logic/math problem

[jwvdcw]

You're on the game show Let's Make a Deal and you’re given a choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, Monty Hall, who knows what’s behind the other doors, opens another door, say No. 3, which has a goat. He then says to you, ‘Do you want to stay with door No. 1 or switch to door No. 2?’

Should you stay or switch, or is it 50/50?

[M2d]

I've seen this before, and, if I remember correctly, people "proved" mathematically that you should switch.
that answer never sat right with me.
I think the rationale was that your original choice had a 1:2 chance of being right while your chances of being right on the second pick (after one choice has been eliminated) is 1:1.
Where I think this line of reasoning breaks down is that people don't realize that you are forced to make a 1:1 choice. If you choose to switch, you have even money to pick correctly. If you choose not to switch, you have even money to be correct. People just look at the 1:2 vs. 1:1 and forget that the real question is "is 1:1 better than 1:1?"

[aloiz]

Always switch. 2/3 of the times it'll be behind the door you didn't pick, and 1/3 of the time it'll be behind the door you picked. The way I like to visualize it is to expand the problem a bit. Say I give you a full deck of shuffled cards and tell you to pick out the As without looking. You pick a card, and then I go through the remaining 51 cards and show you 50 that are not the As. Now I give you a choice between the card you choose and the one remaining card in the deck. Which one do you think is more likely to be the As? Also notice that it wouldn't make sense that the odds of you choosing the As would change after I revealed the cards, because no matter what card you choose I'm always going to reveal 50 that aren't the As.

aloiz

[daryn]

you should always switch. anyone who says anything different doesn't understand.


to see clearer let's change the situation:

theere are now 1000 doors. 1 has a new car and 999 have goats. you pick one out of the 1000, let's say #532. now monty hall opens 998 doors and you see 998 goats.

all that's left is #532 which you originally picked, and #781.

do you switch?

[cardcounter0]

It is called Bayes Theorem

[daveymck]

the 2/3 thing is wrong the odds you have picked the correct door initially were 33.3% recurring.

One choice has been taken away therefore your odds of having picked the correct door have raised to 50%, why change or why not change it is 50-50 however you apply the maths to it, its a choice of 2 at the point of the decision.

With a choice of 1,000 it is clearer I still belive on a choice of 3 it is still 50-50.

[daryn]

what about on a choice of 4 initially? choice of 5?

it's the same. even with the 1000 door scenario, you still end up with 2 doors, why isn't it 50/50 then?

always switch!

[jwvdcw]

[ QUOTE ]
the 2/3 thing is wrong the odds you have picked the correct door initially were 33.3% recurring.

One choice has been taken away therefore your odds of having picked the correct door have raised to 50%, why change or why not change it is 50-50 however you apply the maths to it, its a choice of 2 at the point of the decision.

With a choice of 1,000 it is clearer I still belive on a choice of 3 it is still 50-50.

[/ QUOTE ]

No. If you stay, you will be right 33% of the time;If you switch you'll be right 66% of the time.

[daveymck]

Once Daryn posted the 1,000 way of looking at it I knew I was wrong but didnt want to edit my reply totally.

I do think winning a goat isnt that bad a prize providing it is a nanny. The sale of milk might make it +ev over a car depending on the car.

/images/graemlins/wink.gif

[jwvdcw]

[ QUOTE ]
Once Daryn posted the 1,000 way of looking at it I knew I was wrong but didnt want to edit my reply totally.

I do think winning a goat isnt that bad a prize providing it is a nanny. The sale of milk might make it +ev over a car depending on the car.

/images/graemlins/wink.gif

[/ QUOTE ]

lol...I'm surprised so many people see this right away. Usually, most people will adamently say that its 50-50.

[SossMan]

This same post comes up on here about once every three months. Usually after a poster or someone the poster knows has taken their first statistics class. It's a pretty famous stats puzzle.

[Beavis68]

The simple act of choosing not to choose is making a new choice, and is a 50/50 proposition.

[jwvdcw]

[ QUOTE ]
The simple act of choosing not to choose is making a new choice, and is a 50/50 proposition.

[/ QUOTE ]

/images/graemlins/confused.gif So what do you win by doing this?

[Six_of_One]

For the skeptical, there is a website out there that allows you to test this. I don't have the address, but if you do a search on google for 'Monty Hall problem' or something, it'll turn up. You can prove to yourself that switching is best.

[Mucking Idiot]

it's a freaking 50/50 shot...flip a coin...ask monteys wife...whatever...there are still 2 doors and one of them randomly will have the car.

If montey showed you the car was behind door 3, what do you think the f'ing odds of picking a goat next would be? Should you switch?

jw, i hate u for posting this as im now stupider for having read all replies. /images/graemlins/mad.gif

[jwvdcw]

[ QUOTE ]
it's a freaking 50/50 shot...flip a coin...ask monteys wife...whatever...there are still 2 doors and one of them randomly will have the car.

If montey showed you the car was behind door 3, what do you think the f'ing odds of picking a goat next would be? Should you switch?

jw, i hate u for posting this as im now stupider for having read all replies. /images/graemlins/mad.gif

[/ QUOTE ]

Ummmm..the problem specifically says that the host knows which door has the car. He then shows you a door that has a goat. He knows it will have a goat! He is not randomly picking one door. Rethink it and get back to me.

[jwvdcw]

[ QUOTE ]

lol...I'm surprised so many people see this right away. Usually, most people will adamently say that its 50-50.

[/ QUOTE ]

Thanks for proving me right mucking idiot.

[Mucking Idiot]

NEVER!!!!!!!!!!!

[jwvdcw]

[ QUOTE ]
NEVER!!!!!!!!!!!

[/ QUOTE ]

are you serious or not...please don't make me type out a lengthy explanation if you're not serious.

[Mucking Idiot]

Are YOU serious? 1 will always be half of 2...always(unless the bellboy takes some)

go ahead and try

[jwvdcw]

k..gimme a few minutes..down to 4 left in a tourney...will do it in a little bit

[cardcounter0]

NO! Don't explain it. We'll do it the scientific way.

Deal out two red cards and one black card face down.
Pick a card with the idea trying to get the black card.
jwvdcw will reveal one of the red cards.
At that point, you don't switch your original pick, and I will switch and choose the other card.
We will play this 20 or 30 times, or however long it takes, person who ends up with the black card gets $100 from the person with the red card.

You will either figure it out, or go broke.
/images/graemlins/grin.gif

[jwvdcw]

Ok...got 3rd in the tourney for $900 /images/graemlins/grin.gif

First, of all, if you really think its 50/50(even after this), then I got a proposition for you: We'll get a trusted person here to moderate this(perhaps we'll say that the winner will give him some $$ to entice him to participate a little). I'll randomly pick a number from 1-3 and tell that person. Then you choose 1,2,or 3. Afterwards, I'll tell you one number that it is NOT. Then you can either keep your initial number or switch to the other remaining number. Since you think its 50/50 to stay with your number, you stay with your number. I'll bet you $20 each time. In fact, I'll put up $25 and you only put up $20 each time! Thats amazing odds for you if it truly is 50/50.



Anyway,here you go:

Let's say you pick 1.
If the number is 1, you win.
If the number is 2, then 3 is eliminated, and I get 2, I win.
If the number is 3, then 2 is eliminated, and I get 3, I win.

Let's say you pick 2.
If the number is 1, then 3 is eliminated, and I get 1, I win.
If the number is 2, you win.
If the number is 3, then 1 is eliminated, I get 3, I win.

Let's say you pick 3.
If the number is 1, then 2 is eliminated, I get 1, I win.
If the number is 2, then 1 is eliminated, I get 2, I win.
If the number is 3, you win.


Heres another example:

Car is behind Door #1. You choose Door #1. You win by staying.
Car is behind Door #1. You choose Door #2. You win by switching.
Car is behind Door #1. You choose Door #3. You win by switching.
Car is behind Door #2. You choose Door #1. You win by switching.
Car is behind Door #2. You choose Door #2. You win by staying.
Car is behind Door #2. You choose Door #3. You win by switching.
Car is behind Door #3. You choose Door #1. You win by switching.
Car is behind Door #3. You choose Door #2. You win by switching.
Car is behind Door #3. You choose Door #3. You win by staying.


These quotes were taken from Jericho and Maurile Tremblay at another site.

[daryn]

again, i really have found that the easiest way explain this to people is to use an extreme example, like the 1000 door example i used above.

you still get down to 2 doors, but there's no way in hell it's 50/50.

[razor]

Let's Make a Deal (http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html)

have fun...

[jwvdcw]

Any reply MI?

[Mucking Idiot]

wow...i feel like a mucking idiot!
that's a really good one




















































i hate you even more now

[Mucking Idiot]

you be quiet

[andyfox]

I completely flunked this test a few years ago when it was presented here. The way I thought about it that helped:

Let's say you're on a driving range. Your friend says he's picked one of the ten thousand balls there and you have to identify it. You pick one. He now eliminates 9,998 balls and says, OK, it's either the one you picked or the other one that's left, do you want to switch?

When you picked, the ball you picked had a one in ten thousand chance of being the right one. Now he's told you that 9,998 of them are not it. So the one he's left has a 1 in 2 chance of being it. That's a lot better odds than 1 in 10,000.

[Mano]

An easy way of looking at it, is that if you picked the correct door to begin with and switch, you will loose. If you picked the wrong door to begin with and switch, you will win. Since the odds of picking the correct door to begin with are 1 in 3, the odds of winning if you switch are 2 in 3.

[daryn]

sometimes i wonder if people read my posts at all /images/graemlins/frown.gif


guess not

[razor]

Best explanation yet!

[andyfox]

Sorry, sir, just read it now. Same point as mine.

[mosta]

I didn't read all the replies, because I've been over this one (or, way) too many times, but the easiest way, I think, to see that always switching is optimal to notice that the always switching strategy is the inverse of the original 1-in-3 choice (and therefore is a 2-in-3 favorite).

to be a little formalistic, note 1. it is always possible for the host to reveal a loser. 2. once a loser is eliminated, all that is left are one winner and one loser. 3. (because of 2) switching always turns a winner into a loser and a loser into a winner. then, if your original choice was 1-in-3 of winning, being able to always switch is 2-in-3 of winning. (ie, all of your, 2-in-3, losers become, 2-in-3, winners.) QED

[jwvdcw]

[ QUOTE ]
I completely flunked this test a few years ago when it was presented here. The way I thought about it that helped:

Let's say you're on a driving range. Your friend says he's picked one of the ten thousand balls there and you have to identify it. You pick one. He now eliminates 9,998 balls and says, OK, it's either the one you picked or the other one that's left, do you want to switch?

When you picked, the ball you picked had a one in ten thousand chance of being the right one. Now he's told you that 9,998 of them are not it. So the one he's left has a 1 in 2 chance of being it. That's a lot better odds than 1 in 10,000.

[/ QUOTE ]

No, you're still not correct. The one thats left has a 9999 in 10000 chance of being correct. Just as in this problem the door you switch to has a 66% chance of being correct.

[Duke]

switch. It's pretty obvious, no?

Anyhow, this problem presents itself at the end of one of the bonus stages in a slot machine called "Atlantica."

Like, you have to pick 1 of 3 doors, and after you pick they give you a hint. They show you the "goat." I've always been switching and winning about twice as much as I lose. Anyhow, I figured if anyone worked for Atronic they could confirm whether or not that will work. Like... if it just waits until you make a final selection and then picks again randomly you have no edge either way. If I don't know, then of course I benefit to always switch because it can't hurt and might help.

And no, I don't think this makes the slot machine beatable either way.

~D

[Duke]

[ QUOTE ]
I've seen this before, and, if I remember correctly, people "proved" mathematically that you should switch.

[/ QUOTE ]

Oh god I love that you put proved in quotes.

It goes well with your "I'm just a caveman" line of reasoning.

~D

[Cubswin]

This is such a trick question. Everyone knows that you should hold out as long as you are able and then go for the box that carol marol points to. You guys are all suckers. /images/graemlins/grin.gif

cubs

[nicky g]

I know you should switch becausesome maths professor on TV said so /images/graemlins/tongue.gif. But I don't understand your explanation.

"So the one he's left has a 1 in 2 chance of being it. That's a lot better odds than 1 in 10,000. "


OK, I see this in that it's incredibly unlikely that you picked the right one. But how does this one have a 50% chance of being it. But what's happened to the other 50% probability? How can one choice have a 1/10000 chance of being right, and the other a 1/2 chance of being right, when there are only two choices? That's what confuses me. If one has a 1/2 chance of beig right, how is it that the other one doesn't? Does the other one in fact have a 9999/10000 of being right? /images/graemlins/confused.gif

[mosta]

if it helps, note that the switching process is non-stochastic, completely determinate. host always reveals an (unchosen) loser. switching always turns a winner into a loser and a loser into a winner (because there's only one of each left). so think of "switching" as merely flipping your original choice, whatever it was (winner to loser or loser to winner). Now, since two-thirds of the time you pick a loser, applying "switching" (after the host has revealed an unchosen loser) means those two-thirds become winners.

once you see a loser revealed, you know there's only one loser and one winner left, and you can switch to reverse your choice and thus invert the distribution of the original choice.

[jwvdcw]

[ QUOTE ]
I know you should switch becausesome maths professor on TV said so /images/graemlins/tongue.gif. But I don't understand your explanation.

"So the one he's left has a 1 in 2 chance of being it. That's a lot better odds than 1 in 10,000. "


OK, I see this in that it's incredibly unlikely that you picked the right one. But how does this one have a 50% chance of being it. But what's happened to the other 50% probability? How can one choice have a 1/10000 chance of being right, and the other a 1/2 chance of being right, when there are only two choices? That's what confuses me. If one has a 1/2 chance of beig right, how is it that the other one doesn't? Does the other one in fact have a 9999/10000 of being right? /images/graemlins/confused.gif

[/ QUOTE ]

Nicky G, you're right. As I responded above, andyfox is wrong about the 1/2. It would be 9999/10000 and 1/10000 in that example.

[Cyrus]

The easiest way to figure out the (goddam) Monty Hall Paradox (I have seen it been posted every month on some or other message board!) is this:

Instead of 3 doors, imagine that there are 1000 doors, behind 1 of which is the car and behind the 999 others there's nothing or just goats. Monty knows where the car is. As soon as you pick 1 door, Monty throws open 998 empty doors! That must be quite a sight, btw.

It should be obvious that you should switch. If it isn't obvious, picture in your mind several successive trials, whereby you pick 1 door and 998 others are opened, time and time again. it should dawn on you that that little un-opened door is HOT!...

[nicky g]

Thanks - sorry I missed your previous post (seems to be a lot of that going on in this thread).

[daryn]

</font><blockquote><font class="small">In risposta di:</font><hr />
The easiest way to figure out the (goddam) Monty Hall Paradox (I have seen it been posted every month on some or other message board!) is this:

Instead of 3 doors, imagine that there are 1000 doors, behind 1 of which is the car and behind the 999 others there's nothing or just goats. Monty knows where the car is. As soon as you pick 1 door, Monty throws open 998 empty doors! That must be quite a sight, btw.

It should be obvious that you should switch. If it isn't obvious, picture in your mind several successive trials, whereby you pick 1 door and 998 others are opened, time and time again. it should dawn on you that that little un-opened door is HOT!...

[/ QUOTE ]




oh come on.. now you're just trying to piss me off /images/graemlins/grin.gif

[Rick Nebiolo]

just skimmed the other replys. it is essential that you make it clear that every time you play the game monty hall will offer you the same choice. in this case the answer is clear that you should switch (since you pick the wrong door in the first place two out of three times).

about two years ago Tom Weideman made an interesting game theory extrapalation on this problem (where Monty was trying to fool you or something to that effect). someone may want to post a link (good luck searching) because it was extremely well done.

~ rick

[Patrick del Poker Grande]

Okay, let's modify it a little bit. First, let me say, I'm all for switching every time and yeah I know why it's 2/3.

Let's now say there are two contestants, each has to pick one of the three boxes, and they can't pick the same one. Of the times that they don't both pick wrong (and Monty's forced to reveal the door with the car), one will have the car and one will have the donkey. They each individually have a 2/3 shot at the car if they switch, but the end result is always that one has a vehicle for the highway and the other has a ride to the bottom of the Grand Canyon, so the results suggest they each have a 1/2 chance.

<font color="white">1/3 of the time, they'll both pick the wrong door and the above situation won't apply. So then I guess they each get half of this 1/3 and that's the difference between the 2/3 and the 1/2.</font>

Text in white.

[jwvdcw]

[ QUOTE ]


Of the times that they don't both pick wrong (and Monty's forced to reveal the door with the car), one will have the car and one will have the donkey. They each individually have a 2/3 shot at the car if they switch,

[/ QUOTE ]

I think you're mistaken because here you are not giving Monty Hall a choice. In other words, in your problem, Month Hall is forced to pick whichever door is left even if he knows that the car is there. THE ENTIRE PUZZLE IS BASED ON THE FACT THAT MONTY HALL KNOWS WHERE THE CARS IS AND HE CAN ALWAYS PICK AN EMPTY DOOR TO SHOW YOU. In your problem, each guy is 33% chance at first but then it goes up to 50% after Monty Hall reveals an empty door since Monty Hall randomly picked a door(or at least he was forced as to which door was opened) instead of opening a door he knows doesn't have the car.

[benfranklin]

The math is counter-intuitive here. The way I convinced myself that it is always right to switch was to restate the problem slightly, but with virtually the same choice and the same outcome.

Suppose you pick door 1, and Monty than says you can keep door 1, or you can have both door 2 and door 3, sight unseen, and if the car is behind either, you win. He always makes this offer, so he is not trying to get you to give up the car if he knows it is really behind door 1. You know that at least one of those doors is a loser, but you don't know which. You always switch in this situation, right?

Your options and your odds in the case where all the doors are closed are the same as in the original problem. Monty is not giving you any more useful information and is not changing the odds by by opening one of the doors. Before he opens the door, the odds are 2-1 that the prize is behind one of the doors you didn't pick. Opening a door doesn't change those odds. You already knew that one of them was a loser. You are still essentially trading one door for two.

[uw_madtown]

[ QUOTE ]
The math is counter-intuitive here. The way I convinced myself that it is always right to switch was to restate the problem slightly, but with virtually the same choice and the same outcome.

Suppose you pick door 1, and Monty than says you can keep door 1, or you can have both door 2 and door 3, sight unseen, and if the car is behind either, you win. He always makes this offer, so he is not trying to get you to give up the car if he knows it is really behind door 1. You know that at least one of those doors is a loser, but you don't know which. You always switch in this situation, right?

Your options and your odds in the case where all the doors are closed are the same as in the original problem. Monty is not giving you any more useful information and is not changing the odds by by opening one of the doors. Before he opens the door, the odds are 2-1 that the prize is behind one of the doors you didn't pick. Opening a door doesn't change those odds. You already knew that one of them was a loser. You are still essentially trading one door for two.

[/ QUOTE ]

While the 1000 doors example is convincing, I think this is the best way to explain this problem to someone and have them instantly "get" it.

- UW

[Cyrus]

"now you're just trying to piss me off"

I had not read your response when I put up mine.

Sorry to have caused you a number one. Hope that you were at least crummily attired. /images/graemlins/cool.gif

[Jman28]

Here's my attempt to make everyone (who doesn't already) understand why it is not 50/50 after Monty takes one away...

You pick a door. You have a 33.33% chance of hitting the car. THIS DOES NOT CHANGE when he shows you a goat, because HE ALWAYS SHOWS YOU A GOAT.

Then you see the door he opens. This door has a 0% chance of having a car behind it.

If the remaining door had a 50% chance of having the car behind it, than there would be a 16.66% chance that the car wasn't behind any doors.

100%-33.33%-0% = 66.66%

[NotReady]

This thread is too long to see if anyone else said the following:

You have a new choice to make. If you stay with D1 it's a new choice, not affected by the previous choice and you have a 50% probability. If you choose D2, same thing.

[NotReady]

Ouch. After I posted this and went to bed, couldn't sleep, then realized you should switch.

Then read daryn's post and found out why.

So, I'm changing my vote.