# Explain the Monty Hall problem to me?

**URL:** <https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632>\
**Category:** Factual Questions\
**Created:** [August 11, 2005, 8:23pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632 "2005-08-11T20:23:17Z")\
**Posts on this page:** 20\
**Page:** 4

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**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:** [August 12, 2005, 4:12pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/61 "2005-08-12T16:12:02Z")

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> [@mazinger\_z](#):
>
> Let me try to simplify things:
> 
> A) If Monty _randomly_ opens a door after you picked, _without any further motivation_ on his part, then you standing pat is a 50-50 chance of winning; likewise, choosing the last remaning door has a 50-50 chance of winning (as others have already correctly pointed out).
> 
> Conversely,  
> B) If Monty _only_ opens a door after you picked _and always reveals a zonker_, then the best way to maximize your odds is to always choose to pick the last remaining door.

What may be tripping up people is that in A, sometimes Monty opens the door with the prize. You lose. But you you never get a chance to switch in those cases. In the cases that you do get the chance to switch, it really is a 50/50 choice. Those times that Monty opens the prize door should be exactly 1/3 [2/3 \* 1/2].

The odds are that you still choose correctly 1/3 of the time. But only 2/3 of the time are you given a choice, and you’ve got a 50/50 chance at that point.

At least I think that’s all correct. I don’t swear by my reasoning.

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**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\
**Post date:** [August 12, 2005, 4:15pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/62 "2005-08-12T16:15:27Z")

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> [@Telemark](#):
>
> At least I think that’s all correct.

As do I.

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**Author:** ![Bytegeist](https://avatars.discourse-cdn.com/v4/letter/b/9dc877/32.png) [@Bytegeist](https://boards.straightdope.com/u/Bytegeist)\
**Post date:** [August 12, 2005, 4:25pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/63 "2005-08-12T16:25:37Z")

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> [@dalej42](#):
>
> Let’s say you zonk and win his and her pigs. … No one in their right mind will take the pigs…

Why not? Pigs are _magical_ animals. (C.f. Lisa Simpson.)

> [@Xema](#):
>
> I believe this is wrong, and that **OldGuy** and **CurtC** are correct.
> 
> **Game** : Monty Hall problem, “Monty’s random choice” variant  
> **Rules** : You chose one of 3 doors. Monty then opens one of the other two at random.  
> **Case to analyze** : Monty’s randomly-chosen door reveals a zonk.  
> **Question** : Given the above, what are your chances of winning by switching doors?
> 
> [analysis deleted …]

Crap. After looking at the problem more carefully, but a little differently, I do see now that Monty’s policy _would_ change your chances of winning. Depending on the probability with which he reveals the prize door (when he has the chance to), a player strategy of always switching will win anywhere from 0 up to 2/3 of the time.

In particular, if Monty opens the prize door with probability P on those trials when he has the choice to (which is 2/3 of the total number of trials), an always-switch strategy on the part of the player will win ( 2(1-P)/3 ) / ( 1/3 + 2(1-P)/3 ) = (2 - 2P) / (3 - 2P) of the time. In the “normal” case of P = 0, that comes to 2/3 of the time. In the case of P = 1/2 (the variant **Xema** describes), it wins half of the time. And for P = 1, it always loses.

Crap.

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**Author:** ![K364](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/k364/32/5_2.png) [@K364](https://boards.straightdope.com/u/K364)\
**Post date:** [August 12, 2005, 4:27pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/64 "2005-08-12T16:27:28Z")

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Someone has probably covered this already…

If you adopt the switching strategy, **the only way you can lose is by initially picking the prize door**. Since the odds of initially picking the prize door are 1/3, the switching strategy will pay off 2/3 of the time.

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [August 12, 2005, 4:28pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/65 "2005-08-12T16:28:42Z")

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Don’t forget, though, that if you have no information about how Monty opens a door, that you can’t take his motivation into account.

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**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\
**Post date:** [August 12, 2005, 6:38pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/66 "2005-08-12T18:38:57Z")

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> [@Ellis Dee](#):
>
> > [@](#):
> >
> > If the 50 losers are uncovered at random (unlikely, but obviously possible) then either of the remaining cards is equally likely to be the winner.
> 
> Patently false. Try a computer simulation and see for yourself.

Sorry, a computer simulation of a random-picking Monty would reveal a 50/50 chance for switching in this scenario. The thing is, in only one trial out of 51 would you ever get to the point where you have the option of switching, but in those that you do, it’s a 50/50 proposition.

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [August 12, 2005, 6:41pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/67 "2005-08-12T18:41:36Z")

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> [@CurtC](#):
>
> Sorry, a computer simulation of a random-picking Monty would reveal a 50/50 chance for switching in this scenario. The thing is, in only one trial out of 51 would you ever get to the point where you have the option of switching, but in those that you do, it’s a 50/50 proposition.

Run that computer simulation and get back to us.

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**Author:** ![saoirse](https://avatars.discourse-cdn.com/v4/letter/s/5daacb/32.png) [@saoirse](https://boards.straightdope.com/u/saoirse)\
**Post date:** [August 12, 2005, 7:25pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/68 "2005-08-12T19:25:02Z")

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> [@Xema](#):
>
> Well, “patently” means “obviously” and I’d submit that even if false, this thread indicates that it’s less than obvious. But I believe my statement is true.

I would certainly agree.

I think of it this way: Monty is essentially asking you to bet on whether you picked right or wrong the first time. Since your chances of picking right were 1 in 3 the first time, you would want to bet that you picked wrong. The way you bet that you picked wrong is by switching.

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**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\
**Post date:** [August 12, 2005, 7:30pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/69 "2005-08-12T19:30:03Z")

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> [@ultrafilter](#):
>
> Run that computer simulation and get back to us.

While he’s doing that, could you tell us why you think that when 50 non-winning cards have been randomly selected and revealed, the remaining two should not have equal probability of being the winner? (If, as it seems, you do think that.)

Suppose instead of chosing a card from the (properly shuffled) deck, I say “I’ll just take the card on the bottom.” You leave it there and start dealing cards from the top, face-up. With 50 cards exposed (probably took many trials to reach this point), the Ace of Spades has not appeared. Which of the remaining two is more likely to be the winner? Why?

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**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\
**Post date:** [August 12, 2005, 7:35pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/70 "2005-08-12T19:35:20Z")

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> [@saoirse](#):
>
> I think of it this way: Monty is essentially asking you to bet on whether you picked right or wrong the first time. Since your chances of picking right were 1 in 3 the first time, you would want to bet that you picked wrong. The way you bet that you picked wrong is by switching.

This is a good characterization of the OMHP.

But we’re currently considering whether and how the problem changes when Monty decides at random which of the remaining doors to open.

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**Author:** ![EllisDee](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ellisdee/32/4531_2.png) [@EllisDee](https://boards.straightdope.com/u/EllisDee)\
**Post date:** [August 12, 2005, 7:54pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/71 "2005-08-12T19:54:36Z")

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I apologize for the line by line quoting. I’m just stopping by quickly, and don’ty have time to form a coherent narrative…

> [@Xema](#):
>
> Well, “patently” means “obviously” and I’d submit that even if false, this thread indicates that it’s less than obvious. But I believe my statement is true.

Very good point; my apologies.

You make a convincing argument, but I still think the deck of cards simulation will show me to be correct. Using your logic:

1. Monty says he will only open a zonk, and he knows where the unpicked zonk is
2. Monty says he will open randomly, and happens to open an unpicked zonk

You are saying the probabilities are different. What if he was lying to you about his motivation? Does the probability then magically change?

> [@](#):
>
> We are presuming a pack of 52 different cards, of which one (say, the Ace of Spades) has been declared the winner. I choose one. Remaining cards are then chosen at random and revealed. If this reaches the point where 50 cards have been exposed and the winning card has not yet been seen, how does the chance that my card is the winner compare to the chance that the one unturned card is?

1:52. Unless, of course, you are saying that your original pick had a 50% chance of being the ace of spades?

> [@](#):
>
> I say they are equal. Indeed, at any point in this game, all unturned cards - mine and all the others not yet exposed - have an equal chance of being the winner.

Correct. They all have a 1 in 52 chance.

> [@](#):
>
> (If the selection process is truly random, how can it be otherwise?) Most trials end before 50 cards have been turned. The probability that the Ace of Spades is the final card is one in 52 - the same as for my original choice.

This sounds very very similar to what I said above about the martingale. ie: 1/52% of the time we picked right, 1/52% of the time we picked wrong, and 50/5% of the time we never got here in the first place. I honestly don’t know who is right. Did anybody simulate this yet?

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<div class="post-metadata">

**Author:** ![EllisDee](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ellisdee/32/4531_2.png) [@EllisDee](https://boards.straightdope.com/u/EllisDee)\
**Post date:** [August 12, 2005, 7:55pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/72 "2005-08-12T19:55:41Z")

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Correction:

“1/52% of the time we picked right, 1/52% of the time we picked wrong, and 50/ **52** % of the time we never got here in the first place. I honestly don’t know who is right. Did anybody simulate this yet?”

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<div class="post-metadata">

**Author:** ![mazinger\_z](https://avatars.discourse-cdn.com/v4/letter/m/5f9b8f/32.png) [@mazinger\_z](https://boards.straightdope.com/u/mazinger_z)\
**Post date:** [August 12, 2005, 8:04pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/73 "2005-08-12T20:04:26Z")

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> [@Xema](#):
>
> While he’s doing that, could you tell us why you think that when 50 non-winning cards have been randomly selected and revealed, the remaining two should not have equal probability of being the winner? (If, as it seems, you do think that.)
> 
> Suppose instead of chosing a card from the (properly shuffled) deck, I say “I’ll just take the card on the bottom.” You leave it there and start dealing cards from the top, face-up. With 50 cards exposed (probably took many trials to reach this point), the Ace of Spades has not appeared. Which of the remaining two is more likely to be the winner? Why?

The odds are the same in both scenarios. (Though, to be pendantic, since you choose the first card, and I choose the remaining card, then I will always have the better odds).

Anyway, to illustrate, for clarity (though the exact nominal odds will be different), instead of 50 cards, use 100 (yes, I know this example has been posted already). The chances in each scenario of you picking the Ace of Spades is 1 in 100, or 1% (no repeating cards in this scenario).

Now, after you pick the card, and without dealing out any more cards and or anything else for that matter, on a scale of 1 to 10 (the scale isn’t important, it’s how you feel how likely is), how likely do you think you picked the A of spaces? Your answer should be very unlikely, or 1 (with 1 being the least likely that you picked it). Remember this last part, it is very important.

Now, let’s continue: you pick your card, I now look through the remaining deck and tell you that these 98 are not the A of spades, and I leave one down. More likely than not (in fact 98% likely, or a 9 out of 10, well 98 out of 100) the last remaining card I have down is the A of spades.

Why? Your initial pick has a 1% chance of being successful. I have just burned or turned over 98 other cards which are also not it. As a matter of logic, do you really think that you picked the A the first time around? 99 times out of 100 you did not. Mind you, we’re not saying that the last card down is the A, we’re just saying that it is 99% certain it is.

Someone with better math skills will come by and illustrate it for you. I just hoped that I was able to show the logic behind it.

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<div class="post-metadata">

**Author:** ![Acsenray](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/acsenray/32/4519_2.png) [@Acsenray](https://boards.straightdope.com/u/Acsenray)\
**Post date:** [August 12, 2005, 8:10pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/74 "2005-08-12T20:10:53Z")

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I’m not buying the arguments here. When Monty opens one of the doors, the probability gets redistributed. The revealed door is no longer part of the problem. It’s 50-50.

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<div class="post-metadata">

**Author:** ![shelbo](https://avatars.discourse-cdn.com/v4/letter/s/3d9bf3/32.png) [@shelbo](https://boards.straightdope.com/u/shelbo)\
**Post date:** [August 12, 2005, 8:13pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/75 "2005-08-12T20:13:40Z")

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> [@PigBoy](#):
>
> I’m not saying anything here that hasn’t already been explained, but what helped the problem make sense to me was simply to visualize the possible situations.
> 
> There are three possible arrangements, shown below (X is the car; O is a zonk). Let’s say you choose Door 1, on the left:
> 
> X O O -Monty opens either 2 or 3; if you switch you lose
> 
> O X O -Monty opens door 3; if you switch to 2 you win
> 
> O O X -Monty opens door 2; if you switch to 3 you win
> 
> You can see that switching wins 2 out of 3 times.

I just wanted to thank **PigBoy** for this post. The scales have fallen from my eyes, and now I can see. A third of the time, you pick the right one initially, so if you switch you lose. 2/3 of the time, you pick the wrong one initially, so if you switch you win.  
Simple!

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<div class="post-metadata">

**Author:** ![mazinger\_z](https://avatars.discourse-cdn.com/v4/letter/m/5f9b8f/32.png) [@mazinger\_z](https://boards.straightdope.com/u/mazinger_z)\
**Post date:** [August 12, 2005, 8:14pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/76 "2005-08-12T20:14:59Z")

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> [@Ellis Dee](#):
>
> I apologize for the line by line quoting. I’m just stopping by quickly, and don’ty have time to form a coherent narrative…Very good point; my apologies.
> 
> You make a convincing argument, but I still think the deck of cards simulation will show me to be correct. Using your logic:
> 
> 1. Monty says he will only open a zonk, and he knows where the unpicked zonk is
> 2. Monty says he will open randomly, and happens to open an unpicked zonk
> 
> You are saying the probabilities are different. What if he was lying to you about his motivation? Does the probability then magically change?

Yes, but it’s not magical.

If Monty was _truly random_ of opening a door, then, as it has been posted before, _he will pick the right door 1/3 of the time._ Monty is a showman, so this is not going to happen. The audience wants to see the contestant sweat it out. If it did happen, and he did pick the right door, then the game would stop there and the contestant would know that he lost. His knowledge of where the prize is influences the odds.\*  
\*- The odds become uncalculable if he is re-bluffing for any particular instance, I believe, but is still 2/3 over the course of the long haul (that is, to stretch the number of times this happens out to infinity).

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<div class="post-metadata">

**Author:** ![mazinger\_z](https://avatars.discourse-cdn.com/v4/letter/m/5f9b8f/32.png) [@mazinger\_z](https://boards.straightdope.com/u/mazinger_z)\
**Post date:** [August 12, 2005, 8:18pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/77 "2005-08-12T20:18:59Z")

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> [@mazinger\_z](#):
>
> …we’re just saying that it is 9 **8** % certain it is.

D’oh! :smack:

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<div class="post-metadata">

**Author:** ![mkl12](https://avatars.discourse-cdn.com/v4/letter/m/b4bc9f/32.png) [@mkl12](https://boards.straightdope.com/u/mkl12)\
**Post date:** [August 12, 2005, 8:59pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/78 "2005-08-12T20:59:02Z")

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> [@acsenray](#):
>
> I’m not buying the arguments here. When Monty opens one of the doors, the probability gets redistributed. The revealed door is no longer part of the problem. It’s 50-50.

Which case are you talking about? Monty chooses a Zonk or he randomly opens a door and it happens to be a Zonk?

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<div class="post-metadata">

**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\
**Post date:** [August 12, 2005, 9:03pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/79 "2005-08-12T21:03:00Z")

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> [@ultrafilter](#):
>
> Run that computer simulation and get back to us.

“events: 100 successes: 49”

Looks pretty close to 50/50 to me.

```auto

program MontyCards;

var
  firstpick,prizecard,trialpick,count :integer;
  events, successes :integer;
  deck :set of 1..52;

begin
Randomize;
events:=0;
successes:=0;
repeat
  deck := [1..52];
  prizecard := Trunc(Random * 52 + 1);
  firstpick := Trunc(Random * 52 + 1);
  deck := deck - [firstpick]; { remove my pick from avail choices }
  { now have Monty guess until only one card is left or he shows the prize }
  count := 51;
  repeat
    trialpick := Trunc(Random * 52 + 1);
    if (trialpick in deck) then
      begin
      count := count-1;
      deck := deck - [trialpick];
      end;
  until (count=1) or (trialpick=prizecard);

  { did Monty not reveal the prize card? IOW, is the price card either
    my card, or the one left? }
  if (trialpick <> prizecard) or (firstpick = prizecard) then
    begin
    { these are the only events that count }
    events := events + 1;
    if firstpick <> prizecard then successes := successes + 1;
    writeln('events: ',events,' successes: ',successes);
    end;

until events=100;
end.

```

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<div class="post-metadata">

**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\
**Post date:** [August 12, 2005, 9:27pm UTC](https://boards.straightdope.com/t/explain-the-monty-hall-problem-to-me/316632/80 "2005-08-12T21:27:34Z")

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> [@Ellis Dee](#):
>
> What if he was lying to you about his motivation? Does the probability then magically change?

It changes, but not through magic. If Monty says he’s picking a door at random but in fact is peeking and basing his selection on what he sees, then things change fundamentally.

> [@](#):
>
> 1:52. Unless, of course, you are saying that your original pick had a 50% chance of being the ace of spades?

I believe and hope I’ve said nothing to that effect.  
If I understand correctly, you seem to be saying that the probability I’ve chosen the winner stays constant throughout this game. Whereas it seems to me that as cards are randomly chosen, exposed and shown to be losers, the chance that my card (and all other unexposed cards) may win is increasing. When 32 cards have thus been exposed, I’d say my odds are up to 1:20.

> [@](#):
>
> I honestly don’t know who is right. Did anybody simulate this yet?

Can I note that this is quite a change (IMO, an improvement) from “patently false”. I’ll also note that this problem (like most others of its type) should be amenable to analysis; simulation need not be seen as the final arbiter of truth.

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