# Another Monty Hall Question

I still do not believe the "correct" solution to the Monty Hall Problem.

Here is my reasoning: The player can pick from $1$ of $3$ doors. The prize can be behind $1$ of $3$ doors. Monty will open $1$ of $3$ doors.

$3 \times 3 \times 3 = 27$ possible sequences of events.

In $15$ of those possible events, Monty either opens the door the player picked or the door with the prize. Since Monty will not do either of those, these $15$ events are removed from the possibilities.

Of the remaining $12$ possibilities, $6$ times the player wins if he stays with his original pick and $6$ times he wins if he switches. It looks to me like the player has the same chance of winning whether he stays or switches.

Could someone please explain the flaw in my reasoning.

• The "27 possible sequences of events" does not make sense at all, as you are assuming that the three doors are distinct (i.e., the one the player chooses, where the prize is, and the one that Monty opens). – Clarinetist Jan 3 '15 at 5:31
• I recommend watching the numberphile video on the topic youtube.com/watch?v=4Lb-6rxZxx0 – JMoravitz Jan 3 '15 at 5:34
• There is nothing to believe here. One has first to agree which events are equally likely (the TV station might put the prize always behind the same door) and the rest follows by simple reasoning... – Fabian Jan 3 '15 at 6:30
• @JMoravitz actually the extended math version is more interesting – Caran-d'Ache Jan 3 '15 at 6:39
• Monty's actions are constrained as follows: (1) He knows what's behind each door. (2) He must open a door. (3) He must open a door hiding a goat. Once you factor in these rules, the puzzle becomes obvious. – user4894 Jan 3 '15 at 6:59

The basic flaw in the reasoning is that it assumes that each of the 12 possible outcomes are equally likely -- because without such an assumption "same number of outcomes" does not translate into "equal probability".

In many simple probability situations we can get a plausible "all outcomes are equally likely" assumption from symmetry arguments. But that won't hold here because the group of outcomes where the contestant chose the prize door are fundamentally different from the group of outcomes where the contestant chose a non-prize door; the host have different amount of options in the two cases.

As a simpler illustration that it doesn't always work to assume that all the possibilities are equally likely, consider this different game:

1. First the contestant flips a fair coin.
2. If the coin came out heads, the host rolls a fair die once; otherwise the host keeps rolling the die again and again until he rolls 6.

Here there are a priori 12 outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}. But some of them are impossible; if we exclude them there are only 7 outcomes: {H1, H2, H3, H4, H5, H6, T6}. In 6 of the cases the coin flip was head; only one cases was the tails.

Can we then conclude the odds of flipping heads is 6 to 1?

Of course not. The coin is still fair, and what happens after it is flipped can't influence the probability of coming up tails. So this shows that comparing numbers of outcomes doesn't always give true probabilities, which ruins your Monty Hall analysis.

The reason that your numbers come out like that is because there are two possible doors for Monty to open if the contestant has already picked the right answer, but only one possible door for Monty to open if the contestant has picked wrongly. In the "two possible doors" case, switching to either of the doors loses, but this only happens in 1/3 of cases, so it is not meaningful to count in this way.

Here is an image. D indicates the door opened, the right side is the contestant's initial choice, and the top indicates which doors Monty can open:

If the prize is at A (D=A above), the contestant picks A, then Monty can choose either B or C doors (as marked in green circles). If the contestant picks B when the correct answer is A, then Monty can open only C, and vice-versa if the contestant picks C (the pink circles are Monty's choices)

Assume that the contestant's pick $P$, the prize distribution $D$, and Monty's choice $M$ are uniformly distributed among the possibilities, then $$P(M=B|P=A,D=A)=P(M=C|P=A,D=A)=1/2$$ but $$P(M=B|P=B,D=A)=0$$ and $$P(M=C|P=B,D=A)=1$$ so since the pick and the prize are independent, \begin{align} P(M=C,P=B,D=A)&=P(M=C|P=B,D=A) P(P=B) P(D=A)\\&=1\times\frac13\times\frac13\\&={1\over9} \end{align} but $$P(M=C,P=A,D=A)=P(M=C|P=A,D=A) P(P=A)P(D=A)={1\over18}$$ In other words the green circles only have probability $1/18$ whereas the pink circles have probability $1/9$.

The multiplicity of your counting just relates to Monty's freedom to choose either B or C when the contestant has the right answer.

As another way to look at this, it might be helpful to think of a situation where it would be appropriate to use your calculation.

Suppose the game is played as follows: the player, the game studio, and Monty each secretly choose a door and write their choice on a piece of paper. A judge looks at the papers, and if Monty's choice is the same as the player's or the same as the studio's (i.e., where the prize is kept), then the papers are destroyed and the process is repeated—this will eliminate 15 of the 27 possibilities, as you have noted, and now the player has he same chance of winning whether he keeps his original choice or switches. Your calculation makes sense here.

This is different from the usual Monty Hall problem, however, because Monty normally makes his choice after the player and studio have already chosen.

Compare to the coin/die game in Henning Makholm's answer: if the coin and die were tossed simultaneously until you got a legal combination, then the odds of flipping heads would indeed be 6 to 1.

• This argument is definitely wrong. – Suzu Hirose Jan 3 '15 at 7:02
• @SuzuHirose Would you care to elaborate? – Théophile Jan 3 '15 at 7:04

I'll try to give a simple answer.
As you note, there are three choices involved:
Where is the prize?
What will the contestant choose?
What door will monty open?
Let's call the doors A B and C, and let's get to write down the possible scenarios in a simbolic manner;
(A,B,C), for instance, would represent the scenario in which the prize is in A, the contestant chooses B and monty opens C (i kept the order of the questions).
(A,B,C) is a possible scenario, as opposed to, let's say, (A,B,A) (where monty would open the prize door), or (A,B,B) (where monty would open the contestant's choice).

So, the possibile scenarios are these:
(A,A,B) (A,A,C) (A,B,C) (A,C,B)
(B,B,C) (B,B,A) (B,C,A) (B,A,C)
(C,C,A) (C,C,B) (C,A,B) (C,B,A),
and only half of these would make the contestant win if he chooses to switch. Now, the question is:
are these scenarios equally likely to happen? Is (A,A,B) equally likely as, let's say, (A,B,C)?

Well, let's take a step back.
The blurry territory comes with monty's choice;
before it, we're sure that there are nine possibilities, and those are equally probable.

List:
(A,A) (A,B) (A,C)
(B,B) (B,C) (B,A)
(C,C) (C,A) (C,B)

So... couldn't we see what happens next like this?

(A,A,(B or C)) (A,B,C) (A,C,B)
(B,B,(C or B)) (B,C,A) (B,A,C)
(C,C,(A or B)) (C,A,B) (C,B,A)

I mean, let's look at (A,A,B) and (A,A,C).
Aren't they just two faces of the same possibility, (A,A)?
What i'm suggesting is that at the end there aren't twelve equally probable possibilities.
There are nine, and three of those have two possibile manifestations. Or, there are... 18, and 12 of those have an identical twin.
Like, maybe monty will open the door either slowly or fast at random if you don't choose the prize at first, and at normal speed if you choose the right door;
and, you wouldn't be able to see the difference between those speeds, because you just see him open the door once, without any other reference; still, what you see must be one of three possibilities: normal, slow or fast.

So you would have

(A,A,(B or C),n) (A,B,C,(s or f)) (A,C,B,(s or f))
(B,B,(C or B),n) (B,C,A,(s or f)) (B,A,C,(s or f))
(C,C,(A or B),n) (C,A,B,(s or f)) (C,B,A,(s or f)),

at least with the schematization i proposed.
With yours, we would have 18 possibilities instead of nine two faced, but by both schematizations, we would come to the conclusion that it's better to switch.
Ok this is getting confusing, I'll stop...

Monty does not have three choices, but only one or two. Initially before Monty opens a door the probabilities are as you expect:

• You choose A, prize behind A: probability $\frac19$
• You choose A, prize behind B: probability $\frac19$
• You choose A, prize behind C: probability $\frac19$
• You choose B, prize behind A: probability $\frac19$
• You choose B, prize behind B: probability $\frac19$
• You choose B, prize behind C: probability $\frac19$
• You choose C, prize behind A: probability $\frac19$
• You choose C, prize behind B: probability $\frac19$
• You choose C, prize behind C: probability $\frac19$

Given this, Monty has to open a door which is not you initial choice and which is not the one with a prize. Sometimes Monty has a choice of door, and sometimes not, leading to $12$ possibilities with varying probabilities. Once Monty has opened a door, you can win or lose by switching and no further chance is involved as there are only two remaining doors. The probabilities become:

• You choose A, prize behind A, Monty opens B: probability $\frac1{18}$, switching loses
• You choose A, prize behind A, Monty opens C: probability $\frac1{18}$, switching loses
• You choose A, prize behind B, Monty opens C: probability $\frac1{9}$, switching wins
• You choose A, prize behind C, Monty opens B: probability $\frac1{9}$, switching wins
• You choose B, prize behind A, Monty opens C: probability $\frac1{9}$, switching wins
• You choose B, prize behind B, Monty opens A: probability $\frac1{18}$, switching loses
• You choose B, prize behind B, Monty opens C: probability $\frac1{18}$, switching loses
• You choose B, prize behind C, Monty opens A: probability $\frac1{9}$, switching wins
• You choose C, prize behind A, Monty opens B: probability $\frac1{9}$, switching wins
• You choose C, prize behind B, Monty opens A: probability $\frac1{9}$, switching wins
• You choose C, prize behind C, Monty opens A: probability $\frac1{18}$, switching loses
• You choose C, prize behind C, Monty opens B: probability $\frac1{18}$, switching loses

So overall switching wins with probability $6\times \dfrac19=\dfrac23$ and loses with probability $6\times \dfrac1{18}=\dfrac13$.

In my opinion the reason why the Monty Hall problem is confusing is because people make the solution unnecessarily complicated. I have a very simple solution that goes like this.

The player would win by switching doors if and only if (s)he chose the wrong door initially, which happens 2/3 of the time.

• Yes, that's a solution in a nutshell. The question then is, when the OP tried a different solution method and came up with the wrong answer, exactly which steps of the "solution" were incorrect? – David K Jul 27 '15 at 20:31
• @DavidK: I guess I don't see the point in sifting through long complicated incorrect solutions, when I have a very easy and correct one. Maybe this is from having to grade too many students' proofs :) In general I'd say it's a failure to consider the additional info from starting with 3 doors. – j0equ1nn Oct 22 '15 at 9:09
• It can be useful to someone to learn from mistakes, but I doubt a student learns much from a mistake when all they find out is that a completely different approach to a particular problem gives a different answer. When I graded papers, I tried at least to identify the first incorrect step; but maybe this just shows I never graded enough papers to learn I shouldn't do that. :) – David K Oct 22 '15 at 13:25
• In this case there's an obvious error in OP's argument: it assumes 27 events are equiprobable, using a model that is appropriate for three independent variables; but these variables are not independent in the problem statement, so the model is not justified. At least two answers have already expounded on this point, however, so I guess we can just leave it at that. – David K Oct 22 '15 at 13:29
• @DavidK: Yeah, the other answers did cover what was wrong with the argument posed. I just like for people to know the solution need not be difficult or counterintuitive. I do think it's meaningful though to discuss what constitutes a valid proof, versus an intuitively sayisfying one (like Thurston's "Proof and Progress" article). – j0equ1nn Oct 22 '15 at 17:46

The doors can be opened in any order. For 3 different elements {$g_1, g_2, C$} (assuming the two goats are indistinguishable) , there are only $3!$ ways of ordering them (without replacement, which is the case in the Monty Hall problem). But Monty narrows it down even more by showing you a door with a goat in it.

• So just what is wrong with this? This starts with the wrong premise that there are $27$ choices; I corrected the false premise, seriously. – user203864 Jan 3 '15 at 6:04

3×3×3=27 possible sequences of events

what are "sequences of events"? a sequence implies to me a strict ordering of said events, with possible dependencies between the ordered events. i am confused about your original premise. see clarinetist's first reply concerning the construction of a 3x3x3 space (not sequences).

i've always found the proof easier to grok if you increase the number of doors. try reasoning about 100 doors instead of 3 doors. 1 door conceals the prize, 99 doors conceal goats.

pick a door. you have a 1/100 chance of picking the prize. simple. monty can't change those odds and he has no idea which door you will pick.

next, monty, knowing which door conceals the prize, proves to you that 98 doors hid goats, by opening those 98 doors, leaving 2 doors still closed: your door and monty's door. do you still believe you chose the prize on the first pick?

i claim that on average switching to monty's door makes you correct way more than 1% of the time, simply because the odds of you picking correctly the first time are slim with 100 doors.

now, redo the math with 3 doors instead of 100.

• In this case though, OP's problem with the poor counting methodology remains, there are 99 ways for Monty to open the 98 doors if you have picked the correct door (just pick one door of all 99 "goat" doors not to open, and there are 99 ways to do this), and there is only one possible way to pick the door not to open in the other 99 cases, because Monty has to open 98 goats, so the miscounting problem OP described remains. – Suzu Hirose Jan 3 '15 at 6:35
• not sure i follow your reasoning. yes, most people would like to believe they chose the correct door on the first pick. at 1/100 odds they clearly won't chose the prize most of the time. with 100 doors those odds are obviously skewed against the participant. with three doors not so clear. what is wrong with my reasoning? – user62627 Jan 3 '15 at 14:52
• If you use the same reasoning as in OP's post, the same fallacy which OP made with three doors also occurs with your method. – Suzu Hirose Jan 3 '15 at 22:17
• yes, a true statement, i suppose, but i'm pretty sure i'm NOT using the same reasoning, so i don't understand how your reply applies to my comment. – user62627 Jan 4 '15 at 3:20
• My reply applies to your answer as follows. Your argument does not solve OP's problem or point out where OP has gone wrong. If you use the same reasoning as OP used, even with 100 doors the identical miscounting fallacy remains, exactly the same as in the original problem. In other words this does not answer OP's question at all. Using 100 doors does not make the fallacy go away. Perhaps you need to read the question or something. – Suzu Hirose Jan 4 '15 at 3:28

If you want a set of outcomes with equal probability, you need to account somehow for the fact that when the contestant chooses the prize door on the first guess, Monty has a choice of which door to open.

This is best modeled as a random variable with two equally likely outcomes. We might as well assume Monty flips a coin and opens the leftmost possible door if the coin is heads, the rightmost possible door if the coin is tails. So:

Prize at A, contestant guesses A, coin is heads $\implies$ open door B.

Prize at A, contestant guesses A, coin is tails $\implies$ open door C.

Prize at A, contestant guesses B, coin is heads $\implies$ open door C.

Prize at A, contestant guesses B, coin is tails $\implies$ open door C.

These are four equally likely events. The difference with your "solution" is you only counted one A-B-C event and here there are two A-B-C events (A-B-heads-C and A-B-tails-C).

If you count all the possible outcomes including the coin flip every time (even when it isn't "needed") then you have 18 equally likely outcomes; 12 are winners and 6 are losers if you switch.

# Without swapping

If the player chooses

• $g_1$, he still chooses $g_1$.
• $g_2$, he still chooses $g_2$.
• $c$, he still chooses $c$.

The sample space $S=\{g_1,g_2,c\}$ and the expected event $A=\{c\}$.

$$p(A)=\frac{n(A)}{n(S)}=\frac{1}{3}$$

# With swapping

If the player chooses

• $g_1$, he will be led to choose $c$.
• $g_2$, he will be led to choose $c$.
• $c$, he will be led to choose $g$.

The sample space $S=\{c,c,g\}$ and the expected event $A=\{c,c\}$.

$$p(A)=\frac{n(A)}{n(S)}=\frac{2}{3}$$