Monty Hall Redux

Comments on Cecil's Columns/Staff Reports
#1

I read that question, and back-and-forth answers/comment section, with much amusement…boy did that create a firestorm! lol…

But in reality, it’s soooo much simpler than all that.

When Cecil said that Monty Hall is no dummy, and that empirical evidence would give rise to contestants figuring out what to do based upon Monty’s choosing whether or not to reveal a door (based upon a contestant’s initial choice being right or wrong) is, in and of itself, wrong. Not that Monty Hall isn’t a dummy (haha), but that it doesn’t matter what Monty chose to do, other than playing head games which can’t be factored in to probability analysis. I’ll explain…

When a game of chance, such as Let’s Make a Deal, involves only 3 choices…one of which is going to be revealed as wrong by someone “in the know” (as Monty obviously is)…it’s a special case. When you only have 3 choices, then any one you pick are 33% apt to be right. This means they’re 66% apt to be wrong (oh quit bickering over that last 1% lol).

When Monty reveals, with foreknowledge, a remaining wrong choice after you’ve chosen…he has effectively removed 50% of the 66% potentially right choices (1 of the other 2 choices you didn’t select…and at least one will always be wrong). That ends up being 33% remaining probability of being right, which were **your odds **from the get-go should you switch. In a nutshell, it makes no difference if you switch choices. QED.

Once again, assuming Monty is always, and without fail, going to reveal a knowledgeable wrong choice after you’ve chosen from only 3 alternatives. Now…this doesn’t hold water for any amount of choices above 3, where the “Monty” type person in the know is going to reveal all the remaining wrong choices except 1 after you pick something. For example:

Let’s say Monty gave you 4 choices…you pick one, he reveals 2 wrong choices…leaving only one other choice besides your own…you originally had a 25% chance, but now you have a 37.5% chance by switching because, out of the 75% potential right answers versus your original one, he just removed 66% of them. (2 outta 3 removed, and those 3 comprised 75% probably of being the right choice…and 2 outta 3 is 66%…66% of 75% is 37.5%…clearly you should switch).

In the case of Marilyn vos Savant’s extreme example…then clearly the odds improve astronomically by choosing the remaining choice other than your own. To wit…she brought up 1,000,000 doors, and you choose 1 then she shows you 999,998 wrong ones except for number #777,777. So now it’s a contest between your initial choice, which we’ll call #1…and her remaining option of #777,777. Your initial choice was 1 out of 1,000,000…which is .0001% chance of being right. Not very hot odds, though Vegas would love them.

Her initial odds were 99.9999% of having the right door…and, nice lady that she is…out of her 999,999 doors, she decided to remove 999,998 of them that are guaranteed to be wrong Which means she removed…well, it’s a big percentage…but suffice it to say 99% of them. lol

So out of your initial .0001% chance of being right, you now have the opportunity to decide to take a gamble (haha) on being 99% right (when the math is done over the percentage that she had, then removed, and vis a vis your original percentage). Yeah…not too tricky, there.

So…to conclude. In the case of Let’s Make a Deal, when you choose something and every other wrong option but 1 is left…and the initial choices are only 3…it doesn’t matter. This is, of course, assuming that Let’s Make a Deal mattered in the first place. :wink:

#2

Welcome to the Straight Dope. Here’s a link to Cecil’s column.

And here are some earlier threads:
Monty Hall Problem Revisited Again Again!
Monty Hall
Monty Hall [edited title]
Monty Hall Question
Monty Hall Problem
Monty Hall Open’s a door…

And, unfortunately, your conclusion is wrong. The easiest way to see that is to run an experiment. Randomly assign a prize to one of three doors (using a die or another random generator of your choice) and randomly choose a door. Keep track of whether you’d win by “staying” or “switching”. Run this test 50 or so times and see how often switching is the best strategy.

#3

For some clarification purposes, and a bit better syntax…I tried to edit that post, but apparently I’m only allowed to do so within the first 5 minutes. lol…

So! Deal with some of the places I wasn’t the greatest enunciator, and that I have to add more to that post in a response. :rolleyes:

After having posted my previous post, I naturally went back and read the original topic to see if I had missed anything. And…one thing I noticed was that it quickly went from a discussion of the nature of Let’s Make a Deal (which deals with only 3 choices, 1 of which is wrong that will be revealed) into decks of cards, and boys and girls and the odds of their sexes…lol…all of which are actually accurate.

The problem is…when it comes to Let’s Make a Deal, and the question originally posed to a scary IQ person (not that IQ tests aren’t scary in and of themselves. lol), that the resulting answer which was considered wrong led off into a field of discussion that didn’t actually address whether the answer was wrong. Only examples of greater quantities than 3 were used…and frankly, I still think Marilyn (hence my post above. lol) about playing Let’s Make a Deal was incorrect, but not that her examples above choices of 3 in that situation were.

As a final thought…I’d like to suggest that some goofy 5 minute rule about editing one’s own post be re-considered. :smiley:

#4

How do I also test if someone in the know is always guaranteed to reveal a wrong choice no matter if I picked right to begin with or not? I did go out of my way to indicate that “inside knowledge” plays a part in it…lol…except for when it’s in choices of 3’s.

50% of 66% still strikes me as 33%…and I read most of the links you gave me. I’m gonna feel really stupid if I’m still missing something! lol

#5

It’s not 50% of 66%, you are given the full 66% chance.
The choice boils down to that you get to choose your one door or choose both of the other doors.
Monty Hall’s actions and which door he opens are basically irrelevant. He’s always going to open a door with nothing behind it. He’s giving you the option to choose BOTH remaining doors.

#6

Hush you. We used to not be able to edit at all. You’ll get five minutes, and you’ll like it!

#7

As a final thought, for tonight…having re-read yet again a lot of the back-and-forth…I wish to tackle what Marilyn said as a reason why in a game of chance with only 3 choices…though I’m not fond of taking on the apparently highest IQ person around. lol

But…let’s give it a try. From the http://www.straightdope.com/columns/read/916/on-lets-make-a-deal-you-pick-door-1-monty-opens-door-2-no-prize-do-you-stay-with-door-1-or-switch-to-3 discussion, Marilyn stated that when given a choice between a car, and 2 goats…that after you make your pick you should definitely switch if a goat was picked by the person in the know to make you reconsider your first pick since, “The first door has a one-third chance of winning, but the second door has a two-thirds chance.”

Can anyone tell me why the door that was initially picked maintains its 1/3 chance of winning, but somehow the remaining door gains a 2/3 chance just because the host knew to pick a goat door out of the 3?

#8

No…they’re not. And here I said I was done with this for tonight! lol

In cases of 3…Once Monty reveals one of the other 2 doors, you’re not choosing from 66% probability of being right…you’re choosing from 66% probability of being right when half of it has been revealed.

Which…your first initial door pick winds up being kind of the same odds. Keep in mind…probability cares not for artificial selections…the odds don’t change just because you put your “mark” on a certain door. The odds remain the same, and in cases of 3 where these rules are followed…it doesn’t matter. Above 3? Yep…they do.

#9

The answer to the puzzle is that Monty, by revealing the door is giving you information. Let’s take the following scenario:

Your friend has two children named Sandy and Chris. You’ve never seen these little sprats, but one day, your friend lets it slip that at least one of them is a boy. What are the odds that both children are boys?

The answer is 1:3 because here are the following possibilities:

1). Both Chris and Sandy are boys.
2). Chris is a girl and Sandy is a boy.
3). Chris is a boy and Sandy is a girl.

Note that possibilities #2 and #3 are NOT the same! If Sandy is the boy and Chris is the girl, Sandy would be very upset if you tried to put a dress on him.

Okay, instead of saying "I have at least one boy, your friend says “Did I ever tell you Chris is a boy?” What are the odds that both Chris and Sandy are boys? The answer is now 1:2 because there are only two possibilities:

1). Chris and Sandy are both boys.
2). Chris is a boy and Sandy is a girl.

Why is there a change? Because you got more information. In the second scenario, you were told the additional info that a particular child is also a boy, and information can change your odds. Think of card counters in Black Jack, by knowing how many face cards are left in a deck, they can actually improve the odds enough to beat the house.

The same with the Monty Hall problem. By knowing which door you picked, and by seeing which door Monty picked gave you more information. Here are the three possibilities you get for picking Door #1:

1). You pick Door #1, the prize is behind Door #2, and Monty shows you Door #3: Switching wins
2). You pick Door #1, the prize is behind Door #3 and Monty shows you Door #2: Switching wins
3). You pick Door #1, the prize is behind Door #1, Monty shows you Door #2/3 You pick the other one; Switching loses.

Therefore, by switching, your odds of winning are 2:3. In fact, the only time switching doesn’t help you if you originally picked the right door.

You can see that the same three choices if you had picked Door #2 or Door #3. Your odds changed because you have more information even though the structure of the problem is still the same.

Let’s take this scenario:

You pick Door #1, Monty reveals Door #2. Right at that time, Mr. Spock beams down, finds this game fascinating and wants to play too. He sees the goat behind the open Door #2, and logically deduces he should either pick Door #1 or Door #3. His odds of winning are in fact 1:2 while your odds are still 1:3. Not hearing your pick, he also picks Door #1.

Now, Monty asks if either of you want to switch: You switching will improve your odds from 1:3 to 2:3 while Spock switching will still have the same odds. This is true even though both you and Spock picked the same door! Why? Because you have more information than Spock. You know which door you picked and you know which door Monty opened while Spock only knows what door Monty opened.

For more fun, checkout Jeff Rosenthal’s Monty Fall/Monty Crawl scenarios.

#10

There is no reason to make a distinction between a case with 3 doors or one with 4 doors. What do you think fundamentally changes?

Anyway, if you actually read through the links, the case with 3 doors has been tested millions of times using computer simulations and your chances are twice as good if you switch to the other door.

#11

moridin5 said:

The 5 minute editing limit is to prevent abuse. We want people’s words to stand. I’ve been on boards where people have made outrageous or stupid remarks, then later (after being chided) deleted the offending remark and claimed innocence. The five minute limit helps keep posters honest.

This is where you are in error. Your mistake is that Monty is not revealing the door by chance, but rather with foreknowledge. It is an equivalent situation to Monty not revealing any doors, but offering the contestant the option to keep their 1 door or swap to the 2 other doors. Revealing the goat behind one of the 2 remaining doors just confirms that there is at least one goat in the two remaining doors. But the contestant knew that already.

#12

Maybe thinking about it like this helps:

There are three cases.

  1. You choose the door with the prize; the probability of that is, rather obviously, 1/3. Monty reveals either goat, and you win by not switching.
  2. You choose the door concealing goat A; the probability of that is also 1/3. Monty reveals goat B, and you win by switching.
  3. You choose the door concealing goat B; the probability of that is also 1/3. Monty reveals goat A, and you win by switching.

Thus, you win by switching in two of three cases. It’s important to realize that your chance of being right from the outset is 1/3, and that thus, the chance that the prize is behind either of the other doors is 2/3; after one goat has been revealed, it still is, since the chance that you guessed correctly in the first place is still 1/3.

#13

It seems to me that the thing which confuses people about the Monty Hall Problem is that the numbers are just too small. When dealing with thirds, it’s not obvious why the probabilities are the way they are.

So forget three doors, and instead imagine a deck of 52 cards, spread out on a table, all face down. You have to find the ace of spades to win a prize, and Monty knows exactly where it is in the deck.

You point to a card at random. Your chances of having picked the ace of spades are obviously 1/52.

Before you can flip your card, Monty offers you a deal. He takes another card out of the deck and flips it, revealing the three of hearts. You can now stick with your original card, or pick a new one. Let’s examine the options:

Stick with original card: we picked this when we had a 1/52 chance.
Put it back and pick a new card: with the three of hearts revealed, there is now a 1/51 chance of picking the ace of spades.

The obvious answer is to switch. And let’s suppose Monty keeps making this deal, revealing a new card every time. Your odds go up to 1/50, then 1/49, and so on, as long as Monty is willing to offer the switch.

Now imagine that you continue this game until there are three cards left. This game is now identical to the three doors in the Monty Hall problem, and you can see why the odds are in your favor if you switch.

#14

This only works if he knows that that card is not the ace of spades.
Powers &8^]

#15

As noted. :wink:

#16

Jeez, is there no ban on further Monty Hall threads? I would support such a thing.

According to you, 1/3 of the time you will win if and only if you don’t switch, and 1/3 of the time you will win if and only if you do switch. Can you tell us what happens the remaining 1/3 of the time (lol).

#17

You are losing, of course. :rolleyes:

That is not where his logic meets some difficulties.

#18

This actually made me laugh out loud. lol.

#19

friedo, your example works better if when after you have selected your card, Monty then turns over 50 cards that are not the Ace of Spades, and then asks if you want to switch.

You can keep your first card at 1/52 chance of being right, or swap to a 51/52 chance of being right.

#20

The key thing I see here - in a totally random world, you pick a door, Monty picks a door, and 1 in 3 times Monty shows you the car not the goat. The other 2/3, he shows a goat.

In that scenario, you have the same odds staying as changing. However, you can eliminate the 1/3 of the scenarios where Monty shows the car; or rather, Monty can. He knows where the car is. So, after Monty flashes his goat, you have eliminated 1/3 of the random scenarios.

Look at it another way. The car has a 2/3 chance of being behind the other 2 doors. Monty has thoughtfully told you, “if you switch and pick the other 2 doors instead, don’t pick this one”.