# Monty Hall Revisited: Winning Both Goats!

Suppose you're at a game show, knowing the winning strategy for the classic Monty Hall problem. Your turn comes up and you take the stage. You confidently choose door number (?).

Just as you are getting ready to tell the host you want to switch doors, the host announces that the rules have been changed for today's show. You groan in disappointment.

The host explains that no doors are going to be revealed or opened prematurely, unlike the classic Monty Hall problem. Instead, you are given the choice of keeping the door you picked, or switching to the other two doors. If you stay, and the car is not behind your door, you lose. If it is, you win. If you switch, and the car is behind one of the two other doors, you win. It can be behind either one. If it was behind the first door you picked, you lose.

Suddenly you feel confident again. You realize that just like in the classic Monty Hall, your chances of winning the car if you switch are 2/3: you now get two doors instead of one. You tell the host you would like to switch to the two doors.

The host then asks you to reconsider, explaining that your chances aren't quite as good as you thought. The odds of a single door hiding a goat are 2/3, since two of the three doors hide a goat. Then, multiplying these together (for two doors) yields a probability of 4/9 that both of your two doors contain goats, yielding you a 5/9 chance of winning the car. The host grins as you realize your anticipated 2/3 odds have shrunk slightly, but you still decide to switch to the two doors. After all, 5/9 is still better than 1/3.

As you leave, goat leashes in hand, it still bothers you that your initial intuitive guess seemed to have been proven wrong by the host. You console yourself with the fact that you won both goats. Who ever needed a lawn mower, anyhow?

What is your true probability of winning if you switch to the two doors? Who was right, and why?

Bonus point challenge: There is yet another way to mathematically prove which is the correct answer, found by dark wanderer (using the sum of all probabilities rule).

• Do we know whether Monty's decision to make/change this offer was contingent on his knowledge of what door was initially picked? – Zomulgustar Apr 12 '18 at 4:54
• @AndyJuell It is not related. Assume it was always going to happen. – Tim Apr 12 '18 at 5:19
• @AndyJuell We know that the real Monty never let the contestant change. He would sometimes show another door, a door that guaranteed did not have the big win, just to heighten tension. But that was just pure showmanship as he did not let the contestant change, so it did noting to influence the chances of winning. – MichaelK Apr 12 '18 at 10:40
• What if the car is a Lada? What if one of the goats is actually the G.O.A.T. ? (New England Pats fans know what I mean) – Carl Witthoft Apr 12 '18 at 17:34

Leaving aside the dubious assumption that Monty is entirely on the up-and-up...

Multiplying the probabilities to get 4/9 would only be appropriate if the two probability-2/3 events were independent, which they are not.

As with the original MHP, knowledge about what is behind one door gives us partial information about what is behind the others. (Like three playing cards shown, then shuffled face down). If we knew that a fair die had been rolled for each door, with a 1-4 stocking it with a goat, and a 5 or 6 yielding a car, then the 5/9 figure would be applicable...but there would be a 8/27 chance that all three doors had goats, and a 1/27 all three cars. In short, not the scenario given.

There are three ways of selecting two of the three doors. One of these selections has both goats, and the other two selections have a car and a goat. Assuming these three alternatives are equally likely (not guaranteed), two of them being successes yields the original 2/3 probability.

• Very nice summary, that’s a clever way to explain it. – Tim Apr 12 '18 at 5:18

The host then asks you to reconsider, explaining that your chances aren't quite as good as you thought. The odds of a single door hiding a goat are 2/3, since two of the three doors hide a goat. Then, multiplying these together (for two doors) yields a probability of 4/9 that both of your two doors contain goats

This doesn't work because

on the condition that one of the doors (say door 2) contains a goat, the probability for either of the other two containing one becomes 1/2. So we actually calculate 2/3 * 1/2 = 2/6 chance that both of the two remaining doors contain a goat, meaning you still have a 2/3 chance of winning if you switch.

• Excellent work, I was hoping someone would reconcile the two results mathematically. I was going to add this anyway as an extra challenge for bonus points (upvote). Upvote earned! – Tim Apr 12 '18 at 14:30

Others have said the answer but I thought I'd chime in since I had a different explanation.

2/3

And the reason is very easy:

When you choose a door you have a one in three chance of choosing the car. You are then told you can have both other doors but you have no new information. With no new information the probabilities can't have changed so there is clearly a 2/3 chance of winning the car if you get to have both other doors.

You could rephrase the question as:

"You are told the car is behind one door and you get to choose either two doors or one door."

• This is simple and correct – Floris Apr 13 '18 at 2:33

Let's do the classical probability tree:

Not switching

                             ,---------´  |  ---------.          (1/3 each)
car behind door           1               2               3
/  |  \         /  |  \         /  |  \    (1/3 each)
first door chosen    1    2    3     1    2    3     1    2    3
|    |    |     |    |    |     |    |    |
you get             Car Goat Goat   Goat Car Goat   Goat Goat Car (1/9 each)


Switching:

                             ,---------´  |  ---------.          (1/3 each)
car behind door           1               2               3
/  |  \         /  |  \         /  |  \    (1/3 each)
first door chosen    1    2    3     1    2    3     1    2    3
|    |    |     |    |    |     |    |    |
switching to doors  2&3  1&3  1&2   2&3  1&3  1&2   2&3  1&3  1&2
|    |    |     |    |    |     |    |    |
you get             Goat Car  Car   Car  Goat Car   Car  Car Goat (1/9 each)


So

the odds of getting a car using the Not Switching strategy are 3/9 = 1/3,
while using the Switching strategy they are 6/9 = 2/3.

• "you always get the car whether you switch or not": no, that would contradict the OP which reads "If you switch, and the car (...) was behind the first door you picked, you lose." You just mixed up G and C in the last line of your second graph. – xhienne Apr 12 '18 at 19:20
• @xhienne Oops, I think I misinterpreted "If you switch, and the car is behind one of the two other doors, you win." (as "other doors than the chosen", not "other than the first"). Guess I need to update the answer, but then the graph becomes really boring… I would have assumed that you have to pick a specific door when switching, not just saying "I'll choose the other two doors". – Bergi Apr 12 '18 at 19:27
• Thank you for correcting your answer. BTW, choosing a specific door when switching is what the original MHP is about. This puzzle is a variation of it. – xhienne Apr 12 '18 at 19:34
• @xhienne Yes, I somehow thought that the variation is only that the first door is not opened, and that you don't get the item behind the second door but the items behind the doors other than the selected. – Bergi Apr 12 '18 at 19:40
• Upvoted since this answer is fairly unique, but no philosophical bonus points. The alternate mathematical proof still has yet to be mentioned, though it's quite simple. – Tim Apr 12 '18 at 23:00

@Andy Juell already mentioned independence and @Lily Potter already mentioned conditional probability, but I thought I'd give a closer look at how these relate in general.

Say we have two events that we're interested in, call them $A$ and $B$. We wonder what the probability is that both will occur, that is, we want to know $\mathbb P(A \cap B)$ where $\cap$ denotes the intersection of the two events. We can always decompose a joint probability (a probability of multiple things) into conditional probabilities: $\mathbb P(A \cap B) = \mathbb P(A) \mathbb P(B|A)$ where $\mathbb P(B|A)$ is the conditional probability that $B$ occurs, given that we already know that $A$ occurred (note that the other decomposition, $\mathbb P(A \cap B) = \mathbb P(B) \mathbb P(A|B)$, is equivalent; the two individual terms may differ, but their product will be the same).

If the two events are independent (i.e. the outcome of one has no influence on the outcome of the other), then $\mathbb P(B|A) = P(B)$ because the probability distribution for $B$ won't change regardless of the value for $A$ that we observe.

Now, on to the problem at hand.

Suppose that we've chosen door 1 and are deciding whether we should switch (the argument holds for any door we choose, just relabel the door names). We want to know what the probability is that there are goats behind doors 2 and 3; call these events $g_2$ and $g_3$, and we want $\mathbb P(g_2 \cap g_3)$. We know that this is the same as $\mathbb P(g_2) \mathbb P(g_3|g_2)$. There are two goats and one car, so if they're distributed uniformly at random, there's a 2/3 chance of a goat behind each door, so $\mathbb P(g_2) = 2/3$. Note that $\mathbb P(g_3) = 2/3$ too, but we're interested in $\mathbb P(g_3|g_2)$. If we know that there is a goat behind door two, there's only one goat left (and one car), so $\mathbb P(g_3|g_2)=1/2$; clearly $g_3$ and $g_2$ are not independent; this was the error that Monty made. Thus we get $$\mathbb P(g_2 \cap g_3) = \mathbb P(g_2) \mathbb P(g_3|g_2) = \frac{2}{3} * \frac{1}{2} = \frac{1}{3} \neq \mathbb P(g_2) \mathbb P(g_3) = \frac{2}{3} * \frac{2}{3} = \frac{4}{9}$$ Since the probability of both other doors having goats is 1/3, the probability of at least one of them having a car is 2/3, as intuited.
Note that this is a common method in probability - if you want to know the probability that at least one of a set of events occurs, it's often easiest to determine this as $\mathbb P(\ge 1 \text{ event occurs}) = 1 - \mathbb P(\text{no event occurs})$.

• Very thorough. The last line inside your spoiler tag is the closest anyone has come to getting the bonus challenge. – Tim Apr 13 '18 at 14:49

There are three doors and one car. The car is placed randomly. Therefore you have a chance of picking the correct door equal to the percentage of the pool you select. Anything else is a distraction.

Monty's argument is a form of 'whataboutery' or 'whataboutism', a variation of the tu quoque fallacy where an arguer proposes unrelated material in order to cast doubt on a position without actually refuting any arguments. Your initial logic is in no way attacked or contravened by Monty's position and you have no need to address the competing argument as you know your argument to be sound. While his argument can be shown to be false (as addressed in other answers), it is more important to trust that mathematics functions reliably-- and thus to not need to evaluate Monty's counterpoint.

To be clear, you and Monty both agree:

• The car is placed at random behind one of the doors
• When you make the final choice to switch or not, because you have been given no information, the odds that an object, whether car or goat, is behind any given door are the same as for each other door.
• There are three doors
• The probability that a car lies behind the door you first chose is 1 in 3
• The probability that a thing doesn't happen is 1 minus the probability that it does

Thusly, you reason "There is a 1/3 chance that the car is parked behind the door I have chosen. Therefore, there is a 2/3 chance that it is not parked behind the door I have chosen, because 1 minus 1/3rd is 2/3rds. 2/3rds is larger than 1/3rd, so I ought to choose to switch".

There are a number of potential points of disagreement to this argument. Monty might argue that 1 less 1/3 is not 2/3rds. Monty might argue that the car not being behind your starting door doesn't mean it is behind any of the other doors. Monty does not do any of this, though. Instead, Monty accepts each of your premises but rejects your conclusion, similarly to the tortoise in What the Tortoise Said to Achilles, putting forth an unrelated argument which he claims provides a different account of the probabilities, and thus shows you are wrong.

• I didn't downvote, but the OP doesn't actually show any math that the player did to determine what his chances of winning were. I presume that you assume he did the correct math in his head, but this answer might be improved if you said what math it was that he did / should have done, and then stated how, given his correct formulation of the problem, it doesn't matter how Monty tries to distract or confuse him (because Monty is also relying on some math that he did, so we need to compare the two methods and see who had a false assumption or arithmetic error). – Nathan Apr 13 '18 at 13:57
• Good job solving the bonus point challenge. Having said that, just because you don't need to refute an argument doesn't mean you shouldn't, especially since that was the very question asked. You could improve your answer further by demonstrating that you understand the flaw in the incorrect part. Multiplying probabilities together is a correct way to calculate statistics, and it should be understood at a deeper level than just, this is the way it is. Why is it that way? That was the main challenge of this question. – Tim Apr 13 '18 at 22:01

the $\frac23\times\frac23=\frac49$ argument is fallacious because the events are not independent
By that argument, there’s a $\frac23\times\frac23\times\frac23=\frac{8}{27}$ probability that all three doors have goats,
and a $\frac13\times\frac13=\frac19$ probability that the “other two doors” both have cars.