The "math for dummies" game show

[jlm110108]

Well, I'm not afraid to admit that I must be worse than a dummy. I still don't get it. I saw the clip where Charlie is using a game show example to teach about probabilities. I read a blog where a mathemetician explained it, and I still don't get it.

If you have a choice of 3 doors, one with a car and 2 with goats, and you select the middle door, and the game show host reveals the right hand door has a goat behind it, I don't see why you're better off switching your choice to the left hand door.

The car is going to be where it is no matter what you choose. If I switch my choice, the car is still where it was when I had 3 doors to choose from. Same if I don't switch my choice.

So, what am I missing? Or is this just the kind of trick experts play on non-experts so they can laugh at them?

Now I see why my enjoyment of math stuff stopped at algebra and geometry. I don't get the theortical stuff.

JOanne

[TTn]

So, what am I missing? Or is this just the kind of trick experts play on non-experts so they can laugh at them?That does happen sometimes.
I have not seen what you are refering to.
I think it's more a pun about the imprecise definition of probabilities.
A sort of word play, that people mistake for meaning something else.
It's humorous that tons of complicated math theory, are equivelant to a guess.

That's my guess from what you've explained.

[jlm110108]

TTN, that makes sense.

Joanne

[Eli06]

I was wondering about that too. Kind of confusing, I was guessing that its because when you take away one card the probability of guessing correctly are higher, but even with that it still doesn't make much since.

[Mad]

If you have a choice of 3 doors, one with a car and 2 with goats, and you select the middle door, and the game show host reveals the right hand door has a goat behind it, I don't see why you're better off switching your choice to the left hand door.

The car is going to be where it is no matter what you choose. If I switch my choice, the car is still where it was when I had 3 doors to choose from. Same if I don't switch my choice.

So, what am I missing? Or is this just the kind of trick experts play on non-experts so they can laugh at them?


To be frank, I thought the explanation in the episode was a bit vague. But no, it's not a trick (or at least, not a trick of that kind). I'll try to explain.

We've got three doors, A, B, C. You're supposed to choose the door with the car behind it, say you choose B. The probability that you chose the right door is one third, since you have absolutely no information that could help you decide, and all doors are equally likely. Conversely, the probability that you made the wrong choice is two thirds. So the probability that you chose wrong is bigger than the probability that you chose right. Keep this in mind.

The moderator opens another door, C, to reveal a goat. For the purpose of this scenario, I will assume that we're talking about the classical goats problem. That means that the moderator already knows which door the car is behind, and he won't open that door. He also won't open your door. So, if the car is behind A, he can only open C. If the car is behind B, he can open either A or C.

You have the option of changing to door A or staying with B.

And this is where the trick -- in so far as there is a trick involved -- lies: for this second decision, you have gained knowledge. You now know that C wasn't the right door, so you're down to two doors. You also know -- and this is the crucial point -- that the chance you got it wrong the first time is higher than the chance that got it right. It is therefore more likely that B was the wrong choice.

And that's why you should change to A.

Hope this helps.

Love and Peace,
Markus

[janet-a]

dude!!!! now i get it!!!! :) :)

[jlm110108]

Markus,

Thank you! It finally makes sense. If you're talking strictly about probabilities and not which card has which picture, it makes sense to change. If you're convinced by some psychic twinge or whatever that your first choice was right, then you're choosing to ignore the probabilities, and making your decision on some other less quantifiable basis, like feelings.

But if you're going strictly by probabilities, it makes the most sense to change.

Of course, if you were right to begin with, then you've fallen onto the other side of the probability if you change -- you're one of the 1 of 2 or 2 of 3 (or whatever of whatever that makes up the difference between 1 of 2 and 2 of 3) that was wrong.

I try to think logically, but my logic isn't usually even in the same universe as a mathemetician's logic.

Joanne

[bluewildindigo]

Thank you! I've been wondering about that since the episode aired! You did a great job of explaining it.

[jlm110108]

Oh, sure! Here I thought I was the only one who didn't get it, and it turns out I was just the only one with enough nerve to ask. :D

But I did have a second thought about this whole thing. Since when the middle card was selected, all 3 cards had the same probability of being the right one, wouldn't the 2 remaining cards still each have a 1 in 3 chance of being right? The cards didn't change, so the choice shouldn't change either.

I know, it's Friday, and my mind has turned to mush. I think I'm going back to the dark side of illogical thinking.

Joanne

[TTn]

mad wrote:The moderator opens another door, C, to reveal a goat. For the purpose of this scenario, I will assume that we're talking about the classical goats problem. That means that the moderator already knows which door the car is behind, and he won't open that door.


Well that makes all the difference of course. That's a trick if they didn't tell you.
Agreed.

[sashapet_02]

I did this with my maths class (13 year olds) when I was teaching them about probability.

It took them a while to believe me, but I got them convinced in the end.

It comes down simply to this:

After your first choice:
If you have already chosen the car, you shouldn't switch.
If you have already chosen the goat, you should switch.

You are twice as likely to have chosen the goat on your first pick, therefore you should switch.

I did a demonstration like Charlie did, asking someone in the class to choose a card. It was kind of difficult to get the demonstration to work quite as well when they actually managed to pick the car first time! (I wonder what Charlie would have done if the girl had picked the car first time - it would have spoiled his 'great reveal' moment!!!!)

Sasha x

[TTn]

I just want to make things clear, now that a teacher is ringing in on this.

There is absolutely no advantage to switch, if the door opener remains un-biased, and doesn't purposely open the wrong door to decieve you.
That's not a probability application. As I said people shoud'nt get the wrong idea.

[sashapet_02]

I just want to make things clear, now that a teacher is ringing in on this.

There is absolutely no advantage to switch, if the door opener remains un-biased, and doesn't purposely open the wrong door to decieve you.
That's not a probability application. As I said people shoud'nt get the wrong idea.

I'm assuming the door opener knows what is behind each door, and obviously will not open the door containing the car. In which case there *is* an advantage to switch.

I think it's called the Monty Hall problem. http://en.wikipedia.org/wiki/Monty_Hall_problem

Sasha

[Mad]

But I did have a second thought about this whole thing. Since when the middle card was selected, all 3 cards had the same probability of being the right one, wouldn't the 2 remaining cards still each have a 1 in 3 chance of being right?


No. Say PA is the probability that A is the correct card, PB the same for B and so on. The probability P that either A, B or C is the right card must be one; we know there is one correct card. On the other hand, it is also the sum of the probabilities for each possible case of one card being the right one, i.e. P=PA + PB + PC. In the first step, PA=PB=PC=1/3, so that works out fine.

If each of the two remaining cards (say, A and B) still had a chance PA' and respectively PB' of 1 in 3 of being right in step two, the chance P' that out of these two cards one is right would be only PA' + PB' =2/3. Obviously there's a problem. P' must be one, because we know that either A or B is the correct card. So PA' and PB' can't be both 1/3 in the second step.

The wikipedia article Sasha linked to suggests another interpretation that may help you. You can think of the second choice as a choice between the one card you originally chose, and the other two cards. If you choose the other two cards and the car is behind one of them, you win because the moderator will open the wrong door (out of these two) and thus tell you which of the two you should take. And as we've seen above, the chance that one of the two other doors is the correct one (out of the three doors) is two thirds.



The moderator opens another door, C, to reveal a goat. For the purpose of this scenario, I will assume that we're talking about the classical goats problem. That means that the moderator already knows which door the car is behind, and he won't open that door.


Well that makes all the difference of course. That's a trick if they didn't tell you.
Agreed.


I would assume the contestants are familiar with the rules of the game. ;)

But you're right; that was what sparked the controversy around the letter to Marylin vos Savant (http://www.vanderbilt.edu/~bednarjt/monty/montyhall.html) (which seems to be the version of the problem most people are familiar with). In this incarnation, the problem wasn't very well-posed, and there were a number of loop-holes (such as that the moderator could decide whether he would open a door at all).

Wikipedia (http://en.wikipedia.org/wiki/Monty_Hall_problem) attributes the following, more precise formulation to Mueser and Granberg:

A thoroughly honest game-show host has placed a car behind one of three doors. There is a goat behind each of the other doors. You have no prior knowledge that allows you to distinguish among the doors. "First you point toward a door," he says. "Then I'll open one of the other doors to reveal a goat. After I've shown you the goat, you make your final choice whether to stick with your initial choice of doors, or to switch to the remaining door. You win whatever is behind the door."

You begin by pointing to door number 1. The host shows you that door number 3 has a goat.

It's this, more precisely stated problem, that the "You should always switch"-solution applies to.


I did a demonstration like Charlie did, asking someone in the class to choose a card. It was kind of difficult to get the demonstration to work quite as well when they actually managed to pick the car first time! (I wonder what Charlie would have done if the girl had picked the car first time - it would have spoiled his 'great reveal' moment!!!!)


Well, it's probably not a good idea to demonstrate a probabilistic problem with just one example. They would have to repeat the experiment to see that overall, they win about twice more often if they switch. Have you considered letting them write a computer simulation of the problem (I don't know if you have the technical ressources for that at your school)? It's pretty cool to see how close to two thirds of the time you get it right if you play the game, say, a thousand times.


There is absolutely no advantage to switch, if the door opener remains un-biased, and doesn't purposely open the wrong door to decieve you.
That's not a probability application. As I said people shoud'nt get the wrong idea.


Just for clarification: by being un-biased, you mean that in the given situation the door opener first just chose a door randomly (and he could as well have chosen the door with the car behind it, or the door chosen by the contestant), and then it just happened to be another door with a goat behind it?

In that case you're right; there is no inherent advantage to switch.

Love and Peace,
Markus

[sashapet_02]

Well, it's probably not a good idea to demonstrate a probabilistic problem with just one example. They would have to repeat the experiment to see that overall, they win about twice more often if they switch. Have you considered letting them write a computer simulation of the problem (I don't know if you have the technical ressources for that at your school)? It's pretty cool to see how close to two thirds of the time you get it right if you play the game, say, a thousand times.



Oh yeah, I didn't mean we just did it that one time. I showed them a couple of times and then they got in to groups and did it between 10 and 20 times. It just got me thinking that because the kid who did it first had managed to pick the car first time, how it would have screwed up Charlie's demonstration if it had happened to him (but of course, it being fictional, that wasn't an issue!)

It was interesting that they actually understood what was going on more when they were playing the role of the gameshow host - because they could see what was going on behind the cards. The kids playing the game could see that it was beginning to approximate 2/3 from the results, but the kid playing host was understanding what was going on.

Sasha x

[TTn]

That means that the moderator already knows which door the car is behind, and he won't open that door. Just for clarification: by being un-biased, you mean that in the given situation the door opener first just chose a door randomly (and he could as well have chosen the door with the car behind it, or the door chosen by the contestant), and then it just happened to be another door with a goat behind it?

In that case you're right; there is no inherent advantage to switch.

Right, in a "fair" game where the moderator has no special interest, the probability is equal. It was not defined in the original post, until you suggested it. I've heard tons of others, so it is a help to be specific.

[DonEppesFan]

I LOVE the Monty Hall problem now! I know this thread is super old, but I saw the "Math for Beginners" clip from the episode awhile back and finally just saw this episode, and I wanted to share my enthusiasm towards this awesome problem!

At first, I'm like no way does that make sense, but then I took out three cards: two jokers and a King. I flipped them over and told my brother to pick one. Then I flipped over the card that I knew was a joker. I asked him if he wanted to switch (knowing what he now knew), and he said he was sticking with his gut because (as it seems) "it doesn't matter." Turns out, he should have switched. I did it with my sister... she should have switched also. I then did it with my father... he should have switched as well. I didn't want to press my luck, so I put the cards away, but I was very impressed by this math. I know it was only a 2/3 third chance that you'd win if you switched, but that's a really great probability considering you start with only a 1/3 chance.

Here's a video that really explained it to me if anyone is still confused:

http://youtube.com/watch?v=mhlc7peGlGg

I would really love to see more Math for Non-Math people on the show because it seriously blows me away. I like it how it's broken down so that people who aren't particularly "math" people can feel smart in the area!

(Back to work I go... Thank goodness for rain and a slower day... :))