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I recently posted a tweet claiming I had encountered a real life Monty Hall dilemma. Based on the resulting discussion, I'm not sure I have.


The Scenario

  • I have 3 tacos (A,B,C) where tacos A and C are filled with beans, and taco B is filled with steak.

  • I have no foreknowledge of the filling of any tacos.

  • My wife only knows that taco C is filled with beans.

  • My wife and I both know that I want steak.

  • After I pick taco A, my wife informs me taco C is filled with beans.

  • I switch my pick from taco A to taco B, thinking the logic behind the Monty Hall problem is relevant to my choice.


Edit for clarity

  • Timing: The contents of taco C were not revealed to me until after I had made my selection of taco A.

  • My knowledge of what my wife knew: When she told me the contents of taco C, I knew that she had previously opened taco C. I also knew that she had no other knowledge of the contents of the other tacos.

Questions

  1. Even though my wife does not know the fillings of all the tacos, does her revealing that taco C is definitively not the taco I want after I've made my initial selection satisfy the logic for me switching (from A to B) if I thought it would give me a 66.6% chance of getting the steak taco?
  2. If this is not a Monty Hall situation, is there any benefit in me switching?
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    $\begingroup$ what if you had chosen taco C. what would your wife have told you then? $\endgroup$
    – zhw.
    Commented May 25, 2018 at 16:42
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    $\begingroup$ It is not a true Monty Hall problem. Monty knows where the prize is and isn't. Your wife has incomplete information. The best you can do is give yourself a $50/50$ chance of getting the right taco. $\endgroup$
    – Doug M
    Commented May 25, 2018 at 16:43
  • $\begingroup$ @zhw. If I had chosen C, the only thing she could have told me was that C was definitively full of beans and not what I wanted. She had no other knowledge of the fillings of any other tacos. $\endgroup$
    – Will Cole
    Commented May 25, 2018 at 16:43
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    $\begingroup$ I don't get it, if you have 3 tacos and any of them have steak - then you've got the taco with the steak. What does your wife have to do with it? She can get her own tacos. $\endgroup$
    – davidbak
    Commented May 25, 2018 at 19:57
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    $\begingroup$ I'd call this a generalized Monty Hall situation. We can imagine Monty having various states of knowledge and strategy (random, haphazard, benevolent, malevolent, etc.). Excellent question, and I'm pretty sure the answer depends upon timing - I think it may matter when she obtains the knowledge of the taco not containing steak - whether before or after you've made your choice, and whether her selection of which taco to inspect included the one you initially picked. A lot of people have gone out on a limb and a lot of people are going to be wrong. $\endgroup$ Commented May 25, 2018 at 21:32

12 Answers 12

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No, this is not a Monty Hall problem. If your wife only knew the contents of #3, and was going to reveal it regardless, then the odds were always 50/50/0. The information never changed. It was just delayed until after your original choice. Essentially, you NEVER had the chance to pick #3, as she would have immediately told you it was wrong. (In this case, she is on your team, and essentially part of the player). #3 would be eliminated regardless: "No, not that one!"

Imagine you had picked #3. Monty Hall never said, "You picked a goat. Want to switch?"

If he did, the odds would immediately become 50/50, which is what we have here.

Monty always reveals the worst half of the 2/3 you didn't select, leaving the player at 33/67/0.

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    $\begingroup$ Does this change at all if Will doesn't know that his wife only knows the contents of one taco? All Will knows is that he picked one taco, and his wife revealed a different one to be the wrong choice. From his point of view, it looks rather equivalent to Monty Hall. There's a 1/3 chance he picks taco C to start with and then it's a 50-50 shot at the other two, but there's a 2/3 chance he doesn't pick taco C to start, in which case we might be back to Monty Hall. $\endgroup$ Commented May 25, 2018 at 19:55
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    $\begingroup$ @NuclearWang I don't think so. The game seems to be 1) Will picks a taco. 2) Wife tells him taco C has beans. 3) Will gets to pick between taco A and taco B. His first choice is inconsequential, regardless of whether or not he knows his wife is playing along. If he thinks she knows all 3, then he might think his odds improve, but they don't actually. $\endgroup$ Commented May 25, 2018 at 20:18
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    $\begingroup$ Yes, if Will thinks his wife is acting like Monty Hall, revealing the worst half of the 2/3 he didn't select, then he is in a Monty Hall problem, and should switch. $\endgroup$
    – GAEfan
    Commented May 25, 2018 at 20:20
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    $\begingroup$ @LordFarquaad is correct, whatever the outcome is, it can only depend on the wife's actual knowledge and not perceived knowledge. The wife's perceived knowledge could possibly affect Will's perceived odds, but not the true odds. If Will chooses his final taco under the assumption of a Monty Hall wife, is about to bite into it, and then learns that the revealed bean taco was pre-determined, it won't change the likelihood of Will biting into a steak taco. It could possibly change Will's own estimate of his likelihood of having the steak, though. $\endgroup$ Commented May 25, 2018 at 21:05
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    $\begingroup$ No change. If she was always going to reveal #3 as beans, then it was always 50/50/0. Monty doesn't always reveal door #3; He reveals the worst of the other 2. Different game. $\endgroup$
    – GAEfan
    Commented May 25, 2018 at 22:29
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tl;dr: Switching in this case has no effect, unlike the Monty Hall problem, where switching doubles your odds.

The reason this is different is that Will's wife knew the content of one, and only one taco, and that it was a bean one, and Will knows that his wife only knew the content of one bean one. (Monty is different because he knew all doors.)

Here's why:

Unlike the MH problem, which has one car, and two goats that can be treated as identical, the Will's Wife Problem has one Steak, One known bean, and one unknown bean, so the beans need to be considered differently.

That's important because MH's reveal gave players a strong incentive to switch, but gave them NO new information by revealing a goat in an unpicked door - whatever the player picked, he could show a goat, so no new info is provided by the reveal. But that's not the case for Will's wife:

Since she only knows the content of the known bean taco, her revealing that it's not one of the ones you picked actually changes what you know about your odds. Because she would have behaved differently if you'd picked the known bean one - she'd have said, "the one you picked is bean".

Without her info, you only had a 1/3 chance of having the steak, but by showing you that you didn't pick the only bean one she knew about, it means you already know you have a 50% chance of having the steak.

And since you also have a 50% chance of having the unknown bean taco, it's irrelevant if you switch or not.

In the MH problem, the key fact is this: Since the reveal in no way changes what you know about your odds, there's only a 1/3 chance that you START with the car. And since you:

  • Win by staying in all cases where you started with the car, and
  • Win by switching in all cases where you didn't start with the car...

In the MH problem, switching doubles your odds (from 1/3 to 2/3), but in this case, switching has no impact (since it's 50% either way).

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  • $\begingroup$ This seems to require that Will knows that his wife knows the contents of only one of the bean tacos, which isn't stated in the OP. If Will doesn't know what his wife knows, does this change anything? $\endgroup$ Commented May 25, 2018 at 19:59
  • $\begingroup$ It does. In this particular case, Will's wife and Monty Hall can both do the same thing, but only because of the specific one Will picked. It's only because Will knows that they're using different rule-sets that he should switch with MH, and that it doesn't matter with his wife. $\endgroup$
    – Jaydles
    Commented May 25, 2018 at 20:18
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    $\begingroup$ So what if the wife lies and says she knows all the tacos, Will selects his final taco and is about to bite into it, and then the wife admits that the revealed taco was pre-determined? I don't see how that can change the true likelihood of Will holding the steak taco, although it can change Will's own perception of his likelihood of having the steak. $\endgroup$ Commented May 25, 2018 at 21:04
  • $\begingroup$ @Jaydles does it actually? Monty Hall works because no matter what you pick, there's a 100% chance that he can give you more information. In other words, regardless of what Will thinks his wife knows, if he picks taco A or B, she'll reveal taco C, and Will has a 50/50 chance whether he switches or not. If Will picks taco C, his wife would just have to say "Ok, that's a bean taco; the game's ruined. Just pick one of the other two at random." Still 50/50. $\endgroup$ Commented May 25, 2018 at 21:04
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    $\begingroup$ @LordFarquaad, I think so: in his specific scenario (where he's picked one where his wife can reveal a different one), if he knows his wife only knew one, he knows that it doesn't matter what he does. While if he knew his wife knew ALL - like MH does - he knows that switching improves his odds. You could argue that in either case, he could switch to be safe, but my point was just that knowing what his wife knows, vs. what Monty does, should impact his decision. (Staying is fine in one case, but irrational in the other.) $\endgroup$
    – Jaydles
    Commented May 25, 2018 at 21:25
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The other posts provide several intuitive ways to think about this. I will simply compute the conditional probabilities to show that indeed this is not a monty hall style problem.

A priori probabilities $$P(\text{meat in A}) = 1/3$$ $$P(\text{meat in B})= 1/3$$ $$P(\text{meat in C}) = 1/3$$

After your wife tells you that C is beans

(The notation $P(A \mid B)$ means probability of event "A" given information "B"

$$P(\text{meat in A} \mid \text{beans in C}) = \frac{P(\text{meat in A} \textit{ and } \text{beans in C)}}{P(\text{beans in C})} = \frac{1/3}{2/3} = \frac 12$$ and similarly $$P(\text{meat in B} \mid \text{beans in C}) = \frac 12$$ $$P(\text{meat in C} \mid \text{beans in C}) = 0$$

So you can see that $A$ and $B$ have the same probability of having meat once you have the additional information of "beans in C". So there is no advantage to switching and the probability is 50%.

For a comparison with the Monty hall case, I refer to Bram28's answer: https://math.stackexchange.com/a/2796175/66711

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    $\begingroup$ The problem with this analysis is that it doesn't really show the contrast with the Monty Hall scenario. In fact, if you're not careful, it is easy for someone to apply this same logic to the Monty Hall scenario: $$P(\text{car in A} \mid \text{goat in C}) = \frac{P(\text{car in A} \textit{ and } \text{goat in C)}}{P(\text{goat in C})} = \frac{1/3}{2/3} = \frac 12$$ ... so switching doesn't help in Monty Hall either?! So again, you need to show how the Monty Hall problem is different. $\endgroup$
    – Bram28
    Commented May 27, 2018 at 19:50
  • $\begingroup$ And by the way, I know very well the shortcomings in applying the equation in my previous comment to the Monty Hall problem. But again, you don't want to make the reader fall into that trap: why don't you want to use that equation for Monty Hall? So: I say that more clarification and analysis is needed: if you do the mathematical analysis ... where and how in that mathematical analysis is the Monty Hall problem different? $\endgroup$
    – Bram28
    Commented May 27, 2018 at 19:53
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    $\begingroup$ In particular, you need to explain to the OP (and readers in general) what the event 'beans in $C$' means. If you use it as 'there are no beans in $C$, then the analogue in the Monty Hall problem would be 'there is a goat in $C$' ... and now switching makes no difference. The crucial difference is between 'your wife tells you that there are beans in C' and 'Monty shows $C$ to have a goat', given the difference in knowledge and intentions between the OP's wife and Monty ... This is not coming out in your analysis, or at the very least you are not clearly defining the event 'beans in $C$'. $\endgroup$
    – Bram28
    Commented May 27, 2018 at 20:44
  • $\begingroup$ @Bram28 Good point. I included a link to your answer at the end, since it's pointless for me to rewrite what you have written :-) $\endgroup$
    – Ant
    Commented May 28, 2018 at 11:48
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Short Mathematical Answer

No. The situation is quite different.

Original Monty Hall problem: Assume you pick door A. Then Monty shows either door B or door C to have a goat. And by symmetry, the chances of that are the same, i.e.

$$P(RevealGoat_B)=P(RevealGoat_C)=\frac{1}{2}$$

In fact, the conditional probability of showing $C$ to be empty if the car is behind door $A$ is the same as well:

$$P(RevealGoat_C|Car_A)=\frac{1}{2}$$

Therefore, if we assume that Monty shows door C to be empty, the chance that the car is behind door A is:

$$P(Car_A|RevealGoat_C) = \frac{P(RevealGoat_C|Car_A)\cdot P(Car_A)}{P(RevealGoat_C)}=\frac{\frac{1}{2}\cdot \frac{1}{3}}{\frac{1}{2}}=\frac{1}{3}$$

which is why you should switch to door $B$.

Your Situation:

Assume you pick taco $A$ and that your wife will reveal contents of taco $C$ no matter what. The chance that she ends up revealing beans in taco $C$ is therefore simply the a priori chance of there being beans in taco $C$, i.e.

$$P(RevealBeans_C) = \frac{2}{3}$$

while the conditional probability chance that she ends up revealing beans in taco $C$ if the meat is in taco $A$ is of course:

$$P(RevealBeans_C|Meat_A) = 1$$

And therefore, after she showed you that there were beans in burrito $C$, the chance that your taco $A$ has the meat is:

$$P(Meat_A|RevealBeans_C) = \frac{P(RevealBeans_C|Meat_A)\cdot P(Meat_A)}{P(RevealBeans_C)}=\frac{1\cdot \frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}$$

So no, not the same situation, and switching tacos does not help.

Long, more Detailed and more Conceptual Answer

The key is to realize that it is not just knowing that $C$ contains the beans or, as in the Month Hall problem, that $C$ contains a goat, but how you found out that information.

In the original Monty Hal problem, you know that Monty Hall knows where the prize is, and that he will show a goat after you pick, and that, if you happen to initially pick the door with the prize, Monty will randomly pick between the two goats to show to you. Indeed, there is no a priori reason for Monty to open door $C$, as Monty could have opened up door $B$ as well (assuming there was no car behind door $B$). So it's with all of that dynamics of the situation, and your knowledge of that dynamics, that you can figure out that it is better to switch. Indeed, see the Variants section in the Wikipedia page on the Monty Hall problem for how different behaviors and intentions on Monty's side will change whether switching is a good idea or not,

None of the dynamics behind the original Monty Hall problem were in place in the situation where you found yourself with the tacos and your wife. Your wife only knew the contents of taco $C$, and was only able to reveal its contents (that is, your wife was not going to just open up taco $A$ or $B$, and even if she did, she might have opened up one with the meat, unlike Monty Hall, who always shows a goat). In fact, it wasn't even clear that your wife was definitely going to 'spill the beans' on $C$! (forgive my pun ...)

So no, this is definitely not like the Monty Hall scenario.

Mathematically, the difference with the Monty Hall problem is as follows:

In the original Monty Hall, we have that:

$C_A$: event that $A$ (the one you picked) has the car

$C_B$: event that $B$ has the car

$C_C$: event that $C$ has the car

$ShG_C$: event that Monty shows $C$ to have a goat

$P(ShG_C|C_A) = \frac{1}{2}$ (if $A$ has the car, Monty will randomly pick between $B$ and $C$ to show a goat)

$P(ShG_C|C_B) = 1$ (if $B$ has car, then Monty definitely shows $C$, given that you picked $A$)

$P(ShG_C|C_C) = 0$ (of course, it is impossible for Monty to show a goat in $C$ if the car is in $C$!)

Hence:

$$P(ShG_C)=$$

$$P(ShG_C|C_A)\cdot P(C_A)+ P(ShG_C|C_B)\cdot P(C_B)+P(ShG_C|C_C)\cdot P(C_C)=$$

$$\frac{1}{2}\cdot \frac{1}{3} + 1\cdot \frac{1}{3} + 0\cdot \frac{1}{3} =\frac{1}{2}$$

(of course! Monty is always going to show one of the other two doors, and given the symmetry between $B$ and $C$, the probability of showing $C$ is $\frac{1}{2}$)

And therefore:

$$P(C_A|ShG_C)=\frac{P(ShG_C|C_A) \cdot P(C_A)}{P(ShG_C)}=\frac{\frac{1}{2}\cdot \frac{1}{3}}{\frac{1}{2}}=\frac{1}{3}$$

i.e. the chance that the one you picked has the car is $1$ out of $3$, and hence you should switch!

But in this scenario, given that all your wife knows is the content of $C$, we have:

$M_A$: event that $A$ (the one you picked) has the meat

$M_B$: event that $B$ has the meat

$M_C$: event that $C$ has the meat

$ShB_C$: event that your wife shows $C$ to have beans

Now, what is the chance that your wife was going to show that $C$ has the beans? We don't really know of course, because this depends on her intentions. Maybe your wife was only going to tell you that $C$ has the beans if it contained the beans, and if $C$ contained the meat, she might have said nothing at all, just to keep you guessing some more. In fact, even if $C$ contained the beans, she may have chosen not to say anything. Then again, she might have told you what's in $C$ regardless of the contents of $C$: "Aww, you picked $A$; that's too bad, since $C$ contains the meat! You lose!"

In other words, in contrast to the Monty Hall scenario, we don't really know $P(ShB_C|M_A)$ or $P(ShB_B|M_B)$ (though like the Monty Hall scenario, we do have that $P(ShB_B|M_C)=0$ of course: it is impossible for your wife to reveal $C$ has beans if it has the meat!).

However, given that your wife knows nothing about the contents of $A$ and $B$, we can assume that

$$P(ShB_C|M_A)=P(ShB_B|M_B)$$

This is the crucial difference with the Monty Hall scenario, as in the Monty Hall scenario we have

$$P(ShG_C|C_A) \not = P(ShG_C|C_B)$$

OK, but do we know if switching was going to help or hurt you? Or, given that you don't know what your wife is going to do, maybe we can't tell?

Well, let's define:

$B_C$: event that $C$ has beans

In your taco scenario, we can say that:

$$P(ShB_C|M_A)=P(ShB_C|M_B)=P(ShB_C|B_C)$$

for in all three cases, your wife would be looking at beans in $C$, and decides to reveal that with probability $P(ShB_C|B_C)$.

So, we now have that:

$$P(ShB_C)=$$

$$P(ShB_C|M_A)\cdot P(M_A)+ P(ShB_C|M_B)\cdot P(M_B)+P(ShB_C|M_C)\cdot P(M_C)=$$

$$P(ShB_C|B_C)\cdot \frac{1}{3} + P(ShB_C|B_C)\cdot \frac{1}{3} + 0\cdot \frac{1}{3} =P(ShB_C|B_C)\cdot \frac{2}{3}$$

Note that in the case where your wife was definitely going to reveal that $C$ has the beans if $C$ has the beans, we have $P(ShB_C|B_C)=1$, and thus

$$P(ShB_C)=\frac{2}{3}$$

again in contrast to the Monty Hall scenario, where we have that

$$P(ShG_C)=\frac{1}{2}$$

Now, from your later edit, it seems like your wife would indeed always reveal the beans when she was looking at the beans, since she wanted you to find the one with the meat, but in fact you don't have to know that, because:

$$P(M_A|ShB_C)=\frac{P(ShB_C|M_A) \cdot P(M_A)}{P(ShB_C)}=\frac{P(ShB_C|B_C) \cdot \frac{1}{3}}{P(ShB_C|B_C)\cdot \frac{2}{3}}=\frac{1}{2}$$

And so, there you have it: in your situation, switching would not have made any difference.

.. and of course it shouldn't have! With revealing the contents of taco $C$ being the only possible move for your wife (as opposed to opening up taco $A$ or $B$), your wife was simply cancelling one of the options, leaving two with equal probability. And indeed, the same would be true if in the Monty Hall scenario Monty Hall would be restricted to revealing what's behind door $C$ from the start as well. Yes, in that scenario you would be helped by knowing that a goat is behind door $C$ (and hence you can increase your chances from $\frac{1}{3}$ to $\frac{1}{2}$) but if Monty had the freedom to reveal a different door, you would be helped even more. And to see that, consider what would have happened if door $C$ would have the car behind it, but Monty was restricted to only being able to reveal what's behind door $C$. Well, then Monty would have revealed the car, and you would have lost. But by having the freedom to open either door, Monty was sure to be able to pick one that is empty, thereby giving you 'maximal' additional information, and thus allowing you to increase your chances from $\frac{1}{3}$ to $\frac{2}{3}$ by switching to the other door.

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I'm finding all of the answers hard to follow.

In your situation, you pick taco A, your wife tells you that she knows taco C has beans but she doesn't know the fillings of tacos A or B. Your wife has the same information that you do. There is a 50:50 chance of steak:beans in taco A.

The original Monty Hall would be you pick taco A, your wife knows all the tacos' fillings, chooses one with beans that you didn't pick and tells you that taco C has beans. Your wife currently has more information than you do and has used that information to influence what she tells you. She has extra information she is hiding from you.

  • Initially, you have a 33.3:66.7 chance of picking steak
  • The chance of steak being in one of the other two tacos is 66.7:33.3
  • If taco A contains steak, your wife has a 50:50 chance of telling you that taco C contains beans
  • If taco A contains beans, your wife has a 100:0 chance of telling you that taco C contains beans
  • The taco not named by your wife now has a probability of 66.7:33.3 of containing steak (the inverse of the first point and equal to the second point)

Thus only when your wife knows the fillings of all the tacos is it advantageous to switch to the taco not named.

Since your wife doesn't know all the tacos fillings, this is not a Monty Hall problem.

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For the probability of getting you want by switching to be different from the probability of getting you want by staying, there must be some asymmetry between them. In the Monty Hall problem, if we assume that Monty Hall never reveals what is behind the door you chose (something that, in a lot of formulations, is not explicitly stated), then that supplies the asymmetry. But in this case, there is no asymmetry between the burrito you chose and the remaining one. If you had chosen C, then you wife still would have told you that C is bean. If you had chosen B, your wife would have told you that C is bean.

If this were the Monty Hall problem, then in the case where you choose A, and A is steak, your wife would have two options: she could tell you that B is bean, or tell you that C is bean. So she has a 50% chance of telling you that C is bean. But in the case where A is bean, your wife has only one option: tell you that C is bean. So the probability of your wife telling you that C is bean is twice as large in the case where A is bean. Therefore, A is twice as likely to be bean as to be steak, and it makes sense to switch.

The previous paragraph presumably does not apply here. The probability of your wife telling you that C is bean is the same regardless of whether A is steak or B is, so each is equally likely to be steak.

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  • $\begingroup$ Actually, the usual MH problem isn't changed by allowing Monty to reveal the door you choose; those times he reveals a different door your odds still remain better by switching. What changes the game is if Monty Hall is allowed to reveal the prize (and still picks uniformly between his options). $\endgroup$
    – user14972
    Commented May 25, 2018 at 20:36
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Without your wife's knowledge, there are three possibilities: SBB; BSB; BBS. So if playing with Monty Hall and you choose A, there are four possibilities (assuming he never opens A if you choose it): SOB; SBO; BSO; BOS. S is the one with the steak; B has beans; O is open (always beans). The chances of each are 1/6; 1/6; 1/3; 1/3. Why? because the chances are even that the steak is behind each and there are two ways to open a burrito if the steak is behind A.

There is a 50% chance that it is either SOB or BOS. There is a 50% chance that it is either SBO or BSO. It is twice as likely to be BSO as SBO and twice as likely to be BOS as SOB. It is still even at this moment whether the steak is behind any of the three.

Now we add additional information. Monty opened C, the third burrito. There are two ways that can happen. Either we have SBO or BSO. But BSO is twice as likely as SBO. Because if the steak is in A, he could have opened either B or C. Given that he opens C, you know it's not BBS. But you don't know whether it is SBO or BSO.

Now, let's get rid of Monty. Your wife knows that C is beans. So the probability that A is steak (SBB) is 1/2 and B (BSB) is 1/2. The probability that C is steak (BBS) is zero. You don't know this. You think the probability is 1/3 for each of A, B, and C. But she does. She shares this information after you pick but before you open. But there are only two possibilities: SBO; BSO. Her information doesn't tell you anything between those two. Those two options remain equally likely.

This is actually the situation that many people thought existed in the Monty Hall problem. But it's different. In the Monty Hall problem, Monty knows the actual location of the prize and never picks it. Your wife doesn't know where the prize (steak) is. She just knows that it isn't C, which is beans.

The magic behind the Monty Hall problem is that SBB turns into two possibilities, SOB and SBO, that are equally likely and together are as likely as each of the other two possibilities. Then when Monty opens a burrito (or door), we pick one of SOB and SBO along with one of BOS and BSO.

Another way of looking at the Monty Hall problem is that it essentially allows you to choose both doors. He opens the wrong one (and you're guaranteed at least one wrong one). So the remaining one is right if either was. And there was a 2/3 chance that one of the two unpicked doors was correct. Opening the door doesn't change that. If he allowed you to switch before opening the door, you would have had a 2/3 chance.

Your wife can't do that. She has one piece of information. She knows that C is beans. That's useful information, but it just tells you that the correct choice is either A or B. This improves your chances to 1/2 of finding the steak. It doesn't give the 2/3 chance that the Monty Hall setup does.

I am ignoring the possibility that Monty might choose to open A if you chose A. The logic is similar but different in that case.

I am relying on symmetry in case you choose something other than A in the Monty Hall problem. In the wife problem, it's not the same because only A and B are symmetric; C is not symmetric with the other two. Her information is different.

It's possible to make a game that uses this information. For example, if your wife wants a beans burrito and you want steak, you can open A. If it's beans, give it to your wife and you take B. If it's steak, you keep it and give your wife B. You have steak and your wife has beans. And you know that C is beans. But that's a different kind of magic than the Monty Hall magic.

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The other answers focus in whether it is Monty Hall (No, because the total knowledge = the wife's knowledge = knowing the ratio, knowing the content of one. This gives a 50% chance. Gaining all of the wife's knowledge therefore gets you 50%. )

So what to do

If Will thinks it is a Monty Hall, but its not: Will will switch, he thinks he is swapping a 33⅓% chance for a 66⅔% change, but is swapping 50% for 50%.

If Will understands the problem (and is correct), then he sticks and has 50% chance.

In both cases he has 50% chance. What has he lost, in swapping?, only the effort of doing the swap. Therefore it is better to assume that it is Monty Hall (even though it is not).

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Actually there's a form of Monty Hall here. If you have free choice and first pick C and your wife informs you it's beans you will change your mind. You came into a problem thinking you had a 1/3 chance but it turns out it's 50/50. C is never steak and if you pick it, you get to discard it. So we know the steak has equal chance (1/2) of being A or B and you always end up on A or B.

Again, like Monty Hall, you made a guess then information was revealed to you and that allowed you to make a better guess.

The trick in Monty Hall is he knows where 'the steak' is and your wife only where it isn't...

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It's good that you broke this down into two questions, because they are actually distinct.

Let's first recap the problem statement:

"One of these (three) breakfast tacos has steak. I want that one. I pick the far left. Wife tells me far right taco is black bean. I switch to the middle taco. I win steak."

Question 1:

Given no more information than this, we assume the Principle of Indifference, name "The left taco is steak" "A," "The middle taco is steak" "B," and "The right taco is steak" "C," giving:

p(A|I) = 0.33... + p(B|I) = 0.33... + p(C|I) = 0.33... = 1.0

Next, you receive new information. Your wife says "The right taco is a bean taco." In conditional probability terms, p(C|I ⋀ W) = 0, "The probability the right taco is steak conditioned on the background information and my wife saying the right taco is a bean taco is 0."

But this can't change p(A|I) = 0.33..., "The probability the left taco is steak on the background information." Crucially, that's all you had when you made the choice. The new information doesn't "propagate backwards in time." Put slightly differently, you had a "degree of freedom" when you chose the left taco you no longer do. The right taco is no longer a choice (if you want the steak). It was when you chose the left taco.

So because you cannot revise p(A|I) = 0.33... and you now know p(C|I ⋀ W) = 0, and mutually exclusive and exhaustive probabilities must sum to 1.0, we're left with:

p(B|I ⋀ W) = 1.0 - p(A|I)  - p(C|I ⋀ W) or
p(B|I ⋀ W) = 1.0 - 0.33... - 0          = 0.66...

0.66... > 0.33...

You should switch to the middle taco.

Question 2:

Many writers have obfuscated the plain Bayesian reasoning using conditional probabilities (which are the only kind that exist) by trying to insert any number of irrelevances: what, exactly, does your wife know? What would have happened if some other event had occurred, e.g. you chose the right taco before your wife told you it was the wrong one? What if she chose one of the remaining tacos to reveal at random?

What Monty Hall knows when he reveals the goat is certainly relevant to whether a problem has the same rules as the actual Monty Hall game or not. No one is disputing that. However, it's still irrelevant to whether the player should switch or not, based on the information they receive and when they receive it.

Players of the actual Monty Hall game should switch, for exactly the same Bayesian reasons you were right to switch, regardless of whether your problem description is a "Monty Hall" problem, as defined by the rules of the TV game show, or not.

Incidentally, the following references are very helpful in understanding the reasoning I've used here:

The Algebra of Probable Inference

Probability Theory: The Logic of Science

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    $\begingroup$ Your post is very confusingly written, so pinning down your error(s) is difficult.Y seem to have an equality within an equality, and probabilities within probabilities. p(B|I ⋀ p(C|I ⋀ W) = 0) is nonsense. That's not valid notation. $\endgroup$ Commented May 25, 2018 at 19:35
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    $\begingroup$ Conditions should be random variables with boolean values, not boolean expressions with fixed value. "p(C|I ⋀ W) =0" is True, and "p(A|B and True)" is at best an unnecessarily convoluted way of saying "p(A|B)". There clearly is an error, because you came to the wrong conclusion (you think OP should switch). Furthermore, you are leaving a bunch of parentheses implicit, e.g. p(B|**(I ⋀ *(*p(C|I ⋀ W) = 0)** )), which means that if properly parenthesized, your expression is a confusing mess. $\endgroup$ Commented May 25, 2018 at 21:28
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    $\begingroup$ @PaulSnively Your logic is mostly sound, but based on the false assumption that taco C was an option. In Monty Hall, there's a 66% chance you pick a goat. When he takes a goat off the table, there's still a 66% chance you picked a goat, so the car's probably behind the other door. In this example, If Will picked taco C, his wife would've said "No, that one." and the game starts over with 2 tacos. So really, p(C|I) = 0, since it was never a choice, so p(A|I) = p(B|I) = .5. In other words, the principle of indifference is only between A and C $\endgroup$ Commented May 25, 2018 at 21:32
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    $\begingroup$ Accumulation: "There clearly is an error, because you came to the wrong conclusion (you think OP should switch)." This is ridiculous: you don't assume the conclusion and then adduce support for it! My conclusion is correct because my use of conditional probability is correct, typographical nits notwithstanding. $\endgroup$ Commented May 25, 2018 at 22:57
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    $\begingroup$ @paul think of it this way: I have a room with 3 light switches. 1 of them works. If I went to try the left one and before I flipped it my wife said “oh the right one’s broken” should I try the middle one? What if, after picking the left one, I noticed the right one was already on (so the wrong choice)? Should I switch there too? Is it different if I noticed before I picked a switch? The point is that someone in the game stops you if you pick a certain choice; that means there were always two choices. It doesn’t really matter that she waits until after you’ve picked to tell you. $\endgroup$ Commented May 26, 2018 at 2:12
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There is no benefit to switching. There is also no benefit from not switching. It's 50-50.

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  • $\begingroup$ I've deleted the comment. However, I reviewed as I did because the answer doesn't, in my mind, address question 1, only question 2 in the OP. If you edit to provide an explanation, I'll gladly upvote. $\endgroup$
    – user88319
    Commented May 25, 2018 at 17:55
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You can turn this into a Monty Hall problem:

Your wife knows all the fillings (say, for example, she prepared them).

You pick A.

She picks C, cuts it open, revealing beans. It does not matter which taco is labled A or B or C. Since she knows all, she can always pick one with beans after your pick, since there are two with beans, just like Monty's goats.

Now you have the situation where initially your probability was 1/3 steak and 2/3 beans.

The new situation is that your original pick is the steak with probability 1/3 and the unpicked taco is the steak with probability 2/3 ... however ...

Probabilities only make sense over many samples (many plays of the same game). Like the Monty Hall game, you only get to play once, so it really does not matter at all - this is the Monty Hall Myth.

The logic only applies to Monty, because he is the only one playing all the games. If every player switched, over time, Monty would lose more often (66.6% of the time), and in consequence every player would have a better chance.

But you do not control the other players. The odds are, many will not switch, based on 'tells' or other non-probabilistic intuition. That impairs your odds, since your play is a singleton - one event of many. In your singleton event, the only thing you know for sure is that one has beans and the other has steak, and this circumstance will always arise, since Monty knows all and there are two goats, so he can always find one.

In your singleton event, Monty shifts your odds from 33:67 to 50:50, and it does not matter whether you switch.

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    $\begingroup$ Your formulation is not precise. Does she always cut one open? Does she always cut a bean one open? Does she always cut C open? Does she always, when C is bean, cut C open? "Probabilities only make sense over many samples (many plays of the same game). Like the Monty Hall game, you only get to play once, so it really does not matter at all - this is the Monty Hall Myth." This is nonsense. $\endgroup$ Commented May 25, 2018 at 19:56

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