the “official problem” (wherever that is written), Monty knows where the car is and gives contestants the choice to switch doors after revealing a goat behind one of the remaining doors. > > In post #26 you stated _"Considering that the contestant has one extra choice in the matter, decide if Mony knows or doesn’t know, I feel compelled to challenge the 2/3 probability for switching. Intuitively I expect it is halfway between 1/2 and 2/3 and should be 7/12. "_ Where are you getting the numbers 1/2 and 1/3 to average out? If the contestant knows that Monty has no idea, the probability is 1/2. If the contestant knows that Monty knows, then the probability is 2/3 for switching. (not 1/3, referring to your typo) If the contestant doesn’t know if Monty knows either way, then… In case it isn’t clear, my claim for a 7/12 probability for switching is based on the problem as presented in the OP which doesn’t specify whether Monty knows or not. You can find the “official” version in the links provided in posts 2 and 3. ---
**Author:** ![x-ray\_vision](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/x-ray_vision/32/351_2.png) [@x-ray\_vision](https://boards.straightdope.com/u/x-ray_vision)\ **Post date:** [July 11, 2010, 8:43pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/35 "2010-07-11T20:43:46Z")
> [@The\_Flying\_Dutchman](#): > > If the contestant knows that Monty knows, then the probability is 2/3 for switching. No, if the contestant knows that Monty knows **and** he will reveal a goat behind one of the remaining doors regardless of what door he originally chose, **then** the probability of winning is 2/3 if he switches. As I keep trying to explain to you, Monty merely knowing something does not affect probability; it’s what he does with that knowledge that affects it (see post #29). > [@](#): > > You can find the “official” version in the links provided in posts 2 and 3. The problem is stated in multiple ways in those links. Does that mean I get to choose which one is official? I choose this one from the Wikipedia article linked to in post 2: > [@](#): > > Suppose you’re on a game show and you’re given the choice of three doors [and will win what is behind the chosen door]. Behind one door is a car; behind the others, goats [unwanted booby prizes]. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one [uniformly] at random. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice? The answer is yes. Your probability increases from 1/3 to 2/3. ---
**Author:** ![Washoe](https://avatars.discourse-cdn.com/v4/letter/w/ecae2f/32.png) [@Washoe](https://boards.straightdope.com/u/Washoe)\ **Post date:** [July 11, 2010, 9:05pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/36 "2010-07-11T21:05:19Z")
Regardless of the logic behind it, the whole thing has never made any sense to me at all anyway. I’m Armenian—I’d rather have the goat. ---
**Author:** ![Bytegeist](https://avatars.discourse-cdn.com/v4/letter/b/9dc877/32.png) [@Bytegeist](https://boards.straightdope.com/u/Bytegeist)\ **Post date:** [July 11, 2010, 9:52pm UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/37 "2010-07-11T21:52:13Z")
> [@Washoe](#): > > I’m Armenian—I’d rather have the goat. Well then, for you we can rewrite the problem to contain one goat and two cars — the cars being the dud prizes. ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [July 12, 2010, 1:05am UTC](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268/38 "2010-07-12T01:05:30Z")
I’ve always found it helpful to imagine Monty actually saying out loud “If you think the prize is in one of the other doors, let me help you decide which of the two you should pick.” or, in other words, “You’ve picked A. If you think it’s in B or C, you should pick B. Here, I’ll prove that by showing you the goat behind C.” The same statement works for the 52-card Ace of Spades version: “If you think the Ace is still in the deck, you should pick card #37. I’ll even show you the other ones just to prove it.” [Previous page](https://boards.straightdope.com/t/math-question-from-the-movie-21/546268.md?page=1)