Maths challenge No. 3

[Duckers]

If you can convince anyone that an initial selection out of a million doors is a 50-50 chance, please provide a mathematical proof.


Sure, but you tell me first; are we assuming that Monty is going to eliminate 999,998 goats?

Because in the first problem, we were told that Monty was going to eliminate one of the goats.

 

[allan120]

Assume he opens 999,998 doors that concealed goats. This will bring the situation down to two "theoretically 50-50" options (according to your argument), door 1 and door 2, much like the three-door example after one door is opened.

Now where's that mathematical proof?

1:1,000,000 = 1:2 (theoretically, according to Duckers)

 

[Xede]

cxim--I've just been at class today. I'm still here.

You still holding to your bet? Just $10, not $100 or $200. I only mentioned the bet to show how much faith I had that I was right, but if you're feeling guilty... I can take paypal

If you want "proof" and don't accept any of the arguments from this thread, I can get a TA or maybe even the professor from 6.041, MIT's intro probability class, or I can email you a PDF file from that class that answers this exact problem.

I'm really surprised that anybody can still not be convinced after reading ALL these posts carefully.

 

[CJM]

I guess i'm a little late to this thread but how about thinking about it like this -
There are 1000 doors. 1 car and 999 goats. You pick a door. Now what are chances that you picked the car? 1 in 1000, not very good, right? So now the host will show you 998 goats. The only two doors that they CANNOT show you are the one you picked and the one containing the car. There are now two doors to choose from. So either the the one you picked originally (still at 1000 to 1 odds) is the car, or the car is behind the other door. Get it? It's the same idea with 3 doors but on a smaller scale it's harder to see.

 

[Argo]

I think I got it. You guys who were claiming that switching is better are right. Cheers.

 

[Duckers]

Switching might be better

You either switch or you don't, so there is a 50/50 chance of winning that CAR independently from the other events!!!!

I am out of this thread!

 

[mahpoh]

eeerrr.. ducky... may i ask how old you are?

i am 22 and i see that switching the door would be the best thing to do.

 

[Duckers]

double post.

 

[Duckers]

Switching the door might be the best thing to do ( Ok, no doubt about it ) but that doesn't guarantee you will win the car anyway.

How old am I?
3 years old

You aren't talking to any stupid, FYI

I passed Calculus AP with a score of A+, a 5 on the Calculus AP test and a 740 on the Math SAT.

Again, you guys are the ones who don't want to see my analogy.

 

[Handle]

I posted something similar to this on the 1st page, but it seems I should repeat it.

Your chances of picking a GOAT on your INITIAL PICK is 2 in 3, correct? Everyone agree? Good.

Now, if you INITIALLY pick a GOAT, then Monty will reveal the ONLY OTHER GOAT. This means that if you SWITCH you have a 100% chance of getting the CAR. If you DO NOT SWITCH, the chances of you getting a GOAT are 100% (because the beginning of this paragraph explicitly stated you picked a GOAT INITIALLY). The chances that your INITIAL PICK is a goat is 2 in 3, you already agreed to that.

Therefore, the chances of getting a CAR if you SWITCH are 2/3. The calculation is 2/3 x 1/1 = 2/3.

Your chances of picking the CAR on your INITIAL PICK is 1 in 3, correct? Everyone agree? Good.

Now, if you INITIALLY pick the car, then Monty will reveal ONE OF THE GOATS. This means that if you SWITCH you have a 100% chance of getting the GOAT (because you have INITIALLY chosen the CAR, if you SWITCH you are abandoning it) If you DO NOT SWITCH you will receive the CAR, EVERY TIME. However, you already agreed that the chance of picking the CAR INITIALLY is 1 in 3. 1/3 x 1/1 = 1/3.

Therefore, the chances of getting a CAR if you DON'T SWITCH are 1/3.


If you can spot an error in the logic, please explain.

 

[allan120]

Okay, Duckers finally qualified his 50/50 stuff with "independently" or something to that effect. Whee.

Btw, this isn't covered in calculus or the math section of the SAT.

 

[Duckers]

(because the beginning of this paragraph explicitly stated you picked a GOAT INITIALLY


Well ok, I am not arguing about the 2/3 chance to pick the car any more.
(no need to ask Mr Math, an MIT professor or write a program ).

I am talking about an independent event here and you are the ones who refuse to see what I am trying to say!

 

[Pretender]

Unless the following is incorrect, the correct answer is: It doesn't matter if you switch or not

Why?
No matter what door the contestant chooses, there will always be at least 1 other door with a goat behind it. Monty WILL DEFINATELY open the door (or one of the doors, if you chose the door with the car) with the goat behind it. All this means is that there are 2 unopened doors left: 1 with a goat behind it, one with a car behind it. Your initial choice means absolutely nothing (I find it amusing when websites put the main thoughts in bold to get the point across, it makes me laugh ), because regardless of what you initially chose, there will always be 2 unopened doors remaining, one with the car behind it, and one with the goat.

Comprende?

 

[Duckers]

Again Handle,

We don't know whether you are going to pick the car or the goat INITIALLY.

Whatever you pick, Month will show the 2nd goat.


And just let it die guys, it all depends on how you want to think about it.

Chances of picking the car from the beginning
1/3

Chances of picking the car after 1 goat is out. ( the whole probability to pick the car )
2/3

Chances of picking the car after 1 goat is out ( think of this as an independent event)
1/2

Handle, can you prove that the last statement is wrong?
if so, please provide mathematical proof

 

[jhu]

is there a programmer in the house? someone could easily write a program that simulates this problem 10000 times to see the actual results. i'm too lazy to write one right now. anyone? oops, guess someone already did that.

 

[Duckers]

Pretendiente
Yo si comprendo.
Ellos no comprenden!

 

[Handle]

Duckers, I'm not saying that last statement is wrong. I am saying, however, that it is irrelevant.

The problem with using analogies or going out of order on a question like this is that it doesn't mirror the real situation. The chance of picking a car between a choice of two doors, one containing a car, the other a goat, is indeed 1/2. However, the gameshow does not offer that situation as an independent event... that situation only arises after initial circumstances and actions.

Your last statement is not wrong, but it has nothing to do with the question posed. You yourself even wrote: "( think of this as an independent event)" The problem is, in the question, it is not independent. The door that Monty opens is not determined by chance, it all depends on your initial choice. Everything is linked together, making something independent is an oversimplification.

 

[Argo]

For all those who are still in doubt look at cjm's example. It explains the whole situation the best. Just think about that. If you chose 1 door out of million, every time monty opens the remaining 999,999 doors your chances don't go up. The thing that confuses everybody, is that when two doors are remaining each has 1/2 chance of having the goat. No, that is not true. The first door has 1/n chance (where n is number of doors), while the second door has (n-1)/n chance. Try thinking about it this way: the second door represents all the other doors that monty opened. Thus when you open it, you're actually opening n-1 doors, or two in our example. Therefore your chances are 2/3. Fhew, I don't believe it took me this long to figure it out.

 

[Duckers]

Handle,

You were able to understand what I was trying to say and that's something I admire!

I liked the way you worded those two last paragraphs, and as you said; the 1/2 situation is not relevant to the whole situation, but it holds true as an independent event. ( which is exactly what I was trying to point out since yesterday )

 

[CJM]

so is there anyone left that still needs to be converted ?

 

[Dexion]

How many chances do we have here? Answer = 2. How many doors do we have? Answer = 3.

Chance 1 = 1/3
Chance 2 = Chance 1 + 1/3 = 2/3

Therefore theres a 66.66666666% odds you can win the car.
Most people are thinking that the first try you would definately get the goat, if that was the case then its 1/2. However, since there is a chance you can get the car of the first chance, then this is not an irrelevant value and ties directly to the 2nd chance when you do pick a goat. If you don't take the 2nd Chance, its 1/3 probability.

Theres actually a much easier way to calculate this.
Number of chances divided by the number of doors.

 

[jmcoreymv]

I completely understand that switching is better, im just curious to see the code that was written to simulate it, mind posting?

 

[Moonbeam]

Yep, I'm back to saying it's 50 50. Everybody says that for one iteration say GGC if you pick door one he shows you door two so you switch and get a car, or if you pick door two he shows you door one and you switch and get a car, and only if you pick door three and switch do you get a goat, two cars to one goat, but in the third case he can pick either door one or door two so there are two ways to get a goat, NOT JUST ONE. Therefore it's two cars to two goats.

If you want tp do a million doors and switch properly, analogously, you would have to eliminate one door at a time and switch and add up all the times you got a car and got a goat one at a time from a million down to two. Eliminating a million doors in one crack is not the same as eliminating one of three doors. We are back to the simple fact that no matter what door you choose, another door will be eliminated, leaving you with two doors and a fifty fifty chance of being right. You have been leaving out the forth way to get a goat, the fact that if you choose a car Monty has two choices of doors to open, not just one possibility.

 

[Viper GTS]

Are you all forgetting that Marty could possibly open the door with the car? Everyone seems to be assuming that he will automatically choose a door with a goat behind it, but that's not the way the problem is presented. In this particular case, he happened to open a door with a goat. The switching method depends on the intial condition of him NOT opening the door with the car. If you've survived the initial cut, then you can boost your odds by switching.

This thread really is rather comical, I've never seen such heated discussion over a math problem.



Viper GTS

 

[Moonbeam]

Here are the possibilities.


G1G2C= G1switch=C G2switch=C Cswitch=G1 or Cswitch=G2 =C,C,G1,G2

G1CG2= G1switch=C Cswitch=G1 or Cswitch=G2 G2switch=C =C,G1,G2,C

G2G1C =C,C,G2,G1

G2CG1 =C,G2,G1,C

CG1G2 =G1,G2,C,C

CG2G1 =G2,G1,C,C

Fifty% goats and fifty percent cars

What has been left out of the reasoning is that monty has two choices of doors if you pick the car and you 2/3 people are only considering one of them.