The Monty Hall Paradox

Quote: The Link

Imagine that the set of Monty Hall's game show Let's Make a Deal has three closed doors. Behind one of these doors is a car; behind the other two are goats. The contestant does not know where the car is, but Monty Hall does.

The contestant picks a door and Monty opens one of the remaining doors, one he knows doesn't hide the car. If the contestant has already chosen the correct door, Monty is equally likely to open either of the two remaining doors.

After Monty has shown a goat behind the door that he opens, the contestant is always given the option to switch doors. What is the probability of winning the car if she stays with her first choice? What if she decides to switch?

It has been argued that the game show saved themselves a bundle by the fact that people tended to stick with their original choice of "which door".

If you want a brain-teaser, try to figure this out on your own, then go to the link by scrolling down and following the explanations. It is not a spoiler to say that the correct thing for the contestant to do is to switch his choice of doors, but see if you don't get all riled up first arguing that it doesnt matter!!

















































For me, I couldn't fathom this until I got to the part where it has you think about it if there 1000 doors instead of 3 doors.

http://mathforum.org/dr.math/faq/faq.monty.hall.html








































































Spoiler above for anyone approaching this as a brain-teaser.

This has already been discussed at great lenght, here.



Spoiler reply...








































In that thread, I was one of the biggest "it doesn't matter" supporters.

Yeah, it was the 1,000 door analogy that convinced me that I was wrong. And I quickly saw the light:

An even easier way to rationalize it is to ask the person, before opening ANY door, if he'd rather switch to BOTH of the other doors. The immediate answer is "Sure!" He will always switch, even though he KNOWS that at least one of them has a goat.

geez, sorry it was already discussed.

I have to admit, still, that when it gets back to 3 doors my faith in the correct answer starts to waiver again!

Okay I'll bite on this thread.
I do not buy the reasoning. It is using twisted logic to try and prove a point which does not exist.

When you picked the first time, you had a 1 in 3 chance.
When Monty shows you one of the goats, and gives you a chance to pick again, you now have a 1 in 2 chance.
Whether you were right or wrong on the first pick is irrelevant. No matter what door you picked, Monty was going to be able to show you a goat, therefore, you have learned NOTHING by Monty showing you a goat.

Here is another favorite of mine to show twisted logic:

Three friends check into a motel for the night and the clerk tells them the bill is $30, payable in advance. So, they each pay the clerk $10 and go to their room. A few minutes later, the clerk realizes he has made an error and overcharged the trio by $5. He asks the bellhop to return $5 to the 3 friends who had just checked in. The bellhop sees this as an opportunity to make $2 as he reasons that the three friends would have a tough time dividing $5 evenly among them; so he decides to tell them that the clerk made a mistake of only $3, giving a dollar back to each of the friends. He pockets the leftover $2 and goes home for the day! Now, each of the three friends gets a dollar back, thus they each paid $9 for the room which is a total of $27 for the night. We know the bellhop pocketed $2 and adding that to the $27, you get $29, not $30 which was originally spent. Where did the other dollar go????

Raliegh -

I too was a non-believer. Look at the 1,000 door analogy. You pick one. Knowing full well where the car is, he opens 998. Now think about it. You has a 1:1,000 chance of being right. Just because Monty showed you 998 other wrong choices, does not increase the odds that you picked right. You always had a 999:1,000 chance of being wrong, so always switch. Does that make sense?

If Monty opened RANDOM doors, and got lucky that he never showed you the car, then his luck is just as good as your luck, and it's equally safe to stay or switch.



Your puzzle is not twisted logic, but bad math.

You add the $2 to the hotel bill, not to what was paid.

The three friends paid $9 each, which is $27. Out of that $27, the hotel got $25, and the bellhop got $2.

There is no 'other dollar'.

That riddle is old. How old? When was the last time three people shared a $25 hotel room?

Quote: DJTeddyBear

Raliegh -

I too was a non-believer. Look at the 1,000 door analogy. You pick one. Knowing full well where the car is, he opens 998. Now think about it. You has a 1:1,000 chance of being right. Just because Monty showed you 998 other wrong choices, does not increase the odds that you picked right. You always had a 999:1,000 chance of being wrong, so always switch. Does that make sense?

I am still not buying the argument. I believe the extra information is not germane to the situation.

I agree I have a 1:1000 chance of picking the right door, on the first pick. HOWEVER,
I also know before any door is exposed that in a few minutes I will be given a choice of keeping my door, or taking the one that remains.
Therefore my initial choice.............

Hmmmm, What light from yon window breaks...... so there was a 999:1000 chance that the car was behind the door I did not pick, and now 998 of the goats have been removed, so it is highly likely the car is behind that door, not the one I picked.

Okay, I give. I should switch my pick.

Quote: RaleighCraps

Hmmmm, What light from yon window breaks.......

Did I just witness a light being illuminated? Cool.

Personally, when my light lit up, I visualized a bare bulb in the laundry room, and a hand pulling on the chain.

But you painted a cool picture too!

My difficulty with this puzzle is that it is (usually?) described that Monte Hall will ALWAYS open a door, revealing a non-jackpot and then give the player the chance to change choice. Suppose he only did that some of the time, perhaps just when he wanted to. Suppose he ONLY did it just when he wanted to try to mislead the player into switching away from an initial selection that was actually the jackpot. (In that case, I think you would NEVER want to change if given the opportunity.) Suppose you don't know when he will choose to do it but that he doesn't ALWAYS do it. Does this affect the set of assumptions and the conclusion that the player should always change the selection if given the opportunity? Consider the scenario that in deciding whether to reveal a non-jackpot, Monte Hall is actually trying to play games with our minds and not give players in general the opportunity to rationalize an optimal strategy. Is is possible to out-game him?

Quote: Doc

My difficulty with this puzzle is that it is (usually?) described that Monte Hall will ALWAYS open a door, revealing a non-jackpot and then give the player the chance to change choice. Suppose he only did that some of the time, perhaps just when he wanted to. Suppose he ONLY did it just when he wanted to try to mislead the player into switching away from an initial selection that was actually the jackpot. (In that case, I think you would NEVER want to change if given the opportunity.) Suppose you don't know when he will choose to do it but that he doesn't ALWAYS do it. Does this affect the set of assumptions and the conclusion that the player should always change the selection if given the opportunity? Consider the scenario that in deciding whether to reveal a non-jackpot, Monte Hall is actually trying to play games with our minds and not give players in general the opportunity to rationalize an optimal strategy. Is is possible to out-game him?

If there is a variable that sometimes he will, sometimes he won't, then you cannot come up with a legit equation. If he is intentionally showing you so you WILL switch (and will ONLY show when you are right as you suggested above) then obviously you'd stay and win 100% of the time. If he RANDOMLY sometimes shows and sometimes doesn't (but still is sure to show a goat on his reveal), it still is better to change. You are still getting 2 possible doors for 1. If it's not random and Monty is a rational actor with either a desire to make you win sometimes and lose sometimes, then, again, there would be no plan of action.

Doc -

You're stating arguments that apply ONLY to the TV show version of this problem.

Read the original post. There are always two, identical, bad choices. And 'Monty' ALWAYS opens one of the other doors, and ALWAYS gives you the option to switch.

On the show, "Doors" are only used for the final deal, and there always were three reasonable prizes, but of differing values. Before the final deal, curtains were used, and only then did a curtain sometimes conceal a 'zonk'. And it was only before the final deal that Money would sometimes show the contents of the third curtain, and then sometimes give the person the option to switch.

Hell, maybe Monty only did that when the player had picked the expensive prize. If ANY players switch, they saved a bunch of money!

So, yeah, based upon all that, if you're on the show, I think you're right that you can't say to always switch.


But this paradox has been around a lot longer than the TV show. It goes back to at least the late 1800s. It merely got named "The Monty Hall Problem" because of the popularity of the show.

And it is this specific problem, not the variations introduced in the TV show, that we are discussing, with the correct answer being to switch.

For more on the topic, go to: http://en.wikipedia.org/wiki/Monty_Hall_problem

Yes, I understand that the original post included the "always" issues. I was trying to explore the subject of whether there is a strategy for the "sometimes" scenario, if you happen to be a player who is offered the option of changing your initial choice. So yes, I am asking a different but related question.

To somewhat restate my question, if Monte Hall sometimes does and sometimes doesn't offer the chance to switch and might be trying to mess with the player's mind and trick them, is there a preferred strategy? This might be similar to the question of whether there is a preferred strategy for rock, paper, scissors. On his WOO site, the Wizard says:

Quote: Wizard

In the Playtech game the odds are the same for all three throws. However, when playing a novice human opponent always throw paper the first time! This is because most people tend to prefer rock, especially on the first throw. You can thank me later after after you see how well this works! Just don't employ this strategy against Lisa Simpson.

So if Monte gave you the option to switch on your one and only time to play the game with him, is there a preferred strategy for switch/don't, given the case that he doesn't offer the opportunity to every player and might have deception in mind?

I don't get it. I've never had a hard time accepting this (I was a little skeptical when I first heard it... but that was in 9th grade, and once I heard an explanation I never looked back). I mean, it's basically betting that you guessed the wrong door initially. I'm sure it's been said a million times, but once more, for all us simpletons, without adding more doors:

The way the rules are set up, you win by default if you guessed the wrong door and switched, because the correct door is the only option (so you can ONLY lose if you picked the right door). What are the odds of picking the wrong door initially? 2 in 3. So what are the odds of winning if you always switch? 2 of 3. No paradox, no crazy math or logic, just a simple misunderstanding that people seem to never want to let go of for some reason. One last note: The door eliminated is not random: it is always an incorrect door, meaning if an incorrect door was eliminated, and you picked an incorrect door (which we know will happen 2 of 3 times), all the incorrect doors are accounted for, so the door you switch to must be the correct one!

By the way, it was fun as heck convincing my girlfriend (not mathematically blessed) that this worked through trial and error. She saw it with her own eyes and still refuses to believe it :) Try winning some money/favors off your loved ones with this by using a black card and 2 red cards!

Quote: FootofGod

I don't get it. I've never had a hard time accepting this (I was a little skeptical when I first heard it... but that was in 9th grade, and once I heard an explanation I never looked back). I mean, it's basically betting that you guessed the wrong door initially. I'm sure it's been said a million times, but once more, for all us simpletons, without adding more doors:

The way the rules are set up, you win by default if you guessed the wrong door and switched, because the correct door is the only option (so you can ONLY lose if you picked the right door). What are the odds of picking the wrong door initially? 2 in 3. So what are the odds of winning if you always switch? 2 of 3. No paradox, no crazy math or logic, just a simple misunderstanding that people seem to never want to let go of for some reason. One last note: The door eliminated is not random: it is always an incorrect door, meaning if an incorrect door was eliminated, and you picked an incorrect door (which we know will happen 2 of 3 times), all the incorrect doors are accounted for, so the door you switch to must be the correct one!

By the way, it was fun as heck convincing my girlfriend (not mathematically blessed) that this worked through trial and error. She saw it with her own eyes and still refuses to believe it :) Try winning some money/favors off your loved ones with this by using a black card and 2 red cards!

It's like the two coins problem...One is double heads, the other is standard (heads on one side, tails on the other). If you pull a coin at random, and observe a heads on one side, what are the odds the other side is also a heads? I find this one slightly easier to convince people of the right answer.

Quote: cclub79

It's like the two coins problem...One is double heads, the other is standard (heads on one side, tails on the other). If you pull a coin at random, and observe a heads on one side, what are the odds the other side is also a heads? I find this one slightly easier to convince people of the right answer.

75% ?

66.6% ?

Crap. I don't know.

Quote: DJTeddyBear

75% ?

66.6% ?

Crap. I don't know.

One of those is right. Just think about how when you draw a coin, there are 4 possible "sides" that you could see. By seeing a head, you are gaining information and eliminating a possibility of which coin it is.

Quote: cclub79

One of those is right.

And the other is the most common wrong answer, right?


Man, I don't know. Between the posts about the Monty Hall Problem, and Croupier's Roulette Bias discussion, (and the new relavation that THAT might be an April Fools joke), my brain hurts. I'll think about it tomorrow....

Hmmm....

I suggested 75% as an answer because some people would think that, since there's a 75% chance that you'll pull a coin and see heads, then there's a 75% chance that the reverse is heads.

But thinking about it, I know that ain't right.

I'm thinking that 50% is the most common response, since you have a 50% chance of pulling the double header out of the bag meaning it's still 50% that the other side of the coin is heads.

That ain't right either, because if you pull a coin and see tails, then you're at 100% chance the other side is heads.

Some people reason it out differently.

They are either looking at the head side of the two sided coin, or one of the head sides of the double header.

That ain't right either, because they forgot about the OTHER SIDE of the double header.

Add the other side, and my original guess of 66.6% is correct.


Another way to prove it:

There are THREE possible coin faces you're seeing. The head of the two sided coin, side 'a' of the double header, or side 'b' of the double header. Two of those three possibilities have a head on the other side. So the odds of the other side being heads is 66.6%


Thanks for giving my brain muscle a workout.

I really needed to see the wizard's webpage on this too, the layout of possible outcomes if the contestant switches or doesnt switch is certainly absolute proof.

Quote: FootofGod

I don't get it. I've never had a hard time accepting this...

I really was struck by the fact that FootofGod's intuition now accepts the answer readily, and I don't doubt it.

I have to say that my intuition accepts the answer when it is 1,000 doors and fights it when it is 3 doors. I predict what happens to me eventually is that my intuition will also accept the answer for 3 doors and I won't fully appreciate why this changed myself. I wonder if this is a known studied phenomenon? Should be a new thread!

Quote: odiousgambit

I have to say that my intuition accepts the answer when it is 1,000 doors and fights it when it is 3 doors.

See if it will fight this scenario:

You pick a door. Then, without opening any door, Monty asks you if you'd like to switch to BOTH of the other doors. Do you switch?

Most (all?) of the time, people switch.

It's the same as the original question. You KNOW at least one of the other doors has a goat. You KNOW that if Monty opened one, it would show a goat.

Yet, if he shows the goat first, people suddenly think the odds become 50/50, and, because of the emotional attachment they made to their choice (not the chosen door, but the choice itself), they refuse to switch.

Arguments for this often get wordy and hide what is a simple idea. When arguing the following two
statements, you can assume you know what’s behind each door, you might as well use cgg.

p(c | stay) = p(c) = 1/3 because the only way to win is to pick the car on your first selection. Everybody
agrees with this without hesitation.

p(c | change) = p(g) = 2/3 because, if your policy is to change doors, then just convince yourself that picking
either one of two goats on your first selection wins the car and picking the car on your
first selection loses.

Thus your chance of winning doubles under the "change policy".

Quote: rjs357

...you might as well use cgg.

What's cgg ?

What would you do in this situation? There's 100 doors. Before you pick, the host says he is going to open door 1, 2, or 3 after you pick to reveal a Zoink, you will then have the option to switch or stay. A prize is behind one door. You pick door #25. Host opens door #1 to reveal the Zoink. You have the option to switch. Do you?

Quote: RaleighCraps

When you picked the first time, you had a 1 in 3 chance.
When Monty shows you one of the goats, and gives you a chance to pick again, you now have a 1 in 2 chance.

The 1 in 2 chance is true, if you pick randomly between the 2 remaining doors.
That means you pick 50% of staying, and 50% of switching.

But that doesn't mean it is the best strategy for the contestant. It is the best strategy if you know *nothing* about if and how Monty picks his doors. You can use that strategy if you suspect Monty would sometimes try to talk you out of your previous choice, you are not sure if he always shows a goat, you name it. With that strategy (picking a random door between the 2 remaining doors) you pin a 50% success rate for whatever hidden strategy Monty is doing.

However, if you *know* that Monty will always show a goat (and is not showing a goat by chance), you can perform better than 50% by switching (66%). Of course if Monty decides to be a badass (and always shows the car unless you picked it before, in which case he shows you a goat), your success rate by always switching is 0%.

In short: randomly picking gives you always a 1 in 2 change, whatever stragtegy Monty is following. If you know Monty's strategy, you can improve. If you compare to poker, the random pick (50%) would be game theory optimal play, the switch (knowning Monty always shows a goat) would be best exploitive play.