the game doesn't make any sense in switching. the odds don't change just because they open a door. now if there were more than three doors, i might agree, but your only other choice is one door. how is that a choice? if i stay with the same door, my odds are still 2 out of three. i understand what the reasoning behind what they are saying, i just think it doesn't apply when there are only 3 doors, then there are only 2 doors, you don't really have a choice.

as a math major, i call bull shit on this. there is no right or wrong on this game.

This is a tricky problem. Switching is ALWAYS right. If you don't switch, you only win when you were right, 1/3. If you DO switch, you lose whenever you were initially WRONG. Subtle, but true.

Snoop -

It is because you only have a 33% chance of picking the right door the first time, and a 66% chance of picking the wrong door. The reason this works is because the host knows a "wrong" door to open, so it becomes statistically smart to change your door. You only had a 1 in 3 chance of picking the "right" door the first time, so by switching (Only when the guy opening the door knows which one has the goat) you get on the better side of that equation. Note that this is NOT the case if the host doesn't know where the prize is, and just opens one randomly.

@snoop-blog:

Yup, no right or wrong, just what gives you the highest statistical chance of getting the prize. It is still a gamble, but a smarter gamble is one with the odds in your favor, no?

This game was actually in the movie 21. Here is some more info:
[en.wikipedia.org]

I can't figure out why he's right, but he certainly seems to be from my trial. I switched 15 times and got the car 87% of the time. I stayed 11 times and got the car 36% of the time. I realize I didn't have enough trials, but it was very obvious from the beginning. Mathematically, I can't figure it out, even after reading the article twice.

PLEASE read the article before you "call bull shit". PLEASE think before you write your 1-line, no-capitalization recap of why a long-proven mathematical and statistical concept is false. This is an interesting, if not instinctively obvious theorum that has stood the test of time.

When you first selected a door, you had a 1/3 chance of being right and a 2/3 chance of being wrong. After the other door is opened to reveal the "goat", you now know that your door still has a 1/3 chance of being right, but now instead of there being 2 wrong doors (which total the other 66.7% chance of correctness), there is only 1 other door - which by itself has a 66.7% chance of being the right door.

The math not only works, but if you had READ the article, OR tried the exmaple, you would see that it is statistically and mathematically true.

I hate people who dismiss things like this without giving it some thought - use your brain for a minute.

/rant

@BloggyMcBlogBlog: ha you beat me to it

@TylerE: Don't you mean you lose when you were initially right?

@MrFalcon: So if you had a 33% chance of picking the right door, the first time, what is your chance of switching and winning? It's obviously not 50/50 per this article, so what is it?

@mdkiff: the one line was an add-on to the first comment dick. and i refuse to capitalize on blogs. this isn't a gd term paper, nor a book. attacking commenters on here is a troll move. you notice how@MrFalcon: manage to show a little class, something you'll never have.

Wow, I never thought I'd see the Monty Hall problem on Consumerist.

@snoop-blog: As a math major myself... you're wrong.

@snoop-blog: Too much wake n bake maybe?

@snoop-blog: Did you fail math class?

@qwickone:

Stated above by mdkiff, your chance at picking the "right" door by switching is 66% As the article states, the trick in all this is that people exclude the option that is shown to not be right, and then they don't think it matters anymore. Again, so long as the person opening a door to show the "goat" knows that it doesn't include the "prize" then you have a better chance at winning by switching. If you stay with your first pick, you only have a 33% chance at winning. Can you still win by staying? Yes, but not as often as by switching.

Zonk

The Monty Hall game is flawed in one important way; it assumes that you want a car over the goat. For me, I am a big goat fan.

The best way to think about this problem is not with doors but with 'groups of doors'. Say the door you pick is Group A, and the other 2 doors are Group B. Now if I ask you which group probably has the car, you would automatically think Group B since there are more doors there and thus a higher probability. When I reveal some the doors behind Group B, it doesn't change the fact that Group B is still better statistically no matter what I reveal about the items within that group.

Reader Jim's response for the case with 10 doors (say you pick door 1), Monty reveals 7 goats, and you have the option of switching to 5 or 8:

2. Switch to door 5 or 8, though each has a 55% chance of having a goat [tierneylab.blogs.nytimes.com]

If my math is correct, that means Jim's initial door has a -10% chance in order for total to add to 100%.

@snoop-blog: there is no right or wrong on this game.

Ding! Ding! Ding! First poster out of the gate and we already have a sucker!

@qwickone: It's 33.33333% vs 66.66666%. If you switch, you are actually picking the other 2 doors without fear of losing if one of those 2 happens to be wrong.

even if your chances of success are %66, there's still a chance of being wrong was all i was saying. i understand the stats.

@snoop-blog: @MrFalcon: Yup, no right or wrong, just what gives you the highest statistical chance of getting the prize.

Ooh maybe I'm the sucker. I take back my snarkyness, snoop-blog, on account of a linguistic technicality. I'll give you the benefit of the doubt.

Please accept my humble apology. You are a scholar and a gentleman.

@snoop-blog: Your first comment:

the game doesn't make any sense in switching. the odds don't change just because they open a door.

If I recall Monty Hall, though, he would offer and even encourage the switch, no matter what prize was where. He wasn't steering you to the car or the goat, as much as making you worry as to whether you'd get car or goat.

With 3 doors, your odds are ALWAYS 50% after one door is revealed.

Initially, you have 3 doors, you need to pick one. The chance that the door you picked has a car is 1/3 (33%)

Now, you have been shown that one of the doors is NOT a car. So, you have your original choice or you can switch.

At this point, having one door revealed, there are only two doors left. The one you picked, and the other unknown door. The car is behind one of those two, so your original "1 in 3" has been reduced to "1 in 2" (You're not going to pick the door that was revealed - it is taken out of the equation)

If you stick with your original choice now, you have a 50-50 chance.

If you pick the other door, you still have a 50-50 chance.

The car is behind one of the two doors, it doesn't matter what you picked before - you are given a choice to stick with your original choice or pick the other one - in essense you are given a chance to "choose again" in a 50-50 environment.

if you go with the %66, you may be unlucky and still lose, so is this more about luck, in a completely random game, absolutely.

"the game doesn't make any sense in switching. the odds don't change just because they open a door."

I used to agree with the doubters. But the problem I had is thinking that the selection was random, it is not random.

First, when they open that first door they never open your door. Thus, the choice to open that first door was not random.

Second, when they open that first door, they never open a door with the prise in it. Thus, once again, it is not random.

Once you realize, it makes sense. And even if it does not make sense, as qwickone, points out. You can verify it empirically.

@Dooley:exactly the way i understood it, but the article with the game disagrees with you 50-50.

my only point is the whole thing is luck. on a game like this your always better to stick with your gut (put yourself in a real game like this). if you chose right the first time, but changed your answer, then found out your first choice was right, you'd be kicking yourself saying "why didn't i just stick with my gut". i doubt you would be like "oh well, at least i picked the best door statistically."

@Imafish: yeah it wasn't until after my first post that i realized the game was not random.

Have you ever heard - the grass is always greener on the other side! It's not! As dumb as it may sound those old proverbs make sense most of the time...

What if I wanted a goat?

"i doubt you would be like "oh well, at least i picked the best door statistically.""

But if you did pick the "best door statistically" you'd be less likely to kick yourself in the first place. So I'd go with that.

In fact, I can't think of a reason to go against statistics in a game of chance other than a desire to lose.

@snoop-blog:

"my only point is the whole thing is luck. on a game like this your always better to stick with your gut (put yourself in a real game like this). if you chose right the first time, but changed your answer, then found out your first choice was right, you'd be kicking yourself saying "why didn't i just stick with my gut". i doubt you would be like "oh well, at least i picked the best door statistically.""

Hi, I work for the Bellagio. We'd really LOVE to have you come visit our casino. A lot. Really, mortgage your house and come here. Please.

Oh my, this new image feature sure is nifty.

I'm not exactly going into a maths-heavy field, so I took fairly elementary math courses. My professor turned it into a "how to think like a sensible person" course and spent the final month, no exaggeration, talking about probability.

She printed off erroneous Ask Marilyn columns from Parade magazine and spent entire lectures correcting them, made us act out "Let's Make a Deal" in class, and gave us hours of homework that forced us to apply the probability questions to our everyday lives. I hated her as a person, but that professor did us a big favor. She was right, probability makes us look really stupid.

@snoop-blog: Unfortunately when considering probabilities intuition can lead you astray. It is this kind of counter intuitive nature of probability that con artists and hucksters take advantage of.

Here is another example that is similar to the Monty Hall problem that people tend to get incorrect. I have 3 cards, one card is red on both sides, one that is black on both sides, and one that is black on one and white on the other. I take a card randomly and place it on the table, it shows red.

What is the probability the other side is red?

I tell you well it can be the red/red card or the red/black card, so its 50/50 a fair bet. I bet red, you bet black.

In reality the odds are in my favor. In fact the probability the other side is red is 2/3. It could be the red/red card, it could be the red/red card flipped, or the red/black card. The flipped red/red card is a distinct probability but most people don't intuitively catch that.

@Imafish: unless you consider yourself unlucky

@KernelM: math major =/= genius. i'll be the first to admit i didn't get every math problem correct nor did i graduate top of my class. all i was saying is that i understand math naturally unlike english, and to me it was pure luck no matter how you slice it. if it wasn't luck, then you'd be able to be correct %100 of the time.

@snoop-blog: Crazy mathematicians might ;-)

@LS1Andrew: Is that remaining 0.00001% chance reserved for the goat terrorist group that will bust in, kill everyone and burn the studio down to release the goats into the forest?

i understand if you played over and over again what would happen, but if you only have one shot at it, it's pure luck.

@Dooley: This is where the person who opens the door having knowledge of where the prize is comes into play. If one door was randomly removed from the choosing, and nobody knew whether it had the car or the goat in it, then yeah...your gut call would be just as good to stay as to switch. You wouldn't have any statistical advantage to switching, or to staying. However, because someone with knowledge of the results is showing then eliminating a "wrong" choice, that is what keeps it scenario where you're choosing 1 out of 3, and not a new scenario where you're choosing 1 out of 2. Knowledge, as is so often the case, is the differentiating factor here.

The explanation is pretty simple. You have a 2/3's probability of picking a door with a goat behind it (when you start the game).

Therefore, you should assume that you will have picked a loser door each time, because you have 2/3 probability of picking a loser door and only 1/3 probability of picking a winner.

The host will always reveal a door with a goat behind it.

If you picked a loser door and the host reveals the other loser door, then it follows you should switch to the remaining door because it must be the winning one.

This only works if you picked one of the losing doors. However, because you have a 2/3's chance of picking the losing door, you will, on average, win 2/3's of the time with this technique.

@snoop-blog: See my above post ;)

@Dooley: Note that Monty can only open either of two of the three doors: he will never open the one you reserved.

So if pick a goat to start with, Monty will always reveal the other goat. By switching, you'll win every time you picked a goat to start, i.e. 2 out of 3 times.

a rather common Bayes' Law example

@Dooley: Actually, that other door does matter.

Let me give you an analogous situation, and see if you still think that.

Instead of playing Lets Make a Deal, I take out a deck of cards, a standard deck of 52, and I ask you to pick out the Ace of Spades from the deck. You choose a card.

Now that you've chosen a card, I show you 50 of the 52 cards, all not the Ace of Spades. And now I leave you with a proposition: stay with your card, or switch.

The probabilities on that decision isn't anywhere near 50/50. In fact, it's a near certainty(51/52) that the card I'm allowing you to switch to is the Ace of Spades. This is because the choice isn't really "this card or that card", but how confident you were of your initial guess. You chose that initial card out of 52, and your final choice is whether you think you chose the right card initially. In Monte Hall, your probability of being right is 1/3. In this game, it's 1/52.

If you alter the game slightly, if you make it to where the person revealing the doors doesn't know where the prize is, or you start with 3 doors, you show them a loser, and then you let them pick, you'll end up back at 50/50.

But the choice in the Lets Make a Deal game isn't Door 1, which you chose earlier, or Door 2, but is instead Door 1 or Not Door 1. Staying with your door is saying you think you made the right decision on your initial guess, which had a 33% probability that you were right.

It sounds counter-intuitive, I know, but run a simulation of it, and you'll see it work out. You win by switching 2/3rds of the time, and win by staying 1/3rd of the time.

"i understand if you played over and over again what would happen, but if you only have one shot at it, it's pure luck."

Nope, each and every time you have a better chance to win by switching than to staying put. That does not mean you will win every time you switch, only that your chances double every time you switch.

Of course the problem most of us have with this is that Monty never had 2 bad doors....maybe one bad door or two okay ones.