Monty Hall Problem

[Tarsier]

The Monty Hall probability problem is nothing new. Many of you have probably encountered it. However, I was talking about it with one of our junior high classes and enjoyed the debate. If you haven’t encountered it, the problem appears below, so please don’t look it up - try to reason and comment based on your own analysis. The fun in the problem is less trying to figure it out, and more watching the firm determination on both sides of the argument. It’s like watching people argue about color of a dress, only with the Monty Hall problem, a real answer does exist.

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice? To retain your original choice? Neither?

 

[Trishie]

It is in the show’s interests to maintain and promote drama, therefore the host’s suggestion may simply be a ploy.
I would cheerfully stay with my choice, rather than drive myself crazy with second-guessing. As the show requires suspense, I might leave a pause before a final reply

 

[Tarsier]

But remember, the host didn’t make a suggestion.

He simply opened one of the two doors you didn’t choose.

I’m not being nit-picky - this has a bearing on the answer.

 

[spiderweb]

Stick with the door you chose. Your odds just increased that you initially picked the right door as the host has eliminated one of the other possibilities. At this point it’s still a 50% chance.

 

[Tarsier]

If you run simulations of this problem (they are available all over the internet, and you can make your own version with paper doors at home), the outcome will come out very close to 66% probability of a car if you switch doors. Why would you suppose that is?

 

[SMHW]

Simply mathematics of probability of course.

As mentioned earlier, the opening of the door with a goat was NOT a random act. It introduces a new mathematical component into the computation of the probabilities.

The best way I can think of to explain this is to tell people to look at the 3 cases for the original pick (correct, incorrect, incorrect) and then compute the probability of success for someone who always switches versus the probability of success for someone who always stays with the first choice.

 

[Tarsier]

Simply mathematics of probability of course. As mentioned earlier, the opening of the door with a goat was NOT a random act. It introduces a new mathematical component into the computation of the probabilities.

EXACTLY right! If you choose door A, there is a 66.7% chance the car is instead behind doors B or C.

Once the host opens either door B or C, the entire 66.7% probability transfers to the other door, so you are left with a 33.3% probability if you stick with A or a 66.7% probability if you switch to whatever door is left unopened.

 

[SMHW]

It might help to look at it this way:

Suppose someone plays the game and doesn’t switch from the original choice.​

If they pick the correct door and don’t switch then what happens?
If they pick the first wrong door and don’t switch, then what happens?
If they pick the second wrong door and don’t switch, then what happens?

Suppose someone plays the game and switches from the original choice.​

If they pick the correct door and switch then what happens?
If they pick the first wrong door and switch, then what happens?
If they pick the second wrong door and switch, then what happens?

Now of course if someone actually WANTS a goat then things work a little differently.

 

[Tarsier]

Great way to approach. My answers within the quote:

Suppose someone plays the game and doesn’t switch from the original choice.

A) If they pick the correct door and don’t switch then what happens? He wins.

B) If they pick the first wrong door and don’t switch, then what happens? He loses.

C) If they pick the second wrong door and don’t switch, then what happens? He loses.

Suppose someone plays the game and switches from the original choice.

D) If they pick the correct door and switch then what happens? He loses.

E) If they pick the first wrong door and switch, then what happens? He wins.

F) If they pick the second wrong door and switch, then what happens? He wins.

So, if the player doesn’t switch, he loses two out of three times. If he does switch, he wins two out of three times.

66.7% of the time he switches, he wins the car.

But yes, you are right - the goat is preferable anyway. No more mowing.

 

[HerCrazierHalf]

It seems odd but I’d say switch. When you made your first choice the odds were 33%. When a door is eliminated the odds are 50%. So switching would be making a choice at the 50% odds.

Is choosing to keep your door actually choosing to take advantage of the better odds?

 

[Tarsier]

If your first choice was only 33% likely to be correct, what about the host revealing what is behind one of the other doors makes your original choice a 50/50 shot out of three?

 

[HerCrazierHalf]

Simple. If you don’t choose again after a bad door is removed, then you still have a 66% chance of having the wrong choice. The host doing the reveal them becomes him dramatically revealing of your initial choice was right.

 

[Tarsier]

Simple. If you don’t choose again after a bad door is removed, then you still have a 66% chance of having the wrong choice. The host doing the reveal them becomes him dramatically revealing of your initial choice was right.

I guess I’m confused here. Your first sentence seems to suggest (correctly) that both before and after (“still”) the host selects a door, you are 66.6% likely to have picked the wrong door. I’m with you there, as that’s the correct answer.

But your second sentence seems to contradict your first, as you write that he is “dramatically revealing of your initial choice was right.”

So, was your initial choice “right” or still a “66% chance of having the wrong choice”?

 

[HerCrazierHalf]

There are 2 events.

1. choose 1 of 3 doors. 66% chance of failure
2. choose 1 of 2 doors. 50% chance of failure

The odds are determined when you make a selection. So, if you stick with your initial selection it would be the same as the host revealing a bad door without giving the chance to change. You are in event #1.

If you switch, then what you effectively cancel event#1 and move to event #2.

The whole thing assumes that switching is the only way to demonstrate that you’ve made a choice in event #2 instead of continuing in event #1.

 

[Tarsier]

There are 2 events.

choose 1 of 3 doors. 66% chance of failure
choose 1 of 2 doors. 50% chance of failure

The odds are determined when you make a selection. So, if you stick with your initial selection it would be the same as the host revealing a bad door without giving the chance to change. You are in event #1.

If you switch, then what you effectively cancel event#1 and move to event #2.

The whole thing assumes that switching is the only way to demonstrate that you’ve made a choice in event #2 instead of continuing in event #1.

Except that simulations show that your logic doesn’t hold here. You can set these simulations to run crazy numbers of time. Pick a high number to avoid the occasional streak. I just ran the scenario 1000 times. By switching, I won the car 68% of the time.

The problem with what you’ve proposed is that you’ve predicated probability upon your decision to switch: “If you switch, then what you effectively cancel event #1 and move to event #2.”

But probability isn’t based upon your thought process. It is based on the reality of odds.

As you said, your first choice put your odds of having picked the right door at only 33.3%. That never changes.

Look at it another way. Pretend there are 1,000,000,000 doors, and all of them but one have goats. The other has a car. You pick a door. The odds that you picked the wrong door are 999,999,999 out of a billion. Odds are that you picked a goat.

Now, the host, who knows what is behind these doors, closes all of them one at a time, except for door number 312. He then offers that you can stay with your original pick or change to door 312.

Would you say your odds are now 50/50 simply because you have two closed doors in front of you, even though the odds were 999,999,999/1,000,000,000 against you before the host (who knew what was behind the doors) eliminated several of the wrong choices?

 

[HerCrazierHalf]

Would you say your odds are now 50/50 simply because you have two closed doors in front of you, even though the odds were 999,999,999/1,000,000,000 against you before the host (who knew what was behind the doors) eliminated several of the wrong choices?

If you make a choice after removing bad doors, then yes 50/50.

The problem with what you’ve proposed is that you’ve predicated probability upon your decision to switch

This is the paradox. If you don’t switch then did you make a separate choice or simply continue with your original choice?

 

[SMHW]

As Tarsier states, the probability that the originally chosen door is the winner stays the same. The probability of the neither-previously-chosen-nor-opened door basically has it’s original probability of being a winner additively increased by the probabilities of any doors opened by the host.

So if there were three doors, the probability of the originally chosen door being the winner is 33.3% and remains so even after the host reveals the goat in one of the other doors. The un-chosen door now has a (33.3% + 33.3%) chance of being the winner.

It’s very confusing to try and figure out such probabilities because we are used to probability being random. This is not a random scenario; it is highly manipulated.

The way I presented of viewing the problem (a few posts back) makes it clear what the probabilities must be.

 

[Tarsier]

If you make a choice after removing bad doors, then yes 50/50.

Not at all. The host knows what hides behind the doors. You chose one door out of a billion. You almost definitely chose a door with a goat. The host knows that the doors he is opening contain goats according to the premise.

Again, run the simulations with the three doors. 66.6% of the time you’ll win the car if you decide to switch. If, as you said, it is 50/50, that wouldn’t be the case. If you don’t trust computers, make three paper doors and try it with a friend. Same result.

 

[Tarsier]

If anyone is struggling with this, it might help to know you are in good company. From Wiki on the occassion that this problem appeared in the Marilyn vos Savant column:

"Many readers of vos Savant’s column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating the predicted result (Vazsonyi 1999).

The problem is a paradox of the veridical type, because the correct choice (that one should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true. The Monty Hall problem is mathematically closely related to the earlier Three Prisoners problem and to the much older Bertrand’s box paradox."

 

[SMHW]

People suppose that the host eliminating a door increases the probabilities of both the previously chosen and the un-chosen doors being winning choices.

This is not the case.

The chosen door maintains its original probability of being a winner. The opening of an un-chosen door only impacts the probability of the remaining un-chosen door being a winner. The total of the probabilities of the un-chosen doors (both opened and unopened) being winners remains the same.

If there are more than three doors you still have the case that the chosen door retains it’s initial probability of being a winner and the total of the probabilities of the un-chosen doors (both opened and unopened) being winners remains the same.