The Monty Hall problem

[VicenteMMOS]

Hello all!


I’m writing you here to ask for some help with the Monty Hall problem. After learning about it, and listening to the explanation of the proposed solution to the problem, I must say I felt like it wasn’t accurate… even though I myself don’t know how to prove or disprove my hypothesis. Well, let me describe it for you:


The Monty Hall problem is a guessing game. Monty Hall, the TV show host, invites a guest to participate in a game that consists of three numbered closed doors, one of which has a prize behind it, and which only the host knows of. The guest then at first picks one door. The host then reveals one door which does not contain a prize behind, and asks the guest if he wants to change his first choice or stick with it. The guest decides whether or not to change doors, and then the host opens the last door picked by the guest.


The proposed solution I heard of consisted in the host always choosing to switch doors, as he would have a probability of success of 2/3. Now, here’s where I’m not convinced of this statement: under my reasoning, if the guest chooses to switch doors, he won’t have a 2/3 probability of success, he’ll have a 50% probability of success. In other words, given that the host of the program will always reveal an empty door, the probability of success of that door will collapse from 1/3 before its opening to 0 after it’s opened. So this remaining 1/3 probability will not be summed to the other door (the one that the guest didn’t pick), but will equally distributed between the 2 unopened doors. So, 1/3 + (1/3 : 2) = 1/3 + 1/6 = 0,5.


This is my reasoning. Of course, I could be wrong about it, but the way I see it, I don’t understand why would 1/3 probability of success compound itself onto an unpicked arbitrary door, that’s in theory completely equal to the unopened picked door that the guest chose at the start. This reasoning leads to a far different result to this problem than the reasoning I was presented. Let me expand on this difference by exaggerating the problem.

If we suppose there are not only 3 doors, but 1000, the reasoning I was presented (let’s call it Hypothesis A) would go as this: the guest picks one door, the host reveals 998 empty doors, then asks if guest wants to switch. At this point, the probabilities of the last unpicked door would have a 0,999 probability of containing a prize. I think that might not be the case! I don’t really know how to prove or disprove this hypothesis, that’s why I’m asking for your help, but I believe on a Hypothesis B, where the successive collapsing of probabilities of all the open doors by the host will be redistributed equally among the 2 remaining doors, leading to a 50-50 chance of success.


Can someone please help me prove/disprove my hypothesis?


Thank you.

 

[JeffM]

Consider doors A, B, and C.

Let's say that the contestant picks door A, which is not opened.

The initial probability that door A is correct is 1/3. If so, the probability that door B will be opened next is 1/2, and the probability that C will opened next is 1/2. Therefore, the probability that door A is correct and that door B is opened is 1/6, and the probability that door A is correct and that door C is opened is 1/6.

The initial probability that door A is incorrect is 2/3. In that case, door B will be correct 1/2 the time and door C will always be opened. And door C will be correct 1/2 the time and door B will always be opened. Therefore, the probability that door A is incorrect and door C will be opened is 1/3, which is when B is correct and equals 2/3 * 1/2 * 1, and the probability that door A is incorrect and door B will be opened is 1/3, which is when C is correct and equals 2/3 * 1 * 1/2.

To summarize

A correct, B opened: 1/6
A correct, C opened: 1/6
A incorrect, B opened: 1/3 (meaning C correct)
A incorrect, C opened: 1/3 (meaning B correct)

With me so far?

Now let's say Monty opens door B. That will occur if A is correct (1/6 of the time) or if C is correct (1/3 of the time). But 1/3 is double 1/6. C will be correct twice as often as A will be. The contestant should switch choices.

Maybe it is clearer to think about it with numbers.

Suppose we look at 1998 occasions when the contestant picks A.

666 times that will have been the correct choice. 666 times B will have been the correct choice. And 666 times C will have been the correct choice.

Of the 666 times when A is the correct choice, Monty will open door B 333 times.

Of the 1332 times when A is the incorrect choice, Monty will open door B 666 times (every time C is the correct choice.)

So, given that Monty will have opened door B 999 times and only 333 of those times will door A have been the correct choice, how many times will C have been the correct choice?

 

[VicenteMMOS]

Consider doors A, B, and C.

Let's say that the contestant picks door A, which is not opened.

The initial probability that door A is correct is 1/3. If so, the probability that door B will be opened next is 1/2, and the probability that C will opened next is 1/2. Therefore, the probability that door A is correct and that door B is opened is 1/6, and the probability that door A is correct and that door C is opened is 1/6.

The initial probability that door A is incorrect is 2/3. In that case, door B will be correct 1/2 the time and door C will always be opened. And door C will be correct 1/2 the time and door B will always be opened. Therefore, the probability that door A is incorrect and door C will be opened is 1/3, which is when B is correct and equals 2/3 * 1/2 * 1, and the probability that door A is incorrect and door B will be opened is 1/3, which is when C is correct and equals 2/3 * 1 * 1/2.

To summarize

A correct, B opened: 1/6
A correct, C opened: 1/6
A incorrect, B opened: 1/3 (meaning C correct)
A incorrect, C opened: 1/3 (meaning B correct)

With me so far?

Now let's say Monty opens door B. That will occur if A is correct (1/6 of the time) or if C is correct (1/3 of the time). But 1/3 is double 1/6. C will be correct twice as often as A will be. The contestant should switch choices.

Maybe it is clearer to think about it with numbers.

Suppose we look at 1998 occasions when the contestant picks A.

666 times that will have been the correct choice. 666 times B will have been the correct choice. And 666 times C will have been the correct choice.

Of the 666 times when A is the correct choice, Monty will open door B 333 times.

Of the 1332 times when A is the incorrect choice, Monty will open door B 666 times (every time C is the correct choice.)

So, given that Monty will have opened door B 999 times and only 333 of those times will door A have been the correct choice, how many times will C have been the correct choice?

666 times, twice as often as A.

Wow, finally someone actually explained it to me! Thank you! I was not considering this as a multiplication of probabilities of opening 2 successive doors! That’s why my results were always wrong. Thank you.

 

[Steven G]

Consider the lottery. There are many possible numbers that can be chosen, say 1 million.
Now you pick a number (ie buy a ticket). Now I know the winning ticket. I tell you the numbers of 999,998 tickets that are losing tickets which you did not pick. So now there are two tickets left--your ticket and one other ticket. Since there are only two tickets left and one is the winning ticket you are saying that you have a 50% chance of winning. No only question is how did you get so good at picking winning tickets?!!!
Think about it this say, the other ticket being the winning ticket is like you picking 999,999 numbers for the lottery. In the Monty Hall problem, switching is like picking two doors instead of one door. THINK about this!

Here is a nice video on the Monty Hall problem

 

[VicenteMMOS]

Consider the lottery. There are many possible numbers that can be chosen, say 1 million.
Now you pick a number (ie buy a ticket). Now I know the winning ticket. I tell you the numbers of 999,998 tickets that are losing tickets which you did not pick. So now there are two tickets left--your ticket and one other ticket. Since there are only two tickets left and one is the winning ticket you are saying that you have a 50% chance of winning. No only question is how did you get so good at picking winning tickets?!!!
Think about it this say, the other ticket being the winning ticket is like you picking 999,999 numbers for the lottery. In the Monty Hall problem, switching is like picking two doors instead of one door. THINK about this!

Here is a nice video on the Monty Hall problem

I did think about it, quite a lot actually. I was incorrect in the way I thought about it, sure, but I spent by now about 20 hours thinking about this problem. After the proof given by Jeff, the last thing I was thinking is ‘why am I thinking wrong about this’? In other words, I committed that mistake because my way of thinking around the problem was wrong to begin with, and so it’s likely that I’ll keep thinking wrong about the solution to future similar problems. I never thought of calculating the combined probabilities of guest picking a number *and* host picking all other wrong numbers but one. To be completely honest, I’m still unsure as to why this is the solution, even though now i understand it is the solution, as your example also clarifies. Thank you both.

 

[Steven G]

I apologize that I wrote about the lottery as I did not realize that you included looking at 1000 doors in the Monty Hall problem.
This problem initially caused many fights between mathematicians!

Please understand that if the host randomly picks the door(s) to open than everything changes. So what will be the probability of winning if you switch?

 

[VicenteMMOS]

No need to apologize, you didn’t say anything bad

After Jeff’s explanation, I understand that I did not consider that the actual calculation of the guest’s last pick should account for the host’s picks as well, which would in turn increase chances of the last remaining door of being the correct one.

Then again, I recall how many times throughout school and college I would listen to the correct explanation of an exercise, and yet wouldn’t still be able myself to solve new problems on my own. It is as I was saying, that I don’t yet understand why a certain solution to a problem involves the specific steps that they do. In this problem, I’m not sure that if I had another similar problem that I would be able to solve it correctly in the future. I know I’m rambling, but this is basically how I felt all my life about mathematics and physics lessons, despite my love for those subjects. I feel that in order to understand how to solve a problem, first I need to understand why that way of thinking offers me the adequate solution.

 

[JeffM]

In my experience, probability theory is frequently counter-intuitive. In fact, the first time I took a course in it, I dropped it because it initially made no sense to me at all.

I did not previously know what Jomo said (namely, that even some mathematicians had initially fought about the Monty Hall problem), but I was not surprised because the answer is so severely counter-intuitive.

Your experience about following a verbal explanation of an example of some mathematical technique and then being unable to apply that technique is quite common. Ironically, the better the explanation of the technique is, the more probable is the inability to apply it because each step of the logic is presented so smoothly that the listener or reader never has a sudden and surprising enlightenment that is memorable. This is why exercises are so important. Doing something over and over fixes it in the memory.