View Full Version : And another puzzle
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you: 'Do you want to pick door #2?' Is it to your advantage to switch your choice of doors?
Dysparunia
20-02-07, 06:37 PM
Switch doors every time. Better chance
PS-have you been reading The Curious incident of the Dog in the Nighttime?
Id be happy with a goat. At least you wont have to pay for it to be on the road...and i'll have a nice lawn!!. Sorry, noy helping much am i.
Id be happy with a goat. At least you wont have to pay for it to be on the road...and i'll have a nice lawn!!. Sorry, noy helping much am i.
Happy with a goat?
You must be KIDDING :rolleyes:
Taxi !!
Bluepete
20-02-07, 07:22 PM
Bet it can't take off!
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you: 'Do you want to pick door #2?' Is it to your advantage to switch your choice of doors?
Two doors left. 50:50 chance. Doesn't matter what you picked before the host opens a door.
wyrdness
20-02-07, 11:58 PM
Two doors left. 50:50 chance. Doesn't matter what you picked before the host opens a door.
Er, not quite.
The host knows what is behind each of the three doors. If your original choice was the door with the car, then he can pick either of the other two doors to open, which gives you a 50/50 chance. However, if your first choice was a door with a goat (which was a 2/3 probability) then that leaves only one door that he can open - that which has the other goat behind it. Therefore the chances are that the remaining door hides the car.
Does this make sense? I've had a few beers on top of strong painkillers, so might be spouting complete rubbish.
no difference, I'm with sinbad on this.
Assuming you pick the car. The probability of switching is 1/2. You switch, or you don't switch.
We know the host has to pick a goat. That leaves two possible doors. IF you switch, there's a 1/2 probability that the door you switch to has the car (we know it doesn't, but bare with me).
If you don't switch, you have a 1/2 probability of getting the car. We know the 3rd door is a goat because the host picked that one.
OK, so 50:50 there, no matter if you switch or not. There is no advantage in switching.
Let's now assume you pick a goat (but you wouldn't know this yet). The probability of switching is constant, at 1/2. You do, or you don't.
If you do switch, you either started with the car, or you didn't. So if you switch to the other door, there's a 1/2 probability that you get the car. If you don't switch, there's a 1/2 probability that you get the car too.
So to summarise, picking the car first.
You switch to the car, 1/2
You stick with it & have the car, 1/2.
And picking a goat first:
You switch to the car, 1/2
You stick & get the car, 1/2.
Hmm. :confused:
The only way the probabilities could be affected, is if you knew what you had already chosen, which would defeat the purpose of the game.
Of course, back in the real world, if the host knows, then you could read their body language etc, and find out. However, this could be countered by the production team knowing, and only telling the host which door to open when the time arose.
Dysparunia
21-02-07, 09:49 AM
Lets look at it like this,
1) The contestant first chooses the door with the car behind it. She is then shown either door A or door B, which reveals a goat. If she changes her choice of doors, she loses. If she stays with her original choice, she wins.
2) The contestant first chooses door A. She is then shown door B, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.
3) The contestant first chooses door B. She is then is shown door A, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.
Each of the above three options has a 1/3 probability of occurring, because the contestant is equally likely to begin by choosing any one of the three doors. In two of the above options, the contestant wins the car if she switches doors; in only one of the options does she win if she does not switch doors. When she switches, she wins the car twice (the number of favorable outcomes) out of three possible options (the sample space). Thus the probability of winning the car is 2/3 if she switches doors, which means that she should always switch doors - unless she wants to become a goatherd.
Lets look at it like this,
1) The contestant first chooses the door with the car behind it. She is then shown either door A or door B, which reveals a goat. If she changes her choice of doors, she loses. If she stays with her original choice, she wins.
2) The contestant first chooses door A. She is then shown door B, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.
3) The contestant first chooses door B. She is then is shown door A, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.
Each of the above three options has a 1/3 probability of occurring, because the contestant is equally likely to begin by choosing any one of the three doors. In two of the above options, the contestant wins the car if she switches doors; in only one of the options does she win if she does not switch doors. When she switches, she wins the car twice (the number of favorable outcomes) out of three possible options (the sample space). Thus the probability of winning the car is 2/3 if she switches doors, which means that she should always switch doors - unless she wants to become a goatherd.
Probability of picking the right door first time, 1/3
Probability of picking a goat first time, 2/3
Probability of picking the car, and swtching to a goat, (1/3)*1 (guranteed you pick a goat, if you picked the car first) = 1/3
Probability of picking the car & switching to the car (ie, not changing your mind), 1/3
Probability of picking the goat, and switching to the car, 2/3*1/2 = 0.6667 * 0.5 = 0.3333 = 1/3.
Probability of picking the goat, and switching to the other goat, 2/3*1/2 = 1/3 (proven above).
So does changing your mind carry any advantage? Look at the lines that give the probability including switching. No, it doesn't, still an equal chance (aka 50:50, even though it's a 1 in 3 probability).
(Probable outcome of choosing another door depends on what you chose before that, hence the multiplication of probabilities).
wyrdness
21-02-07, 11:51 AM
So does changing your mind carry any advantage? Look at the lines that give the probability including switching. No, it doesn't, still an equal chance (aka 50:50, even though it's a 1 in 3 probability).
Sorry Baph, I think that you should give up on these puzzles :D:D
You've got four options. Let's look at them and the probabilities of you getting the car.
Pick the door with the car and don't change mind - 1/3 x 1 = 1/3
Pick the door with the car and change mind - 1/3 x 0 = 0
Pick the door with a goat and change mind - 2/3 x 1 = 2/3
Pick the door with a goat and don't change mind - 2/3 x 0 = 0
So changing your mind gives you a 2/3 chance of getting the car, whilst staying with your first decision gives you a 1/3 chance. So the answer is to change your mind.
Which is what Dysparunia said.
Whatever... I would be happy with the goat as the car is probably a KIA anyway so the goat would be more valuable.:smt002
Sorry Baph, I think that you should give up on these puzzles :D:D
You've got four options. Let's look at them and the probabilities of you getting the car.
Pick the door with the car and don't change mind - 1/3 x 1 = 1/3
Pick the door with the car and change mind - 1/3 x 0 = 0
Pick the door with a goat and change mind - 2/3 x 1 = 2/3
Pick the door with a goat and don't change mind - 2/3 x 0 = 0
So changing your mind gives you a 2/3 chance of getting the car, whilst staying with your first decision gives you a 1/3 chance. So the answer is to change your mind.
Which is what Dysparunia said.
The probability of changing your mind is 1/2, not 1 or 0.
If you change your mind, you have only one door to choose. That may, or may not have the car.
You're too stuck on the assumption that you know what you have chosen, before the opportunity to change your mind is presented.
wyrdness
21-02-07, 12:45 PM
The probability of changing your mind is 1/2, not 1 or 0.
If you change your mind, you have only one door to choose. That may, or may not have the car.
You're too stuck on the assumption that you know what you have chosen, before the opportunity to change your mind is presented.
Can you program a computer? If so, write a simple simulation of the problem and see what answer you get. There are some on the net already.
http://math.ucsd.edu/~crypto/Monty/monty.html
http://en.wikipedia.org/wiki/Monty_hall_problem
Can you program a computer? If so, write a simple simulation of the problem and see what answer you get. There are some on the net already.
http://math.ucsd.edu/~crypto/Monty/monty.html
http://en.wikipedia.org/wiki/Monty_hall_problem
Taken from the first link you posted:
How does this problem change if Monty Hall does not know where the car is located? We must decide what it means if Monty should happen to open the door with the car behind by accident. The problem says only that Monty opened a door with a goat behind it so we interpret this to mean that if the car is revealed then the game is over and the next contestant plays the game. If this is the senario then the wheel looks almost the same, the inner wheel represents the door that the car is behind, the middle wheel represents the number of the door that is selected by the contestant, and the outer wheel represents the door number Monty Hall will show the contestant. This time however, Monty Hall has the option of opening a door with a car behind it, but by chance he didn't. Playing the game under these assumptions is equivalent to spinning the roulette wheel to the right except that if the blackened area of the wheel comes up then it is spun again. Once again the red area means that in order to win the contestant will need to switch doors, and the blue means that the contestant should not switch. Notice that there is the same amount of red area as blue. In other words, it doesn't matter if the contestant switches in this case.
It all hinges on the basis that the host (at least) knows where the car is.
PS. I'm a java developer, so no, I don't know how to program a computer :D
I've avoided looking at these puzzle threads so far - they're not good for the blood pressure. But wyrdness is right.
The host opening the door is a bit of a distraction. You're not getting any new info. Essentially you are being offered a choice of
A) one door. or
B) two doors.
True, one of the doors in option B has a goat behind it, which the host shows you, but you already knew that.
So :
1) pick a door.
2) keep this , or swap to get both other two.
I'd swap.
I've avoided looking at these puzzle threads so far - they're not good for the blood pressure. But wyrdness is right.
The host opening the door is a bit of a distraction. You're not getting any new info. Essentially you are being offered a choice of
A) one door. or
B) two doors.
True, one of the doors in option B has a goat behind it, which the host shows you, but you already knew that.
So :
1) pick a door.
2) keep this , or swap to get both other two.
I'd swap.
I agree completely.
You're getting no new information. It's just the same as the host not opening a door at all.
Still an equal chance, unless the host knows you haven't picked the car. The host then knows the only door left (that you haven't chosen & haven't been shown) has the car. You don't know anything. Except that you only have a choice of two doors:
1) The door you originally chose.
2) The remaining door.
That makes it a 50:50 from the point the door is opened. It still stays 1/3 from the outset.
All of this, so long as the host knows where the car is.
wyrdness
21-02-07, 01:23 PM
Taken from the first link you posted:
It all hinges on the basis that the host (at least) knows where the car is.
PS. I'm a java developer, so no, I don't know how to program a computer :D
Did you read the original question properly?
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you: 'Do you want to pick door #2?' Is it to your advantage to switch your choice of doors?
The host knows what's behind the doors. Therefore you have a 2/3 probability of getting the car if you change your mind.
You fell into the trap set by the writer of the question, which was to assume that you have a 50/50 chance, rather than working out what would really happen.
Did you read the original question properly?
The host knows what's behind the doors. Therefore you have a 2/3 probability of getting the car if you change your mind.
You fell into the trap set by the writer of the question, which was to assume that you have a 50/50 chance, rather than working out what would really happen.
Nope, I fell in to the trap of reading the question too fast, obviously.
My bad, in that situation, yes, the advantage is in switching.
By your logic, you're more likely to pick the car by changing your mind, even if you actually picked the car first time. It doesn't matter what he says to you, you still have two doors to pick from.
Doors A-B-C
A and B have a goat, C has a car. (Why not).
You pick A, Host opens B. You choose between A and C (50:50) why would the car be behind C?
You pick B, Host opens A. You choose between B and C (50:50) why would the car be behind C?
You pick C, Host opens A. You choose between B and C (50:50)
You pick C, Host opens B. You choose between A and C (50:50)
So why would it be ever be more likely to be C and so a good idea to change your mind?
Am I missing something? How can you get extra information if the host is going to open a door with a goat behind whatever you do initially? You pick a goat first time, you see a goat, you pick the car first time, you see a goat.
By your logic, you're more likely to pick the car by changing your mind, even if you actually picked the car first time. It doesn't matter what he says to you, you still have two doors to pick from.
Doors A-B-C
A and B have a goat, C has a car. (Why not).
You pick A, Host opens B. You choose between A and C (50:50) why would the car be behind C?
You pick B, Host opens A. You choose between B and C (50:50) why would the car be behind C?
You pick C, Host opens A. You choose between B and C (50:50)
You pick C, Host opens B. You choose between A and C (50:50)
So why would it be ever be more likely to be C and so a good idea to change your mind?
Am I missing something? How can you get extra information if the host is going to open a door with a goat behind whatever you do initially? You pick a goat first time, you see a goat, you pick the car first time, you see a goat.
If the host doesn't know where the car is, there's a possibility of him opening a door & revealing the car. You loose. No choices.
If the host doesn't know where the car is, there's a possibility of him opening a door & revealing the car. You loose. No choices.
But the host does know, it says so.
But the host does know, it says so.
Yes, and the probabilities only stay equal if the host doesn't know. That's the only way he's guranteed not to reveal the car, and so the odds are tipped.
Check out the first link by wyrdness, play the game, and click to have it explain. Read about the probability wheels etc. You'll get it.
Ahh I get it now. It still is 50:50 in isolation because there's two doors, but if you count on hitting a goat, and the chances are 2/3 you will, then whichever door is left after the one you picked and the one he opens are eliminated (by yourself) is the car.
vBulletin® , Copyright ©2000-2025, Jelsoft Enterprises Ltd.