Sound advice for winning on a game show

by Paul Michael on 25 February 2008 13 comments
Photo: DeBaird

I've been a closet fan of game shows for years; most likely because I love the idea of getting something for nothing (hey, this is a frugal writer after all). I recently stumbled across something called "The Monty Hall Paradox" and it's opened my eyes to the laws of probability in the seemingly innocent world of game shows. And armed with this knowledge, you've got a much better chance of winning something.

Monty Hall was the host of a show called "Let's Make A Deal," which was very popular in the 60's and 70's. One of the cunning and cruel games he liked to play was the "3 doors dilemma." It worked like this: You would be presented with 3 doors. Behind one door is a fabulous prize, like a car or vacation. Behind each of the other two doors is a dud, like a goat or toilet paper. 

Monty would ask you to choose a door. Then, before you opened it he would open one of the two remaining doors to reveal one of the dud prizes. Now he asks you a simple question..."Would you like to stick with your original choice, or switch to the other door?"

At this point you're rubbing your head, like I was, and you come up with the only logical solution; it doesn't matter. There are now two doors, one has the star prize, one doesn't, so it's 50/50, right?

Not so. Not so at all, as I found out when I delved deeper into the paradox and discovered that dozens of very smart mathematicians have been trying to explain this one for eons.

I have posted several videos and links below so that these boffins can explain it for you, much better than I could here. But, in a nutshell, here's the solution; you should ALWAYS switch doors.

doors

The reason is simple...at the start of the show you have a 33% chance of picking the star prize, and a 66% chance of picking the dud prize. So, the odds are you will pick a door with a dud prize. That means that when Monty Hall reveals a dud prize behind one of the other doors, the remaining door is much more likely to be the star prize. See? Well, I had to watch the videos to be convinced, so give at least the first one your attention.

But if you ever do make it onto a game show, maybe like The Price Is Right, or someone tries the same trick using 3 cups and a ball, you now know just what to do to put the odds firmly back in your favor. 

Video 1 - The 5-minute explanation

 

Video 2 - The TV show explanation

 

Video 3 - The math boffin 20-minute exploration

 

Other links:

http://www.youtube.com/watch?v=1uS4Sa3nIFc&feature=related

http://www.wiskit.com/marilyn/gameshow.html

http://en.wikipedia.org/wiki/Monty_Hall

http://tierneylab.blogs.nytimes.com/2008/02/20/play-the-door-game/

http://www.math.uni-bielefeld.de/~sillke/PUZZLES/monty-hall

 

 

Additional photo credit: Bzo
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Xin Lu's picture
Xin Lu

You just took me back to my randomized algorithms class.  That was the hardest class ever. To this day, I find probability and combinatorics to be really really hard, but the Monty Hall problem is really a classic simple tool that people can use in their daily lives.

Paul Michael's picture

I just found most of what you said to be really, really hard! Like I said, certainly not a math whiz here. But it's a great lesson in lateral thinking.

Guest's picture
Blaise Pascal

Monty Hall was asked about the "Monty Hall" problem once. He pointed out that on the actual show once a contestant turned down a choice they were never offered it again -- so after choosing door 1, 2, or 3, they would never be asked if they wanted to switch to a door they turned down.

The puzzle's a classic, and sounds very much like the show, but it wouldn't actually happen on the show.

Guest's picture
cendare

fyi, the tierneylab.blogs.nytimes.com link is actually not about this Monty Hall problem -- it really is a totally different 3-door puzzle. (A fun and intriguing one, but nonetheless not relevant here.)

Paul Michael's picture

it was just further reading, that was all. Although I think the tierneylab blog did actually cover the problem as well.

Guest's picture
Warbo

I just have to laugh at this. Another trendy mathematical fallacy destined to land in the scrap heap with the hidden dollar problem.

The fact is, once the first door with the goat is opened the sampling of doors is reduced from 3 to 2. meaning that the door you've already chosen has a 50% chance of being the other goat, not a 66% chance.

The tallys regarding 'switching' and 'non switching' do show some results, though. Here's my guess about why that is.

When pressed with the choice of three doors, people tend more commonly to choose the same door. (I'm no statitian, but I would guess the middle door is probably the door people pick most often). The proprietors of the test know this, and based on how often the middle door gets picked, they will hide the star prize in one of the side doors more often.

When the middle door is picked a goat will be revealed on one of the side doors. Since the odds of the prize being hidden in a side door is greater, and since there is only one side door left, in this case, which will be the most common case, your odds are increased by switching doors.

Unless someone decides to start hiding the prize in the middle door more often.

The boffin explanation makes the problem seem more convincing when it takes the problem into three scenarios, the prize being in door 1, door 2, or door 3 and you chose door 1 initially. The problem with this is that there are really only TWO scenarios that matter. Either you picked the right door initially or you didn't. If the prize is in door 2 or 3, those are collectively the same scenario; You didn't pick the right door. In the boffin explanation they are simply counting the same scenario twice. The other door will be elimanated and in the end your chances really are 50/50.

Guest's picture
Guest

Warbo, the middle door being chosen the most often is NOT the reason for this seeming paradox. In fact, I would venture to guess the web interface with the screenshot in the blog posting selects a door randomly for the money. That is certainly the assumption in the puzzle.

The KEY to this puzzle is that the game show host is aware of what is behind each door.

Warbo, if the game show host was unaware of what is behind each door and selects a door at random, it happens to have a goat behind it, and he asks you if you want to switch, then yes, you have a 50% chance of having the money behind your door.

But run that scenario a few times. 33% of the time, the game show host will select a door and the money will be behind it! AND THAT NEVER HAPPENS IN THIS SCENARIO. THE HOST WILL SELECT A GOAT 100% OF THE TIME BECAUSE HE KNOWS WHAT IS BEHIND THE DOORS.

THAT is why your door has a 33% chance of having the money behind it.

Think of it like this - instead of 3 doors, there are 100 doors. You select one door. The game show host opens 98 other doors and each has a goat behind it. He asks you if you want to switch.

Now, what's more likely - you selected the right door the first time? Or is it more likely that the game show host knowingly opened all the other doors except the one with the money in it? You had a 1% chance of being right the first time. And that does not change throughout the puzzle.

In this scenario there is just a 1% chance you have the right door - and a 99% chance that if you switch you win!

Guest's picture
Courtney I

I forgot to leave my name on that last post, this is Courtney I (in case anyone wants to reply to me).

On a side note, to extrapolate from my post above...

This means in the Monty Hall puzzle, if the game show host knows which door contains the money, you have a 33% chance of your door being right. But if he does NOT know which door contains your money, you have a 50% chance of being right.

How wild is that!! Knowledge is a dangerous thing!!

Guest's picture
DivaJean

Regardless of what you pick, its still more than you left the house with in the morning-- even if its a goat.

Granted, big money is always nicer, but there is no evil in a year supply of Rice a Roni (frequently given as a parting gift back in the 70's from gameshows).

Guest's picture
Guest

Regardless of what those videos say, the choice is 50/50. No matter what door you pick, one goat door will be removed. You will always be left with a goat door and a car door. Doesn't matter if you pick goat 1, goat 2, or the car first.

Guest's picture
Guest

its 67% chance of winning because the host has knowledge of which door had the winning prize behind it and which had the dud and is forced to open the door with the dud.

Consider this. After your first decision there are only three possible outcomes.

You chose the prize right the first time.

You chose dud 1.

You chose dud 2.

Since a dud must be removed by the host. Its easy to see that by switching you increase your chances of winning (the only way you could switch and lose is if you picked the prize the first time, if you chose either dud the first time you'd be better off switching)

The following is a list of all possible outcomes of the game
D = first choice is dud
P = first choice is prize
S = switch
NS = no switch
W = win
L = loss

D-S-W (first dud)
D-S-W (second dud)
P-S-L
D-NS-L (first dud)
D-NS-L (second dud)
P-NS-W

As you can see, by switching, you attain a 67% chance of winning the game.

Guest's picture
Operagost


Regardless of what those videos say, the choice is 50/50. No matter what door you pick, one goat door will be removed. You will always be left with a goat door and a car door. Doesn't matter if you pick goat 1, goat 2, or the car first.

It's amazing how some people, even when faced with overwhelming evidence to the contrary, are so confident in their preconceived notions.

Guest's picture
Guest

I know I'm late on responding, just found this because I was trying to google why the playatmcd website was down. I'll avoid putting anyone down, but it's just like Deal or No Deal. If you pick 1 case out of 26 cases, your odds of having the $1 million (or any specific amount) is always 1 in 26, just because you find out what the other cases are later in the game doesn't change the odds. So in the Monty Hall question, you always have a 1 in 3 chance of picking the winner and a 2 in 3 chance of picking a dud, despite having a door opened. Think of it this way, the odds of getting heads in a coin flip is always 1 in 2 (ok if you're too technical 1 in 2.0000000000000009 because the coin could land on the side and be neither head nor tails, but ignore that.) But the odd of getting heads 20 in a row is over 1 in a million. So if you flipped the coin and got 19 heads in a row, the odds of the 20th coin flip being head doesn't change, it's still 1 in 2.