statistics problem

[Whatever 1]

The three doors dilemmaSo you think you understand gambling odds, eh? Well, try this...Imagine you're on a game show, and you're given the choice of three doors. Behind one door is a car; behind each of the other two, a goat. You pick a door, say No. 1, and the host opens one of the remaining two doors, say No. 3, to reveal a goat. He says to you, "Do you want to stick with the door you've chosen, or do you want to change your mind and go for door No. 2 instead?"Question: Is it to your advantage to switch your choice of doors?

[fmlycar 0]

gonna go w/ no...........your first pick had a 33% chance to be right, now that one door is eliminated you have a 50% chance to be right.???

[celtic020 0]

No

[bdc30 0]

Question: Is it to your advantage to switch your choice of doors?

I don't see how it would make any difference.A 33% chance just turned into a 50% chance,but I don't see how changing doors would matter.(as long as you're not Ron Mexico, and actually ASKto switch to door 3 to get the goat that is...)

[Whatever 1]

Hello. You are wrong. Do you see why?

[bdc30 0]

Do you see why?  

What are you Aseem now?!

[JBradburn6 0]

The three doors dilemmaSo you think you understand gambling odds, eh? Well, try this...Imagine you're on a game show, and you're given the choice of three doors. Behind one door is a car; behind each of the other two, a goat.  You pick a door, say No. 1, and the host opens one of the remaining two doors, say No. 3, to reveal a goat. He says to you, "Do you want to stick with the door you've chosen, or do you want to change your mind and go for door No. 2 instead?"Question: Is it to your advantage to switch your choice of doors?

Yes it is, but I forget why. My friend's brother told me about doing this exercise at the Wharton School of Business, but I forget the reasoning behind switching.

[screech 0]

Question: Is it to your advantage to switch your choice of doors?

It depends if the host knows what door the car is behind and whether or not it is better for him if you don't win the car. It also depends on whether or not the door he always opens one of the doors you never choose, and gives you a choice to switch.If it is good for him that you don't win the car, you should stick with the door you chose, if he doesn't always off a choice to switch.If you chose a door with a goat behind it, he would simply open that door, and you lose.If you pick the door with the car behind it, he will open any one of the remaining 2 doors, and give you the choice, hoping you change doors.If he always offers the choice to switch, you should switch.You will pick the door with a goat behind it 67% of the time. Therefore, when he opens one of the doors you never choose and shows the goat, the other door must be a car. So 67% of the time, the other door is the car. 33% of the time, the car is behind the door you initially picked.

[screech 0]

What do I win?

[econ_tim 0]

there was a topic on this four months agoswitch doors

[martinmc 0]

It dosen't matter of you switch, you have a 50% chance. The original odds don't matter because you now have a new choice, door 1 and door 2. it is the same as if there never was a door 3. Unless there is some information we don't have its 50 -50. If it were a farmer on the show and somehow knows that goats can't be put side by side then the new information would mean something, otherwise its 50 -50.

[RhinestoneCowboy 2]

What do I win?

A goat

[dead money 1]

It dosen't matter of you switch, you have a 50% chance. The original odds don't matter because you now have a new choice, door 1 and door 2. it is the same as if there never was a door 3. Unless there is some information we don't have its 50 -50. If it were a farmer on the show and somehow knows that goats can't be put side by side then the new information would mean something, otherwise its 50 -50.

Only stupid people are this closeminded.

[screech 0]

What do I win?

A goat

[econ_tim 0]

It dosen't matter of you switch, you have a 50% chance. The original odds don't matter because you now have a new choice, door 1 and door 2. it is the same as if there never was a door 3. Unless there is some information we don't have its 50 -50. If it were a farmer on the show and somehow knows that goats can't be put side by side then the new information would mean something, otherwise its 50 -50.

If you stick with your original choice, the odds of being right on your first pick are 1/3rd. Go from there.

[sholden 0]

Of course it matters if you switch. And of course you should switch.If you picked a door without the car then switching will win you the car and not switching will win you the goat.If you picked the door with the car then switching will win you the goat and not switching will win you the car.Since there were three doors to start with, there is a 33% chance that you picked the car.So if you switch you have a 66% chance of winning the car, if you don't switch you have a 33% chance of winning the car. Hence, you switch. Unless you prefer goats of course.[update - no they don't...]But this is just obvious. If you switch you win if you didn't pick the car to start with, and obviously you had a 66% chance of not picking the car. What sort of retard can't see that?

[martinmc 0]

You have a new choice. A 50 - 50 choice. with you first choice you had a 33% chance to win. Now that you know its not door 3 you have a new choice between door 1 and door 2. one has a car, one has a goat. so in your new choice you have a 50% chance to win. if door 3 had a car behind it you would have a 0% chance to win a car and 100% chance to win a goat.

[screech 0]

You have a new choice. A 50 - 50 choice. with you first choice you had a 33% chance to win. Now that you know its not door 3 you have a new choice between door 1 and door 2. one has a car, one has a goat. so in your new choice you have a 50% chance to win. if door 3 had a car behind it you would have a 0% chance to win a car and 100% chance to win a goat.

It's not 50/50 if they always offer you a new choice.There's a 67% chance you pick a door with a goat. Therefore, when he opens the door to another goat to give you a second chance, there is a 67% chance that the car is behind the door.

[shedrackle 0]

THE MATHEMATICAL PROOF:At the beginning of the game, the probability of the contestant's choosing the correct door is 1/3, since there are three doors and they're all the same. In other words, for X = {1,2,3}, the probability that Door X contains the car is P(Door X) = 1/3 (1)or, writing the expression in terms of the probability of each door's having the good prize: P(omega) = P(Door A) + P(Door B) + P(Door C) (2) = 1/3 + 1/3 + 1/3 = 1Where omega is the summation of possible events, and P(omega) is always equal to unity (from probability theory). Note that "P(omega)" can be thought of as "The probability that the good prize lies somewhere within the system of doors."After the host opens one of the unchosen doors (assume he opened Door C), equations (1) and (2) no longer hold. We know for certain that the good prize still lies behind one of the remaining doors, so we can write P(omega) = P'(Door A) + P'(Door B) = 1 (3)but 1/3 + 1/3 != 1so either P(Door A) != P'(Door A) (4)and/or P(Door B) != P'(Door B) (5)are true.Now, when the host opens one of the unchosen doors, he conveys no new information about the chosen door. This is true because for any particular door chosen by the contestant, the host can open at least one other door, effectively removing it from the game. Therefore, the probability that the car lies behind the chosen door is constant. If we assume that Door A was chosen, we can write P(Door A) = P'(Door A) = 1/3 (6)and P(omega) = 1/3 + P'(Door B) (7)(If it is difficult to accept equation (6), remember that the contestant chose the door from a field of three. If he decides to stick with his choice, it doesn't matter if the host opens a door or not -- he's going to stick with his choice, and that's that, so his chances of winning are the same as if Monty does *not* open a door -- that is, 1/3.)Now we can rearrange equation (7) to express the probability that the prize lies behind Door B: P'(Door B) = P(omega) - 1/3 (8)Since the value of P(omega) is always, by definition, unity, this becomes P'(Door B) = 1 - 1/3 (9) = 2/3Now, since the probability that the prize lies behind Door B is 2/3, and the probability that the prize lies behind Door A is 1/3, and the contestant chose Door A, it's in his best interest to switch.

[shedrackle 0]

SET THEORY:Partition the set of doors as follows: A = {1}, which the contestant has chosen, and B = {2,3} which the contestant has NOT chosen.Clearly, the probability that the car can be found in A is 1/3, andthe probability that it can be found in B is 2/3.When the host opens one of the doors in the B partition, he is in effect removing that door from the game, reducing the cardinality of B from 2 to 1. He cannot, however, affect the likelihood that the car is contained within that partition simply by reducing its cardinality.At this point, the host is giving the contestant the option of trading A for ALL the doors in B. And since the probability that the car is in B is still 2/3, it's a good trade.

[rog 0]

Only stupid people are this closeminded.

You sir, are an ass. It's a tough problem, and the solution is counterintuitive. You dont have to be stupid to get it wrong.1/3rd of the time, your first choice is right, and switching loses the prize.1/3 of the time, the prize is one door down from your choice. Since they eliminate the third door, switching gets you the prize.1/3 of the the time, the prize is 2 doors down from your choice. Again switching gets you the prize.In essence, by switching, you get to piuck 2/3 doors. The first pick is really an "anti-pick". If the prize is behind either of the other 2, you're going to win it. The mistake people make is in thinking that with one door eliminated, the prize will be behind your picked door half the time. It wont. It'll still be back there 1/3 of the time. Switching = +EV.

[dead money 1]

You sir, are an ass. It's a tough problem, and the solution is counterintuitive. You dont have to be stupid to get it wrong.

I didnt say he was stupid for getting it wrong. I said he was stupid for being closeminded. He obviousle doesnt know the answer, yet makes his blanket statement with absolutely no mathematical proof to back it up. This is being closeminded, and only stupid people are closeminded.

[martinmc 0]

I completly understand what your saying... but i disagree with it. your odds have CHANGED do to eliminating one of the doors. when i see the goat my odds IMPROVE to 50%.

[dead money 1]

I completly understand what your saying... but i disagree with it. your odds have CHANGED do to eliminating one of the doors. when i see the goat my odds IMPROVE to 50%.

Nope. Lets look at the 3 winning scenarios. First lets say the prize is behind door #1. Also, the host never reveals the prize, only the goat.Scenario #1:You pick door #1. You have a 1 in 3 shot of being right. There 2 things that can happen here.Door #2 is revealed. You stay and you win. This occurs 1/2 the time.Door #3 is revealed. You stay and you win. This occurs 1/2 the time.So the math is: (1/3*1/2)+(1/3*1/2)=1/3.Scenario #2:You pick door #2. You have a 1 in 3 shot of being right. Door #3 is revealed. You switch and win.Scenario #3:You pick door #3. You have a 1 in 3 shot of being right.Door #2 is revealed. You switch and win.Now when we add up the scenarios of switching vs. not switching we get:1/3(Switch)+1/3(Switch)=2/3(Switch)vs. 1/3(No Switch)So clearly switching is the correct option.

[Nickoli 0]

For anyone who doesn't understand the math of this... Go browse through the new book that just came out call "The Curious Incident of the Dog in the Nighttime" or something like that...It's about an autistic kid who is really good at math. He proves this exact problem both mathematically and then also draws a very easy to understand little diagram with goats and cars and why you should always change doors when asked.If you've got some time it is a pretty fast read and not too bad of a book.