# Freedonia Monty Hall Puzzle **URL:** \ **Category:** The Game Room\ **Created:** [October 24, 2016, 4:34am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566 "2016-10-24T04:34:36Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![Biotop](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/biotop/32/321_2.png) [@Biotop](https://boards.straightdope.com/u/Biotop)\ **Post date:** [October 24, 2016, 4:34am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/1 "2016-10-24T04:34:36Z")
In Freedonia they play the Monty Hall game show a little differently. There are still three closed doors. Behind each door is either a goat or a new car. Thing is, there is no rhyme or reason to the number of goats or cars. Sometimes it’s all goats. Sometimes it’s all cars. Sometimes it a mix, 2 goats and a car, or two cars and a goat. It is random. Anyway the game always has two final contestants and Monty, our MC, knows what is behind each door. There is a coin flip and the contestant that wins selects a door. The other contestant then gets a choice from the remaining two. At this point, **if he is able to** , Monty always shows one of the contestants that he has won a goat (ZONK!). He then gives the other player the opportunity to switch his door with the remaining one. So if you go on the show, and you are the player who gets the opportunity to switch after the other contestant is shown to have a goat, _should you switch?_ Does it matter? Note: In the cases where it is all cars, or both contestants have chosen cars and the remaining door is goat, Monty just congratulates the players and they both win cars. If both contestants have goat, the player of the two whose door is first revealed is chosen at random. ---
**Author:** ![kk\_fusion](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kk_fusion/32/15276_2.png) [@kk\_fusion](https://boards.straightdope.com/u/kk_fusion)\ **Post date:** [October 24, 2016, 4:56am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/2 "2016-10-24T04:56:54Z")
**Don’t switch.** I think this is a two-boys situation. If you know that one (random) gate has a goat, the probability of two goats is only 1/3. Whereas the probability of the third gate having a goat remains unchanged at 1/2. ---
**Author:** ![Some\_Call\_Me.Tim](https://avatars.discourse-cdn.com/v4/letter/s/439d5e/32.png) [@Some\_Call\_Me.Tim](https://boards.straightdope.com/u/Some_Call_Me.Tim)\ **Post date:** [October 24, 2016, 5:23am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/3 "2016-10-24T05:23:41Z")
> [@Biotop](#): > > I > There are still three closed doors. Behind each door is either a goat or a new car. Thing is, there is no rhyme or reason to the number of goats or cars. Sometimes it’s all goats. Sometimes it’s all cars. Sometimes it a mix, 2 goats and a car, or two cars and a goat. It is random. The problem is underspecified. What is the relative probability of the four possibilities you mention? All cases 1/4, or a coin flip for each door resulting in 1/8-3/8-3/8-1/8, or something else? ---
**Author:** ![Some\_Call\_Me.Tim](https://avatars.discourse-cdn.com/v4/letter/s/439d5e/32.png) [@Some\_Call\_Me.Tim](https://boards.straightdope.com/u/Some_Call_Me.Tim)\ **Post date:** [October 24, 2016, 5:35am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/4 "2016-10-24T05:35:38Z")
If there are two goats, you get a 1/3 chance of winning by switching. Monty is in a restricted choice situation. If there is one goat[and you’re given a choice], your choice doesn’t matter. You win. If there are three goats, your choice doesn’t matter. You lose. So the right call is to not switch, as in only one of the scenarios does your choice matter. ---
**Author:** ![Left\_Hand\_of\_Dorkness](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/left_hand_of_dorkness/32/7156_2.png) [@Left\_Hand\_of\_Dorkness](https://boards.straightdope.com/u/Left_Hand_of_Dorkness)\ **Post date:** [October 24, 2016, 11:00am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/5 "2016-10-24T11:00:53Z")
I think switching doesn’t matter. Here’s a game with similar odds that makes it clearer: Coins A, B, C. Choose a coin and flip it. If it comes up heads, you win. Your opponent chooses A, you choose B. Your opponent flips her coin, comes up tails. Do you switch from B to C? I don’t think it matters. Unlike original Monty Hall, there’s no connection between your coin and the others. ---
**Author:** ![Left\_Hand\_of\_Dorkness](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/left_hand_of_dorkness/32/7156_2.png) [@Left\_Hand\_of\_Dorkness](https://boards.straightdope.com/u/Left_Hand_of_Dorkness)\ **Post date:** [October 24, 2016, 11:08am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/6 "2016-10-24T11:08:52Z")
Wait. I think I’m wrong, I think you stay. Possibilities of the coin flips: HHH HHT HTH HTT THH THT TTH TTT If you’re A, and Monty reveals B as tails, the remaining possibilities HTH HTT TTH TTT It looks like 50/50 odds of switching leading to success–but if you also had tails, Monty had a 50% chance of revealing yours instead, so I think the last two possibilities should be halved. 2/3 chance of success for staying? ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [October 24, 2016, 11:28am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/7 "2016-10-24T11:28:54Z")
In four distributions you will not be shown a goat, so they don’t matter. In the remaining four you will always be shown a goat if you have a car, and shown a goat 50% of the time if you both have a goat. (The other 50% you’ll be the unlucky If the odds of being in either of those four distributions is even you should never switch. You’ll have picked a car 2/3 of the times you are shown a goat. ---
**Author:** ![That\_Don\_Guy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/that_don_guy/32/538_2.png) [@That\_Don\_Guy](https://boards.straightdope.com/u/That_Don_Guy)\ **Post date:** [October 24, 2016, 3:14pm UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/8 "2016-10-24T15:14:37Z")
[SPOILER]The six possible combinations other then when both 1 and 2 have cars, listed as (Player 1’s door, Player 2’s door, remaining door) are: 1. Goat, Goat, Goat - one player is shown a goat; it doesn’t matter what the other player does 2. Goat, Goat, Car - one player is shown a goat; the other should switch 3. Goat, Car, Goat - P1 is shown the goat; P2 should keep his door 4. Goat, Car, Car - again, P1 is shown the goat, and P2 should keep his door 5. Car, Goat, Goat - this is the same as 3, but with doors (and players) 1 and 2 switched; P1 should keep his door 6. Car, Goat, Car - this is the same as 4, but with doors (and players) 1 and 2 switched; P1 should keep his door In the five situations where it matters, keeping the door is correct 4 times. [/SPOILER] ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [October 24, 2016, 3:30pm UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/9 "2016-10-24T15:30:53Z")
> [@That\_Don\_Guy](#): > > [SPOILER]The six possible combinations other then when both 1 and 2 have cars, listed as (Player 1’s door, Player 2’s door, remaining door) are: > > 1. Goat, Goat, Goat - one player is shown a goat; it doesn’t matter what the other player does > 2. Goat, Goat, Car - one player is shown a goat; the other should switch > 3. Goat, Car, Goat - P1 is shown the goat; P2 should keep his door > 4. Goat, Car, Car - again, P1 is shown the goat, and P2 should keep his door > 5. Car, Goat, Goat - this is the same as 3, but with doors (and players) 1 and 2 switched; P1 should keep his door > 6. Car, Goat, Car - this is the same as 4, but with doors (and players) 1 and 2 switched; P1 should keep his door > In the five situations where it matters, keeping the door is correct 4 times. > [/SPOILER] If the six scenarios you list are equally likely, you mess up any evaluation of probability by excluding scenario one for being one where it doesn’t matter if you switch. ---
**Author:** ![Biotop](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/biotop/32/321_2.png) [@Biotop](https://boards.straightdope.com/u/Biotop)\ **Post date:** [October 24, 2016, 6:58pm UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/10 "2016-10-24T18:58:57Z")
We are still debating this at our house. I \*think \* the don’t switch people are right. ---
**Author:** ![That\_Don\_Guy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/that_don_guy/32/538_2.png) [@That\_Don\_Guy](https://boards.straightdope.com/u/That_Don_Guy)\ **Post date:** [October 24, 2016, 7:48pm UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/11 "2016-10-24T19:48:35Z")
> [@naita](#): > > If the six scenarios you list are equally likely, you mess up any evaluation of probability by excluding scenario one for being one where it doesn’t matter if you switch. I don’t think so, at least in terms of comparing probabilities, but let me explain it this way: If the player switches, then only case 2 (and the two “automatic car” cases) wins a car If the player keeps, then cases 3, 4, 5, and 6 (again, plus the two “automatic car” cases) win a car ---
**Author:** ![Some\_Call\_Me.Tim](https://avatars.discourse-cdn.com/v4/letter/s/439d5e/32.png) [@Some\_Call\_Me.Tim](https://boards.straightdope.com/u/Some_Call_Me.Tim)\ **Post date:** [October 24, 2016, 7:58pm UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/12 "2016-10-24T19:58:21Z")
> [@That\_Don\_Guy](#): > > [SPOILER]The six possible combinations other then when both 1 and 2 have cars, listed as (Player 1’s door, Player 2’s door, remaining door) are: > > 1. Goat, Goat, Goat - one player is shown a goat; it doesn’t matter what the other player does > 2. Goat, Goat, Car - one player is shown a goat; the other should switch > 3. Goat, Car, Goat - P1 is shown the goat; P2 should keep his door > 4. Goat, Car, Car - again, P1 is shown the goat, and P2 should keep his door > 5. Car, Goat, Goat - this is the same as 3, but with doors (and players) 1 and 2 switched; P1 should keep his door > 6. Car, Goat, Car - this is the same as 4, but with doors (and players) 1 and 2 switched; P1 should keep his door > In the five situations where it matters, keeping the door is correct 4 times. > [/SPOILER] In scenario 2 there’s a 50% chance that Monty shows your (P1 or P2)'s goat initially. This matters. In scenarios 4 and 6 it doesn’t matter what the remaining player does, he’ll always win a car. ‘should keep his door’ is thus incorrect, or at least no more correct than ‘should switch’. It doesn’t matter. Let’s throw numbers in this. Scenarios 2 or 3 or 5 are equally likely, and in no other scenarios is there a meaningful choice. So let’s say we play the game enough for one of these three to be played 300 times, each happening 100 times. 100 times in scenario 2, P2 is shown a goat and loses before getting a choice 50 times(as randomly selected by Monty) and should switch the other 50 times. 100 times in scenario 3, P2 gets a choice all 100 times, and P2 should keep 100 of those times. 100 times in scenario 5, P2 never gets a choice and always loses. So, in our set of games, P2 is presented with a choice 150 times, and 100/150 of those times he should choose to keep. 2/3 chance to win by keeping if we’re in the two goat scenario. In practice the odds are even better than that as you might be in the two car scenario where your choice doesn’t matter, it’s all cars for you. ---
**Author:** ![RTFirefly](https://avatars.discourse-cdn.com/v4/letter/r/c77e96/32.png) [@RTFirefly](https://boards.straightdope.com/u/RTFirefly)\ **Post date:** [October 25, 2016, 12:41am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/13 "2016-10-25T00:41:40Z")
_If you think this game is bad enough now, just wait 'til I get through with it!_ ---
**Author:** ![Johnny\_Ace](https://avatars.discourse-cdn.com/v4/letter/j/6bbea6/32.png) [@Johnny\_Ace](https://boards.straightdope.com/u/Johnny_Ace)\ **Post date:** [October 25, 2016, 1:32am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/14 "2016-10-25T01:32:39Z")
This was gone over specifically in the movie 21, where Kevin Spacey as the corrupt professor posits the same puzzle to the wunderkind (I forget who the actor was). If you stay with the door you originally chose as the surviving player, you have a 2/3 chance of winning. ---
**Author:** ![Isosleepy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/isosleepy/32/479_2.png) [@Isosleepy](https://boards.straightdope.com/u/Isosleepy)\ **Post date:** [October 25, 2016, 5:19am UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/15 "2016-10-25T05:19:22Z")
Since the distribution of cars and goats is random, the question really is: do I have a goat? The content of box 3 is unknowable (unlike the “original” version), and so irrelevant. After picking, there are 4 options: I have a goat, p2 has a car I have a car, p2 has a car I have a car, p2 has a goat I have a goat, p2 has a goat. They show p2’s goat. Immediately, 1 and 2 are impossible. Also, if option 4, there were 50/50 odds of my or p2’s goat being revealed. So, I have a 2/3 odds of having a car, given p2’s goat is revealed. Choosing wether to switch, therefore, depends on wether winning a car or a goat is more desirable in Freedonia. If the former, keep your box, if the latter, switch and hope like hell the 3rd box doesn’t happen to also be a car. ---
**Author:** ![Some\_Call\_Me.Tim](https://avatars.discourse-cdn.com/v4/letter/s/439d5e/32.png) [@Some\_Call\_Me.Tim](https://boards.straightdope.com/u/Some_Call_Me.Tim)\ **Post date:** [October 25, 2016, 3:55pm UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/16 "2016-10-25T15:55:14Z")
That’s a really good point. The problem does not specify whether a goat or a car is the preferred outcome in Freedonia, which is essential to decide whether or not to switch. ---
**Author:** ![Wheelz](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/wheelz/32/5718_2.png) [@Wheelz](https://boards.straightdope.com/u/Wheelz)\ **Post date:** [October 25, 2016, 8:50pm UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/17 "2016-10-25T20:50:12Z")
If the distribution of goats and cars is truly random, then it seems to me that I always have a 1 in 2 chance of getting a car, _regardless of what is behind either or both of the other doors._ Therefore, might as well just make a choice and stick with it. Or am I over-simplifying? ---
**Author:** ![newme](https://avatars.discourse-cdn.com/v4/letter/n/e0b2c6/32.png) [@newme](https://boards.straightdope.com/u/newme)\ **Post date:** [October 25, 2016, 10:52pm UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/18 "2016-10-25T22:52:28Z")
There are six equally likely possibilities where Player 1 can choose. The list goes Player1, Player 2, Other Door 1. Goat1, Goat2, Goat3, Monty _chooses_ Goat2 Stay: LOSE Switch: LOSE 2. Goat1, Goat2, Car, Monty _chooses_ Goat2 Stay: LOSE Switch: WIN 3. Car, Goat2, Car, Monty _chooses_ Goat2 Stay: WIN Switch: WIN 4. Car, Goat2, Car, Monty _forced_ to show Goat2 Stay: WIN Switch: WIN 5. Car, Goat2, Goat3, Monty _chooses_ Goat2 Stay: WIN Switch: LOSE 6. Car, Goat2, Goat3, Monty _forced_ to show Goat2 Stay: WIN Switch: LOSE Stay: win 4/6 = 2/3 Switch: win 3/6 = 1/2 ---
**Author:** ![Some\_Call\_Me.Tim](https://avatars.discourse-cdn.com/v4/letter/s/439d5e/32.png) [@Some\_Call\_Me.Tim](https://boards.straightdope.com/u/Some_Call_Me.Tim)\ **Post date:** [October 25, 2016, 11:00pm UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/19 "2016-10-25T23:00:28Z")
> [@Wheelz](#): > > If the distribution of goats and cars is truly random, then it seems to me that I always have a 1 in 2 chance of getting a car, _regardless of what is behind either or both of the other doors._ Therefore, might as well just make a choice and stick with it. > > Or am I over-simplifying? Over-simplifying. Though the initial set-up was random, Monty knew the answer and then took an action based on his knowledge. This can (and in some of the scenarios does) give you additional information. ---
**Author:** ![Some\_Call\_Me.Tim](https://avatars.discourse-cdn.com/v4/letter/s/439d5e/32.png) [@Some\_Call\_Me.Tim](https://boards.straightdope.com/u/Some_Call_Me.Tim)\ **Post date:** [October 25, 2016, 11:12pm UTC](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566/20 "2016-10-25T23:12:11Z")
> [@newme](#): > > There are six equally likely possibilities where Player 1 can choose. > > The list goes Player1, Player 2, Other Door > > 1. Goat1, Goat2, Goat3, Monty _chooses_ Goat2 Stay: LOSE Switch: LOSE > 2. Goat1, Goat2, Car, Monty _chooses_ Goat2 Stay: LOSE Switch: WIN > 3. Car, Goat2, Car, Monty _chooses_ Goat2 Stay: WIN Switch: WIN > 4. Car, Goat2, Car, Monty _forced_ to show Goat2 Stay: WIN Switch: WIN > 5. Car, Goat2, Goat3, Monty _chooses_ Goat2 Stay: WIN Switch: LOSE > 6. Car, Goat2, Goat3, Monty _forced_ to show Goat2 Stay: WIN Switch: LOSE > > Stay: win 4/6 = 2/3 > Switch: win 3/6 = 1/2 Several things: The original scenario never specified how it was randomized. It might be that “Sometimes it’s all cars. Sometimes it a mix, 2 goats and a car, or two cars and a goat” means that those four options are all a 1 in 4 chance, which seems to be how you interpreted it. It might mean that a fair coin is flipped for each door, and thus there’s a 1 in 8 chance of all goats or all cars, and a 3 in 8 chance of the other two. The problem doesn’t specify. The rules do specify that if he is able to Monty **must** show a goat behind a contestant’s door. Therefore scenarios 3 and 5 don’t exist; Monty has no choice if only one of the contestant’s doors has a goat behind it. Also, scenario 2 is half as likely to occur as scenario 6: the two goats scenario is equally likely to put the car behind door 3 as door 1. However, half the time the car is behind door 3 Monty will show player 1 his goat and he won’t get to choose. [Next page](https://boards.straightdope.com/t/freedonia-monty-hall-puzzle/769566.md?page=2)