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Let's Make A Deal

So you are on the old game show Let's Make a Deal. Monty Hall offers you the pick of door #1, #2, or #3 and says a brand new car is behind one of the three doors and the other two have a goat behind them. You pick a door. Monty doesn't reveal your door but opens one of the two other doors and there is a goat behind it (since there is only one door with a car, one of the unpicked doors has to have a goat and Monty know which one). Monty then asks you if you want to keep your door or switch to the other still-unrevealed door.

What do you do?

Does it make a difference?

Why?

Do yourself a favor and try to reason it out before you try to look it up. The answer is documented online in many places but that's cheating.
I'd stick to the door I've chosen at first.
It would seem that odds were increased from 33% to 50% chances of winning with the original door when the goat behind the other door was revealed. Not really.

The original door still only has a 1:3 (33%) chance that the car is behind it, so you're not improving your odds by keeping it. The other 2 doors had a combined chance of 2:3 (66%) that the car was there originally. Revealing the goat behind one doesn't change the original chance, but it does change the number of doors involved with that chance. So instead of 2 doors with 66% chance of having a car behind one or the other, it's 66% chance that the remaining door has the car. So switch the doors, you'll be more likely to win the car. I'm bored, let's do more math!!
(In some ways you can also apply this theory to that ridiculous new greed show with Howie Mandel... don't know the name... but in the process of lusting for more money, the contestants forget their basic math skills. Kind of funny).

Seems the most important question to ask Monty, though, is what color the car is. If it's not red, I'm taking the goat. . .

. . . get your mind out of the gutter!
Quote:
 The original door still only has a 1:3 (33%) chance that the car is behind it, so you're not improving your odds by keeping it. The other 2 doors had a combined chance of 2:3 (66%) that the car was there originally. Revealing the goat behind one doesn't change the original chance, but it does change the number of doors involved with that chance. So instead of 2 doors with 66% chance of having a car behind one or the other, it's 66% chance that the remaining door has the car. So switch the doors, you'll be more likely to win the car. I'm bored, let's do more math!!
I've heard this before and its some of the worst logic I've ever heard. All you know is there is a goat behind one door and a car behind the other. The odds are now 50/50. Switching isn't going to do you any good.
Quote:
 Originally Posted by Rio I've heard this before and its some of the worst logic I've ever heard. All you know is there is a goat behind one door and a car behind the other. The odds are now 50/50. Switching isn't going to do you any good.
I don't agree and I'll scrap it out with you.

Ok, so change it to 30 doors instead of 3. And say you picked one out of those 30 doors originally. That's 1:30 chance there's goodies behind this one. The other doors combined have 29:30 chances the goodies are behind them, right?

Then let's say you're allowed to open 28 more doors out of the 29 that are left. You find no goodies there, either. The one door left out of that 29 is probably where the goodies are, not behind the one door you picked, which still has odds of 1:30. The probability that the goodies were there in the first place - among the 29 doors you DIDN'T pick - is what matters. It's not logical and it's also not 50/50.
I just went and read the explanation. The odds are 2/3 that the remaining door has the car.
That's pretty convoluted logic. I haven't read the solution, so I can not point out exactly where their error is, but I am sure they have made an error. The probability is no longer the probability of a car being behind one of three doors, but the probability of a car being behind one of three doors given that it is not behind ONE of the other doors.
chances of A =1
Chances of B = 1
Chances of C = 1
Total chances =3
probability = chances/total chances
PA=1/3
PB=1/3
PC=1/3
Probability of door number one = 1/3

P not C = 2/3 is a red hearing; once he shows you the door it is 1
so PA given not c is 1/3 /(1/3 +1/3) = 50%
"So if group B had 2/3 of the chances at the start, it *still has* 2/3 of the chances, but those all ride on the one remaining door in group B. So, the remaining door in group B has a 2/3 chance fo winning, whilst the door you originally picked, in group A still has only a 1/3 chance."
"still has" my donkey!
That's like betting a coin will turn up heads on the second toss because it turned up tails on the first toss.
The chances are still 50/50.
The second choice is not dependent upon the first choice.
The odds are reset each time a choice is made.
It's just like babies if you already have two boys there is still a 50/50 chance your next kid will also be a boy.

Maybe it's time for a refresher staistics course
http://www.dartmouth.edu/~chance/tea...book/book.html
Quote:
 Originally Posted by UTpowder The chances are still 50/50. The second choice is not dependent upon the first choice. The odds are reset each time a choice is made. It's just like babies if you already have two boys there is still a 50/50 chance your next kid will also be a boy. Maybe it's time for a refresher staistics coursehttp://www.dartmouth.edu/~chance/tea...book/book.html
I agree. Once Monte reveals the goat behind one of the doors, you are no longer playing with three doors, but two. It's no different than if he just said, "pick door a, b or c, but there's a goat behind door c". Door c is no longer in the equasion, and you are picking between doors a and b...50/50.

The logic of it still being 1/3 is like playing the lotto, with odds of 1:10,000,000, but the person running the lotto pulled the numbers so that he knew the winning numbers, and said it's either a or b (replace a and b with two sequences of numbers), and you get to pick one. Your odds are no longer 1:10,000,000, but 1:2. Your logic says the odds are still 1:10,000,000.
Quote:
 Originally Posted by UTPowder The chances are still 50/50. The second choice is not dependent upon the first choice. The odds are reset each time a choice is made. It's just like babies if you already have two boys there is still a 50/50 chance your next kid will also be a boy.
Quote:
 Originally Posted by JohnH I agree. Once Monte reveals the goat behind one of the doors, you are no longer playing with three doors, but two. It's no different than if he just said, "pick door a, b or c, but there's a goat behind door c". Door c is no longer in the equasion, and you are picking between doors a and b...50/50.
Um, no.
Comparing the probability with babies, coin tossing, or anything else that is a single unique event is entirely different. There, it is 50/50 each time because you are re-setting the odds with each "toss". But this game involves an established history of probability. The only way for it to be 50/50 odds would be for Monty to switch up the game and re-distribute what's behind each door , behind two new doors.
What really makes the probabilites favor switching is that Monty always opens a door with no (or small) prize after you select. Monty knows where the big prize lies. If you pick the door with a big prize (1/3) then Monty opens up a second door with no big prize. If you switch, you get the third door with no big prize. If you don't, then you win. IF your strategy is to switch always, then you have 1/3 chance of a bad decision. If your strategy is never to switch, then you have 1/3 probability of a good decision.

Alternatively assume you have picked a door with no big prize (2/3 likelihood). If you strategy is always to switch, then you win the big prize with 2/3 likelihood, but if your strategy is never to switch you have 2/3 probablility of not getting the big prize.

It is not 50% on the second selection, because you know with certainty if the prize were behind a door you hadn't chosen (2/3 likelihood), that the prize lies behind the remaining door. You also know with certainty that if the prize were behind the door you had chosen (1/3 likelihood) that the prize still remains behind that door.

What could change this would be if Monty selected the first door to open randomly, and one third of the time the prze were behind the first door opened.
Most people think the probabillity of winning after one door has been revealed is either 1/2 and it doesn't make a difference if you switch, or 2/3 if you switch and 1/3 if you don't switch.

The correct answer is that you should switch. The probability then becomes 2/3 that you win.

The easiest way to look at this is to assume you will switch. You pick a door. There is a 2/3 chance the original door you pick is wrong. Now if you switch, the only time you lose is if you originally pick the right door. Therefore, if you switch, the probability is 2/3 that you win.

There is a thorough treament of what is know as the "Monty Hall Problem" or "Monty Hall Paradox" at wikipedia. See http://en.wikipedia.org/wiki/Monty_hall_problem
Most people think the probabillity of winning after one door has been revealed is either 1/2 and it doesn't make a difference if you switch, or 2/3 if you switch and 1/3 if you don't switch.

The correct answer is that you should switch. The probability then becomes 2/3 that you win.

The easiest way to look at this is to assume you will switch. You pick a door. There is a 2/3 chance the original door you pick is wrong. Now if you switch, the only time you lose is if you originally pick the right door. Therefore, if you switch, the probability is 2/3 that you win.

There is a thorough treament of what is know as the "Monty Hall Problem" or "Monty Hall Paradox" at wikipedia. See http://en.wikipedia.org/wiki/Monty_hall_problem
Quote:
 Originally Posted by learn2turn There is a thorough treament of what is know as the "Monty Hall Problem" or "Monty Hall Paradox" at wikipedia. See http://en.wikipedia.org/wiki/Monty_hall_problem
forgot how much fun (not to mention convulted) game theory is
that's a great write-up in wikipedia!

jinx
So if I had chosen the other door to begin with I should switch to the door I now have:
Using their logic......

Right door 1/3
Left and middle door 2/3
Choose left door.
Middle door is revealed not to have a prize.
Odds of it being the original door is now 2/3 Don't switch!

middle door 1/3
outside doors 2/3
choose left door
right door revealed not to have a prize.
odds of prize being on left door now 2/3. Don't switch!

Hmm, it seems that if you choose the left door you are better off not swithching.:
Quote:
 Originally Posted by Ghost Using their logic...... Right door 1/3 Left and middle door 2/3 Choose left door. Middle door is revealed not to have a prize. Odds of it being the original door is now 2/3 Don't switch! middle door 1/3 outside doors 2/3 choose left door right door revealed not to have a prize. odds of prize being on left door now 2/3. Don't switch! Hmm, it seems that if you choose the left door you are better off not swithching.:
.... but that's not the logic involved here. Draw it out, you'll see:
Three doors with 1/3 odds each
1/3
1/3
1/3

Pick one (for example, the top one)
1/3
1/3
1/3

Overall odds of the other group of two don't change.
1/3

} 2/3 (for group)

Now eliminate one when the goat is revealed, there are still 2/3 odds with that grouping, even though there is only one in that group. It's not which door you pick in the first place that matters, it's the original odds associated with that door that matter.
1/3

} 2/3 (for group - door not picked has the best odds now)
X
I always hated the Parade answer and I don't like the Wiki answer any better. The way I see it there are 24 possible scenarios:
Prize behind door1, you pick door 1, monty reveals door 2, you keep=you win
Prize behind door1, you pick door 1, monty reveals door 2, you switch=you lose
Prize behind door1, you pick door 1, monty reveals door 3, you keep=you win
Prize behind door1, you pick door 1, monty reveals door 3, you switch=you lose
Prize behind door1, you pick door 2, monty reveals door 3, you keep=you lose
Prize behind door1, you pick door 2, monty reveals door 3, you switch=you win
Prize behind door1, you pick door 3, monty reveals door 2, you keep=you lose
Prize behind door1, you pick door 3, monty reveals door 2, you switch=you win
If you put the prize behind doors 2 and 3 and repeat the exercise, you get 8 more outcomes apiece identical to the above 8 (4 scenarios for picking the right door and 4 for picking the wrong door) to total 24 outcomes, with 12 switches winning and 12 losing.
That makes the odds 50-50.
Trouble,
Notice the: in my post.
I am using their logic, except that they CHOSE the group of two to be the two you did not pick, and I CHOSE the group of two to be mine and any other door. It doesn't depend on what door you picked. I think there logic is flawed.

It depends only on total outcomes, as the rusty explained above.
Quote:
 Originally Posted by Ghost Trouble, Notice the: in my post. I am using their logic, except that they CHOSE the group of two to be the two you did not pick, and I CHOSE the group of two to be mine and any other door. It doesn't depend on what door you picked. I think there logic is flawed. It depends only on total outcomes, as the rusty explained above.

there is no CHOOSE in math
combining the groups is trivial and ineffective, regardless how you do it.
if the probability is 1/3 for each door {1,2,3}, then P(1 or 2)=2/3. ok, but also P(2 or 3)=2/3 and P(1 or 3)=2/3. so what?
Quote:
 Originally Posted by therusty I always hated the Parade answer and I don't like the Wiki answer any better. The way I see it there are 24 possible scenarios: Prize behind door1, you pick door 1, monty reveals door 2, you keep=you win Prize behind door1, you pick door 1, monty reveals door 2, you switch=you lose Prize behind door1, you pick door 1, monty reveals door 3, you keep=you win Prize behind door1, you pick door 1, monty reveals door 3, you switch=you lose Prize behind door1, you pick door 2, monty reveals door 3, you keep=you lose Prize behind door1, you pick door 2, monty reveals door 3, you switch=you win Prize behind door1, you pick door 3, monty reveals door 2, you keep=you lose Prize behind door1, you pick door 3, monty reveals door 2, you switch=you win If you put the prize behind doors 2 and 3 and repeat the exercise, you get 8 more outcomes apiece identical to the above 8 (4 scenarios for picking the right door and 4 for picking the wrong door) to total 24 outcomes, with 12 switches winning and 12 losing. That makes the odds 50-50.
the door number/order is irrelevant, basically this is what you said in terms of goats and a car (keeping the same order as you):

pick car, monty shows goat1, you keep => you win
pick car, monty shows goat1, you switch => you lose
pick car, monty shows goat2, you keep => you win
pick car, monty shows goat2, you switch => you lose

pick goat1, monty shows goat2, you keep => you lose
pick goat1, monty shows goat2, you switch => you win

pick goat2, monty shows goat1, you keep => you lose
pick goat2, monty shows goat1, you switch => you win

BUT!!!!
the initial probability of picking the car (first group above) is 1/3
the initial probability of picking goat1 (second group) is 1/3
the initial probability of picking goat2 (thirs group) is 1/3

now lets say you always keep - you will win ONLY if you pick the car first - 1/3 chance
if you always switch you will win only if you pick a goat first (any goat though, it's the interchangeable goats that make this problem ) - so if P(goat1) = 1/3 and P(goat2) = 1/3 then P(either goat) = 2/3

jinx
Quote:
 Originally Posted by jinx there is no CHOOSE in math combining the groups is trivial and ineffective, regardless how you do it. if the probability is 1/3 for each door {1,2,3}, then P(1 or 2)=2/3. ok, but also P(2 or 3)=2/3 and P(1 or 3)=2/3. so what?
So by considering all the possible outcomes, and the outcome that makes you win you have the proper math that gives you the real probability of success, but by "grouping" the two doors you can fool a lot of people into believing whichever outcome you choose to "prove" more likely is more likely, just by the way you group them.
Quote:
 Originally Posted by Ghost So by considering all the possible outcomes, and the outcome that makes you win you have the proper math that gives you the real probability of success, but by "grouping" the two doors you can fool a lot of people into believing whichever outcome you choose to "prove" more likely is more likely, just by the way you group them.
any person who pays attention and knows some basic math would catch on to the sort of arbitrary grouping you're referring to.

yes, this problem allowes (in the end i guess requires even) a grouping, but not an arbitrary grouping of doors. the only grouping that is relevant here is the group of goats. why? because they are identical (mathematically) and interchangeable (biologically )

jinx
When the goat is revealed, it increases the probability that the car is in the door you chose just as much as it increases the probability that the car is in the other door you didn't choose.

Play the game 900 times. Each door gets a car 300 times. If the car is revealed not to be behind door number X, you are now dealing with the other 600 outcomes, 300 of which are behind each of the remaining doors.
Quote:
 Originally Posted by Ghost When the goat is revealed, it increases the probability that the car is in the door you chose just as much as it increases the probability that the car is in the other door you didn't choose. Play the game 900 times. Each door gets a car 300 times. If the car is revealed not to be behind door number X, you are now dealing with the other 600 outcomes, 300 of which are behind each of the remaining doors.
wrong . how do i explain this? read the write-up on wikipedia, it really is quite good.

the fact that a goat is revealed has no effect on "the probability that the car is in the door you chose" any more then it increases the probability that the door you chose contains a goat. the initial probability of choosing the car is 1/3, just liike the initial probability of choosing a goat (any goat) is 2/3.

i don't need to play the game 600 times to know i'm right, this problem has been proved in various ways (there are quite a lot of variations of this problem, personally i was familiar with this one http://en.wikipedia.org/wiki/Three_Prisoners_Problem ).

if you need to prove it to yourself, try it with cards (or try to find some goats ) and let me know how it worked for you.

Gamblers Fallacy

I've been lurking in the background for a couple of years satisfied to learn from the great exchanges between the menbers of this great site. However on this subject I had to chim in.

In Philosphy the problem being discussed is the "Gamblers Fallacy". Deep down we all believe things should react to odds as pridicted. The common use for the Gamblers Fallacy is the roulette table were people will wait for red or black to come up a number of times in a row and then bet heavily on the other, believing it has a great chance of coming up.

The problem of odds is that past occurences have no impact on the future.

As an example the odds of flipping heads 10 times in a row, before you start are roughly 1 in 1000, the odds of flipping heads 11 times in a row is roughly 1 in 2000. After you've flipped the heads 10 times in a row the odds of flipping the next head are 1 in 2, because that is the odds of any single flip.

When you chose the first curtain the odds where 1 in 3. When the goat was displayed the odds became 1 in 2 if you choose the new curtain the odds return to 1 in 3.
Quote:
 Originally Posted by wpg_ski_nut The problem of odds is that past occurences have no impact on the future. As an example the odds of flipping heads 10 times in a row, before you start are roughly 1 in 1000, the odds of flipping heads 11 times in a row is roughly 1 in 2000. After you've flipped the heads 10 times in a row the odds of flipping the next head are 1 in 2, because that is the odds of any single flip. When you chose the first curtain the odds where 1 in 3. When the goat was displayed the odds became 1 in 2 if you choose the new curtain the odds return to 1 in 3.
this is not a case of independent events, it's conditional probability.
as for the changing odds: your odds of picking the car are 1/3, they don't change because he shows you a goat behind another door (you already know there has to be a goat behind one of the other doors regardless of whether you chose the door with the car or a goat)

jinx
Sold!
Monty's choice of which door to open is the key. He is not leaving 600 choices. He is in effect, by CHOOSING which door to open, leaving 900, 600 of which are in the remaining door.
Learn something everyday.
Your odds of picking the right door originally were 1 in 3. Just like the odds of flipping heads twice is 1 in 4.
After one wrong door has been eliminated it doesn't change the odds that existed when you originally chose, like flipping 1 head doesn't change the original odds of flipping 2.

However as soon as one of the doors is eliminated the remaining odds are 1 in 2. Just like after the first head the odds of the next are 1 in 2.

Another way to look at it is if after the first door is open Monty lets you choose again from the remaining 2 doors, because you get to choose most people believes that the odds would be 1 in 2. However choice has nothing to do with the odds, the car is behind 1 of the 2 remaining doors, choice doesn't change the odds.

Finnaly after his first offer Monty opens up the other door you didn't choose and it has a goat. The odds of your door now having a goat are 1 to 1 any other odds would imply that you hadn't won the car, and you have. Even though your odds at this point are now 1 to 1 it doesn't change the fact that when you first made your choice the odds were 1 to 3.

A good definition of insanity is doing the same thing repeatadly and expecting a diferent outcome.
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