Monty Hall Problem Revisited Again Again!

Cecil's Columns/Staff Reports
1

I played out all possible cases and outcomes and I find that the odds are even that the contestant will win if they switch or stay.

I believe that the error in the accepted solution presents itself in the case where the contestant selects the door that also hides the car.

In this case, Monte may reveal the door that hides a goat or the other door that hides another goat.

In the solutions I’ve seen, this is treated as one event when in fact they should be treated as two distinct events even though they both reveal a goat.

I’d greatly appreciate your comments!

EVENTS:
Contestant selects door C
Car behind door C
Monte reveals door B
If contestant switches they LOSE
If contestant stays they WIN

Contestant selects door C
Car behind door C
Monte reveals door A
If contestant switches they LOSE
If contestant stays they WIN

Contestant selects door C
Car behind door B
Monte reveals door A
If contestant switches they WIN
If contestant stays they LOSE

Contestant selects door C
Car behind door A
Monte reveals door B
If contestant switches they WIN
If contestant stays they LOSE

Contestant selects door B
Car behind door C
Monte reveals door A
If contestant switches they WIN
If contestant stays they LOSE

Contestant selects door B
Car behind door B
Monte reveals door A
If contestant switches they LOSE
If contestant stays they WIN

Contestant selects door B
Car behind door B
Monte reveals door C
If contestant switches they LOSE
If contestant stays they WIN

Contestant selects door B
Car behind door A
Monte reveals door C
If contestant switches they WIN
If contestant stays they LOSE

Contestant selects door A
Car behind door C
Monte reveals door B
If contestant switches they WIN
If contestant stays they LOSE

Contestant selects door A
Car behind door B
Monte reveals door C
If contestant switches they WIN
If contestant stays they LOSE

Contestant selects door A
Car behind door A
Monte reveals door B
If contestant switches they LOSE
If contestant stays they WIN

Contestant selects door A
Car behind door A
Monte reveals door C
If contestant switches they LOSE
If contestant stays they WIN

2

The mistake you’re making is assuming that all possible scenarios have an equal chance of occurring. That is not the case.

Here’s a better way of looking at it. In any scenario where you picked the right door initially, you should obviously stay with your initial choice. But in any scenario where you picked the wrong door initially, Monty will eliminate the other wrong choice and allow you to pick the remaining door, which will be the correct choice. So if you picked a wrong door initially, you would always win by switching.

There are three doors to start with and only one is a winner. That means that two out of three times you will pick a wrong door initially. And that means that two out of three times, you should switch doors.

3

The problem with your analysis is you have 12 events, and you’re assuming they’re all equally likely. They’re not. Specifically, take the first 4:

Contestant selects door C
Car behind door C
Monte reveals door B
If contestant switches they LOSE
If contestant stays they WIN

Contestant selects door C
Car behind door C
Monte reveals door A
If contestant switches they LOSE
If contestant stays they WIN

Contestant selects door C
Car behind door B
Monte reveals door A
If contestant switches they WIN
If contestant stays they LOSE

Contestant selects door C
Car behind door A
Monte reveals door B
If contestant switches they WIN
If contestant stays they LOSE

In 2 cases, switching wins, but in the other 2, switching loses. But they aren’t all the same likelyhood. Specifically, case 1 is 1/6, case 2 is 1/6, case 3 is 1/3 and case 4 is 1/3. There isn’t a 50% likelyhood that the contestant picked the correct door in the first place, which is what your analysis depends on.

4

Cecil’s column.

Shouldn’t this go in CCC?

5

Yes. So reported.

6

Moved from General Questions to Comments on Cecil’s Columns.

Gfactor
General Questions Moderator

7

I wasn’t going to respond, because all I could really say was “What muldoonthief said.” But the way these threads go, some emphasis is probably a good thing. So BadgerMountain, read what muldoonthief said over and over until you get it.

8

[satire]I pick Heads, Joe flips a coin and it comes up heads.
I pick Heads, Joe flips a coin and it comes up tails.
I pick Heads, Joe flips a coin but drops it on the floor and can’t find it, so gets another coin from his pocket and flips it and it comes up heads.
Thus, my chances of correctly picking a coin toss are 2/3.[/satire]

9

This has already been explained verbally, but I thought a tabular view might help. Door C is the car, G1 and G2 are the goat doors. The number in parenthesis after each choice is the (dependent) probability of that event.


Contestant    Monty    Remaining        Overall
 Chooses     Chooses     Door         Probability

 C  (1/3)    G1 (1/2)     G2       (1/3)*(1/2) = 1/6
 C  (1/3)    G2 (1/2)     G1       (1/3)*(1/2) = 1/6
 G1 (1/3)    G2 (1)       C        (1/3)*1     = 1/3
 G2 (1/3)    G1 (1)       C        (1/3)*1     = 1/3

So you’re correct that out of the four possible outcomes, half of them involve the contestant choosing door C. However, that’s only because initially choosing door C forces Monty to make a choice between two doors. When the contestant makes the wrong initial choice, Monty’s subsequent choice is forced. He must pick the only remaining goat door 100% of the time that this situation occurs. This affects the number of outcomes unevenly. When the contestant makes the right initial choice, we have to split the outcome into two “half-outcomes”, but when the contestant chooses wrongly, the outcome remains whole.

The crux of the whole problem is that Monty’s hand is forced in some situations but not in others. If Monty chooses a door randomly, the problem collapses such that the odds of winning are the same (1/3) using either strategy:


Contestant    Monty    Remaining        Overall
 Chooses     Chooses     Door         Probability

 C  (1/3)    G1 (1/2)     G2       (1/3)*(1/2) = 1/6
 C  (1/3)    G2 (1/2)     G1       (1/3)*(1/2) = 1/6

 G1 (1/3)    C  (1/2)     G2       (1/3)*(1/2) = 1/6
 G1 (1/3)    G2 (1/2)     C        (1/3)*(1/2) = 1/6

 G2 (1/3)    C  (1/2)     G2       (1/3)*(1/2) = 1/6
 G2 (1/3)    G1 (1/2)     C        (1/3)*(1/2) = 1/6

If you don’t switch, you can only win in scenario 1 and 2, which have a combined probability of 1/3. If you switch, you’ll only win in scenarios 4 and 6, which is again 1/3 all together.

10

I found this to be the clearest explanation if you want a non-math answer.

From the column-

11

Nothing wrong with treating them as two events, so long as you assign them the correct probability. As muldoonthief has clearly described, you failed to do this and thus inevitably reached the wrong conclusion.
IME, by far the easiest way to convince anyone of the correct answer is with three playing cards, one black (representing the desirable prize) and two red (the goats). Shuffle the cards and pick one at random - then reveal that one of the remaining two is a goat. In just a few trials it becomes obvious that the “always switch” strategy wins 2/3 of the time (i.e. when the initially selected card is red).

Somehow, the use of actual cards is much more convincing than merely describing their use.

12

When I was new at the SDMB I fell in this trap and it took me a while to see the light. It is amazing how a very simple problem can mislead us so easily.

13

Suppose Monty give YOU a chance to pick one of three doors. You pick Door #1.

At that point, he has his lovely assistant open Door #2, and there’s a goat behind it.

He asks you if you’d like to switch.

You remain convinced that your odds are exactly the same whether you switch or not, so you say, “No- I’ll keep Door #1.” The Monty picks ME out of the audience, and says, “Okay Astorian, you can have whatever’s behind Door #3.”

Do you STILL think you and I had an equal chance of getting the car?

NOW the unfairness should dawn on you. YOU had a 1 in 3 chance of winning when you made your pick, and you still do. I, on the other hand, have a 50-50 chance of winning!

A 1 in 2 choice always has a better chance of being right than a 1 in 3 choice.

14

Heh. Except that in your scenario, there’s a 2/3 chance that door #3 will contain the prize, not 1/2. :stuck_out_tongue:

Really, the simplest explanation is this:

If you adopt the non-switching strategy, you must always win if your initial choice was right, and always lose if your initial choice was wrong.

If you adopt the switching strategy, you must always win if your initial choice was wrong, and lose if your initial choice was right. This is true if and only if Monty never opens the prize door, which is how the show actually worked (even if they never explicitly said so).

Since your initial choice is twice as likely to be wrong than it is to be right (everyone can agree that there’s a 1/3 chance of picking the right door in the beginning), the switching strategy wins twice as often as the non-switching strategy.

15

Ok, I now see the error in giving equal probability to all events.

But consider this for discussion sake… If Monty said go ahead and pick 2 doors and then opened one of your doors that contained a goat, would you switch?

Isn’t this the same game?

16

It’s the same game, only with “your” door exchanged for the “other” door.

In the *original *game, you want to switch. In this variation, because “your” door is *already *the one you want, you don’t switch.

17

Yes, except this time you have a 2/3 chance of being right on your initial guess, so you stick with your choice.

18

I don’t have the math skills to figure all of this out but thought that the two cases that never get revealed in normal play might be missing. That is Monty never opens the door the contestant has picked and never opens the door with the big prize.

I built a spreadsheet with:
Prize location
Door selected
Door revealed
All using the randbetween(1,3) function in Excel.
I then put either one or zero in columns corresponding to:
Win if switch
Win if stick with original
Not feasible due to the reveal matching the prize
Not feasible due to the reveal matching the selection unless already infeasible.

I copied that to 100,000 rows and summed the columns. I then hit the recalc button several times. The results were that winners for switching and winners for staying with the original were roughly the same at 22%. The other 56% were infeasible due to the way the game is normally played.

What have I missed?

19

The way the game is set up, if the person initially picks the wrong door, he wins, and if he initially picks the winning door, he loses. The way you programmed it, once the “prize location” and “door selected” are chosen, the “Door revealed” is twice as likely to invalidate a run where the “prize location” and “door selected” are different than if they’re the same. Those are the winning cases, so the winning cases are invalidated twice as often as the losing cases. That’s skewing your results relative to the original problem.

20

I guess it comes down to what you think the original problem really is. I used the assumption that a non-winning, non-selected door will always be revealed followed by an offer to switch. The problem really reduces to one winning choice and one losing choice. The prize is never revealed with a “want to switch to the other goat”. The fact that you have a starting point is really just distracting information.

To say that I skewed the results by invalidating more losing outcomes is really my point. Those outcomes aren’t feasible in the game as played and therefore shouldn’t be counted in probability-of-losing calculations.