# Monty Hall Redux **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [August 11, 2009, 1:14am UTC](https://boards.straightdope.com/t/monty-hall-redux/506012 "2009-08-11T01:14:08Z")\ **Posts on this page:** 20\ **Page:** 3
**Author:** ![G.Pie](https://avatars.discourse-cdn.com/v4/letter/g/bb73d2/32.png) [@G.Pie](https://boards.straightdope.com/u/G.Pie)\ **Post date:** [August 19, 2009, 5:52pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/41 "2009-08-19T17:52:33Z")
I must admit I’ve become fascinated with this puzzle. I didn’t understand it all when I watched 21 and it kind of ruined the movie for me. Anyway, the 1/3 versus 2/3 I totally now get. But when I read that the intuitive answer (50:50) is indeed correct IF Monty does not know where the car is I went back to being confused. Is anyone willing to clarify the point? I’d be forever in your debt. ---
**Author:** ![qazwart](https://avatars.discourse-cdn.com/v4/letter/q/5fc32e/32.png) [@qazwart](https://boards.straightdope.com/u/qazwart)\ **Post date:** [August 19, 2009, 7:34pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/42 "2009-08-19T19:34:31Z")
> [@G.Pie](#): > > I must admit I’ve become fascinated with this puzzle. I didn’t understand it all when I watched 21 and it kind of ruined the movie for me. Anyway, the 1/3 versus 2/3 I totally now get. But when I read that the intuitive answer (50:50) is indeed correct IF Monty does not know where the car is I went back to being confused. Is anyone willing to clarify the point? I’d be forever in your debt. Draw up a decision table (See below). Actually, switching only works 1/3 of the time. Only 1/3 of the time does switching help. 1/3 of the time, switching doesn’t help, and 1/3 of the time, Monty himself picks the prize, so switching definitely won’t help. However, most “Monty doesn’t know which door scenarios” simply leave out the 1/3 of the time Monty actually picks the prize. Thus, removing 2 of the six rows, and switching helps becomes a 50:50 possibility. Table is incomplete but it should be obvious that the rest of the table would duplicate the top third. ```auto | Door W/ Prize | You Choose | Monty Chooses | Switching Works | | 1 | 1 | 2 | No | | 1 | 1 | 3 | No | | 1 | 2 | 1 | N/A | | 1 | 2 | 3 | Yes | | 1 | 3 | 1 | N/A | | 1 | 3 | 2 | Yes | ``` ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [August 19, 2009, 8:31pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/43 "2009-08-19T20:31:40Z")
> [@G.Pie](#): > > I must admit I’ve become fascinated with this puzzle. I didn’t understand it all when I watched 21 and it kind of ruined the movie for me. Anyway, the 1/3 versus 2/3 I totally now get. But when I read that the intuitive answer (50:50) is indeed correct IF Monty does not know where the car is I went back to being confused. Is anyone willing to clarify the point? I’d be forever in your debt. **qazwart** ’s explanation is completely fine, but to narrow it down, consider these alternative problems: _In this Monty Hall Game, you pick door A and another contestant (who doesn’t know where the prize is either) picks door B. None of the doors are opened. Should you switch to door B, if offered?_ Clearly it doesn’t matter. You’re randomly picking a door, as is the other contestant. Neithe door is better. _In this Monty Hall Game, you pick door A and another contestant (who doesn’t know where the prize is either) picks door B. None of the doors are opened. Should you switch to door C?_ Again, clearly it doesn’t matter. \*In this Monty Hall Game, you pick door A and another contestant (who doesn’t know where the prize is either) picks door B. Door C is opened, and the prize \*\*is \*_behind it. Should you switch to door B, if offered?_ Again, clearly it doesn’t matter. You lost, period. \*In this Monty Hall Game, you pick door A and another contestant (who doesn’t know where the prize is either) picks door B. Door C is opened, and the prize \*\*is not \*_behind it. Should you switch to door B, if offered?_ This might be a little more tricky, but consider the symmetry: \*If \* it were adventageous for you to switch, then it would \*also \* be adventageous for the other contestant to switch because there’s \*no difference \*between your situations. Clearly it can’t be advantageous for both of you to switch, so it must not be advantageous for either of you to switch. Or, to put it another way, Since you both _randomly_ picked your door, neither of you has any advantage over the other. Likewise, if Monty \*randomly \*picks a door, you have no reason to switch. In the “original” problem, Monty _doesn’t_ randomly pick–he _always picks the wrong door_. That’s the difference. ---
**Author:** ![G.Pie](https://avatars.discourse-cdn.com/v4/letter/g/bb73d2/32.png) [@G.Pie](https://boards.straightdope.com/u/G.Pie)\ **Post date:** [August 19, 2009, 10:01pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/44 "2009-08-19T22:01:03Z")
> [@zut](#): > > In the “original” problem, Monty _doesn’t_ randomly pick–he _always picks the wrong door_. That’s the difference. That’s it! That’s everything I need to know. Thank you, both. ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [August 19, 2009, 10:11pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/45 "2009-08-19T22:11:37Z")
> [@Irishman](#): > > As far as the actual game show went, Monty was not constrained. He didn’t have to offer you the chance to switch. He typically did because that creates the drama, the tension. In his book [Calculated Risks](http://www.amazon.com/gp/product/0743254236/ref=s9_simz_gw_s0_p14_i1?pf_rd_m=ATVPDKIKX0DER&pf_rd_s=center-2&pf_rd_r=0MVQTDTMA10HBE27C870&pf_rd_t=101&pf_rd_p=470938631&pf_rd_i=507846), in which he discusses the Monty Hall problem, Gerd Gigerenzer writes > [@](#): > > Take the assumption that Monty always gave the final guest a chance to switch. When asked whether this was true, the real Monty Hall recalled that he rarely offered guests the switching option and could not say how often it was accepted. However, his longtime production assistant, Carol Andrews, asserted that Monty never offered the last guest an opportunity to switch. Barry Nalebuff, who wrote one of the first srticles on the Monty Hall problem, says that in watching the show he saw the switch option offered, but he cannot recall whether Monty made the offer in every show or whether his making the offer depended on which door the guest had picked in the first place. … The real game show ervolved around Monty Hall, and part of his personality was to make spontaneous decisions rather than to follow a strict protocol. ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [August 20, 2009, 7:36pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/46 "2009-08-20T19:36:14Z")
Thanks, that’s a great cite. For the real game show, it was never about logic. Steve Selvin created the “Monty Hall Problem” as a logic puzzle to explore probabilities, and in doing so he created the assumptions necessary for the probability issue to be valid. Thus the need to constrain Monty to always offer to switch and to always show a non-winning door. The point of the problem is to explore the non-intuitive nature of the probability case _because_ Monty knows where the goats are. The standard intuitive position is when Monty eliminates a door, that leaves 2 doors with a 50:50 chance of being right. But intuition is **wrong** because Monty knows the door he reveals is a goat, and is free to pick either of the non-selected doors that is not a winner. Thus we get the non-intuitive 2/3 probability for switching, and 1/3 for staying. But all of that was Steve Selvin creating a context to explore probability using something familiar, a game show, for context. ---
**Author:** ![KellyCriterion](https://avatars.discourse-cdn.com/v4/letter/k/b3f665/32.png) [@KellyCriterion](https://boards.straightdope.com/u/KellyCriterion)\ **Post date:** [August 20, 2009, 8:03pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/47 "2009-08-20T20:03:50Z")
This took me a long time to grasp as well, here’s another way of explaining it that might be helpful for those who still don’t get it: What are the odds that your initial door selection will be a goat? 2/3. Agree? As long as your initial door selection is a goat, you will always win by switching. **\<-- This is the key point to understand.** Agree? In other words, pick a goat first up, _you’ve won_. Odds of picking a goat first up? 2/3. ---
**Author:** ![G.Pie](https://avatars.discourse-cdn.com/v4/letter/g/bb73d2/32.png) [@G.Pie](https://boards.straightdope.com/u/G.Pie)\ **Post date:** [August 21, 2009, 5:21am UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/48 "2009-08-21T05:21:47Z")
A friend of mine summarized it this way: By showing me that one of the other doors is not a winner in a way adds no new information, as it is already known to me that one of the other doors is not the winner. Thus by switching I am effectively trading one door for two, and thus a one-third chance for a two-thirds chance. ---
**Author:** ![Number43](https://avatars.discourse-cdn.com/v4/letter/n/58956e/32.png) [@Number43](https://boards.straightdope.com/u/Number43)\ **Post date:** [August 22, 2009, 3:47am UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/49 "2009-08-22T03:47:30Z")
Perhaps this explanation will be clearer to some people: Whether you picked the right door or a wrong one to begin with, he will always open up a wrong one other than the one you picked. Thus if you picked the right one to begin with, which happens 1/3 of the time, the door you can switch to will be the other wrong one. If you picked one of the wrong ones, which will happen 2/3 of the time, the one you can switch to will be the right one. You don’t even have to consider the whole system of doors to understand this: all it is doing is switching your initial choice from a winning one to a losing one or vice-versa, but since your initial choice was more likely to lose, the changed choice will be more likely to win. This is because it will never be a switch between one wrong door and the other wrong door. ---
**Author:** ![moridin5](https://avatars.discourse-cdn.com/v4/letter/m/dfb087/32.png) [@moridin5](https://boards.straightdope.com/u/moridin5)\ **Post date:** [February 5, 2010, 5:31am UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/50 "2010-02-05T05:31:15Z")
> [@Number43](#): > > Perhaps this explanation will be clearer to some people: > > Whether you picked the right door or a wrong one to begin with, he will always open up a wrong one other than the one you picked. Thus if you picked the right one to begin with, which happens 1/3 of the time, the door you can switch to will be the other wrong one. If you picked one of the wrong ones, which will happen 2/3 of the time, the one you can switch to will be the right one. > > You don’t even have to consider the whole system of doors to understand this: all it is doing is switching your initial choice from a winning one to a losing one or vice-versa, but since your initial choice was more likely to lose, the changed choice will be more likely to win. This is because it will never be a switch between one wrong door and the other wrong door. I have a feeling, from what I’ve read, people are tired of this discussion. lol…And I have definitely taken some time to return. But here I am…and still curious, those who abhor this topic my apologies. 😉 I 100% understand the concept that you expressed based on probabilities alone…what confuses me is that if it’s a given, no matter what, that a wrong choice will be revealed isn’t it in reality a 50% chance? Monty has foreknowledge, obviously, so from the very get-go you’re not choosing from 1 of 3…you’re choosing from 1 of 2. While the wrong choice Monty chooses to reveal seems like something to be factored into a “1 of 3” choice…it really isn’t. No matter what you choose, he can show a wrong choice…and will. What I’m having a hard time wrapping my head around is that knowing that, is he really creating a scenario where the typical “1/3 of the time you’ll pick right, but after he reveals a choice you’re 2/3 to be right by switching” really apply? Probabilities, by their very nature, don’t incorporate someone with an awareness of right/wrong choices, do they? The nature of probability has to do with that it’s random, and unknowable. Other than by example. Thus over time the probability of what will happen can be estimated. Once you toss in Monty knowing (thus no probability of him not) where a wrong door can always be, it seems you’re no longer dealing with “real” probabilities. In the beginning, your odds of choosing wrong are 66%. But since you know that Monty’s odds are 100% of choosing a wrong one without fail…don’t his odds skew the strict probability analysis and turn it into a genuine 50-50 guess? He can’t help but succeed in always picking a wrong one, whereas you have randomness in how you will succeed or fail. Put another way…It seems the (and I may be wrong. lol) real probability of being right by switching would only apply if Monty _didn’t_ know what was the winning door, and himself had to guess at one of the other 2 doors and was wrong and then asked if you wanted to switch. In that case, it seems a no-brainer. Switch! The odds are definitely on your side. But…he doesn’t eliminate a remaining choice by probability, he does it by knowledge. And no matter what you pick, he can _always_ do that. So this is where I have difficulty believing it’s really a clearly defined probability to switch. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [February 5, 2010, 12:46pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/51 "2010-02-05T12:46:11Z")
> [@moridin5](#): > > I 100% understand the concept that you expressed based on probabilities alone…what confuses me is that if it’s a given, no matter what, that a wrong choice will be revealed isn’t it in reality a 50% chance? Monty has foreknowledge, obviously, so from the very get-go you’re not choosing from 1 of 3…you’re choosing from 1 of 2. No, you’re not choosing from 1 of 2. Monty hasn’t done anything at the time you pick. You very clearly have a 1 in 3 chance of picking correctly when you first choose. ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [February 5, 2010, 2:37pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/52 "2010-02-05T14:37:34Z")
> [@moridin5](#): > > …he doesn’t eliminate a remaining choice by probability, he does it by knowledge. Correct. And in reality, he gives you no information by opening a door after you have chosen. He’s telling you that at least one of the doors you didn’t choose was a loser - something you already knew. Let’s change the rules of the game a bit: You choose a door, then Monty, without opening any door, allows you to stick with your choice or switch to take both of the other two doors. Would you stick or switch? ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [February 5, 2010, 8:01pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/53 "2010-02-05T20:01:49Z")
In addition to what they said, you can try this yourself with a deck of cards and a friend. Pick two Jacks (or whatever) as goats and an ace (or whatever) as the car, then let your friend shuffle the three cars and arrange them so you don’t know which is which, but he does. Then you can try any strategy you like, and record the results. (You don’t have to use cards. Three of anything that can be easily distinguished and easily hidden so you don’t know which is which will work. Three pieces of scrap paper and three books could be used - book covers as doors and the papers with words “goat” and “car”, for instance.) We’ll be here when you get back with actual numbers. ---
**Author:** ![Didactylos](https://avatars.discourse-cdn.com/v4/letter/d/a183cd/32.png) [@Didactylos](https://boards.straightdope.com/u/Didactylos)\ **Post date:** [February 7, 2010, 10:18am UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/54 "2010-02-07T10:18:50Z")
> [@Xema](#): > > Let’s change the rules of the game a bit: You choose a door, then Monty, without opening any door, allows you to stick with your choice or switch to take both of the other two doors. Would you stick or switch? I’ve lurked in alot of the Monty Hall-threads, and I think this is the best and most succinct explanation I’ve heard yet. The problem is, when presented with perfectly fine explanations for switching being the best, a lot of people come up with convoluted arguments, usually boiling down to “there’s two cases, ergo 50% probability for each, so switching doesn’t matter”. But taking up a couple of paragraphs, making it more difficult to make them understand the faulty reasoning. ETA: spelling ---
**Author:** ![Louie\_Mingione](https://avatars.discourse-cdn.com/v4/letter/l/a87d85/32.png) [@Louie\_Mingione](https://boards.straightdope.com/u/Louie_Mingione)\ **Post date:** [February 15, 2010, 3:31pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/55 "2010-02-15T15:31:54Z")
I love this forum. But to all who are still in debate over this question, try playing this simulation put together by some folks at UCSD. [http://www.math.ucsd.edu/~crypto/Monty/Montytitle.html](http://www.math.ucsd.edu/~crypto/Monty/Montytitle.html) It gives you the chance to play when Monty knows what’s behind the doors (you should switch for best odds), or when Monty doesn’t know (50/50 chance, but he also gets it wrong sometimes). The history of all who play is tracked and you can see that the odds really do work out as described so many times on this and other forums. ---
**Author:** ![mickeyhooligan](https://avatars.discourse-cdn.com/v4/letter/m/779978/32.png) [@mickeyhooligan](https://boards.straightdope.com/u/mickeyhooligan)\ **Post date:** [February 26, 2010, 12:14am UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/56 "2010-02-26T00:14:39Z")
this was in sc amer and got huge response. I’m not sure its about probabilities at all. does Monty always open a door and ask if the chooser wants to switch? If not, then sticking to the original door would be the best choice. If he always opens a door than probabilties are higher that you should switch…but original question was unclear. Does he always open a second door? ---
**Author:** ![Bytegeist](https://avatars.discourse-cdn.com/v4/letter/b/9dc877/32.png) [@Bytegeist](https://boards.straightdope.com/u/Bytegeist)\ **Post date:** [February 26, 2010, 4:35pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/57 "2010-02-26T16:35:36Z")
> [@mickeyhooligan](#): > > this was in sc amer and got huge response. I’m not sure its about probabilities at all. does Monty always open a door and ask if the chooser wants to switch? You mean if he doesn’t open a door _at all_? Or he opens a door, but doesn’t say anything? Just leaves you standing there in awkward silence, not knowing what you’re allowed to do next? The usual conditions for this problem are that Monty (A) knows where the prize is, and (B) opens a door that is neither the door you first picked nor the one hiding the prize. > [@](#): > > If not, then sticking to the original door would be the best choice. … Does he always open a second door? I’m not sure what you mean. If Monty doesn’t open a door, then what decision do you even have to make? You have no particular reason yet to prefer, or reject, the door you’ve already picked. ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [February 26, 2010, 6:54pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/58 "2010-02-26T18:54:22Z")
**mickeyhooligan** said: > [@](#): > > this was in sc amer and got huge response. I’m not sure its about probabilities at all. does Monty always open a door and ask if the chooser wants to switch? Monty Hall has stated that in the real game show, he was not constrained by any rules. He did not have to offer the chance to switch, he did not always show a goat. In that situation, the probabilities are not strict. But the “Monty Hall” puzzle is defined with strict rules. It is defined that way because it was created to illuminate the non-intuitive statistics. Ergo, the problem is stated with rules that: 1. you pick a door from 3 2. Monty knows what is behind each door 3. Monty will open a door that you did not pick that does not have the prize 4. Monty offers you the chance to switch or to keep the door you previously selected Change those assumptions, you change the outcome, but under those rules, you are better off switching than staying. If Monty does not open a door but offers you the chance to switch, then you do have not gained any new information and have not changed the probabilities at all, you still have a 1/3 chance of getting correct. ---
**Author:** ![BigT](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bigt/32/12044_2.png) [@BigT](https://boards.straightdope.com/u/BigT)\ **Post date:** [November 18, 2010, 11:19am UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/59 "2010-11-18T11:19:54Z")
I didn’t want to create a new thread for this, but I was bored yesterday, and so I worked up a JavaScript program to test the Monty Hall problem. Included below is the base version, and you can read a version with nearly every line commented [here](http://trlkly.drivehq.com/monty-hall.js.txt). If you don’t know any other way, you can test it by copying-and-pasting everything after the first // into your URL bar in Firefox or Chrome, or saving the whole thing as monty-hall.js, and then opening the file. ```php //javascript: function echo (input){ if (typeof WScript != "undefined") { WScript.Echo(input); } else if (typeof window != "undefined") { alert(input); } } var wins = [0, 0]; var door = [0, 0, 0]; for (i=1;i<10000;i++) { door = [0, 0, 0]; PRIZE = Math.floor(Math.random() * 3); door[PRIZE] = 1; GUESS = Math.floor(Math.random() * 3); if (PRIZE == GUESS) { door[(GUESS + Math.floor(Math.random() + 1)) % 3] = -1; } else { if (door[(GUESS + 1) % 3] = 1) { door[(GUESS + 2) % 3] = -1; } else { door[(GUESS + 1) % 3] = -1; } } SWITCH = Math.floor(Math.random()); if (SWITCH == 1) { if (door[(GUESS + 1) % 3] != -1) { GUESS = (GUESS + 1) % 3; } else { GUESS = (GUESS + 2) % 3; } } if (GUESS == PRIZE) { wins[SWITCH]++; } else { wins[(SWITCH + 1) % 2]++; } } echo ("wins with switch: " + wins[1] + " " + "wins without: " + wins[0] + " " + Math.round(wins[1]/(wins[0]+wins[1]) * 100) + "% better if you switch"); ``` typical output: ```auto wins with switch: 6650 wins without: 3349 67% better if you switch ``` (And, yes, it’s nowhere near the optimal coding. I went as procedurally as possible, to more clearly model how the process would actually be done. But feel free to show me how to clean it up.) ---
**Author:** ![John\_W.Kennedy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_w.kennedy/32/1031_2.png) [@John\_W.Kennedy](https://boards.straightdope.com/u/John_W.Kennedy)\ **Post date:** [November 18, 2010, 5:34pm UTC](https://boards.straightdope.com/t/monty-hall-redux/506012/60 "2010-11-18T17:34:17Z")
And, if you like, I have versions in Ada, C++, Java, Javascript, Objective-C, Perl, and Ruby. [Previous page](https://boards.straightdope.com/t/monty-hall-redux/506012.md?page=2) [Next page](https://boards.straightdope.com/t/monty-hall-redux/506012.md?page=4)