Car or Goat?
In the Monty Hall Quiz show, Monty offers the successful candidate another challenge to determine what prize he/she gets. Monty shows the contestant 3 doors, behind one door there is a new car and behind the other two doors there are goats. The contestant chooses one door and then Monty opens one of the other doors which he knows contains a goat.
He then offers the contestant the chance of changing his choice. Should the contestant switch to the other unopened door?
A. No, he shouldn't because he should trust his first instinct.
B. No, he shouldn't because Monty is trying to trick him.
C. Yes, he should because the odds have changed.
D. Maybe, but his odds haven't changed.
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12 solutions
Solution
At the time of the goat being revealed, he has 50/50. So the odds have increased from his initial standpoint but there is nothing to indicate that he should or shouldn't change his vote. At the initial selection had he chose a the car, switching would result in choosing a goat, and had he chose the goat, switching would result in the car. Either way the host will reveal a goat. So your odds are virtually 50/50 the entire time. It is a gamble. The revealed goat has no effect on your choices probable outcome. The car could be in any of the two remaining doors with equal probability.
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It looks like you are the only one who got it right - meaning the official answer (C) is wrong.
The contestant had 1/3 chances to pick a car. He made his initial choice, which remained unopened.
Then Monty increased the contestant's chances to 50/50 when he opened another door, revealing a goat behind it. So, at this point the contestant's initial choice has already increased from 33% to 50% chances of winning the car, no matter whether he switches! Then, why to switch if the chances would just remain the same (50%)?
In my understanding, both A or B would be correct answers. But a better answer would be rewriting D to "Maybe, but the odds have changed."
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He had a 1/3 chance of picking the car and gave up a 2/3 chance of picking a car if he chose the other 2 doors and when Monty revealed a goat in 1 of the other 2 doors if he then chose the remaining door he would acquire the 2/3 chance of getting the car. Think of this question. Would you prefer to pick 1 door or 2 doors? Switching is equivalent to picking 2 doors.
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@Guiseppi Butel – It's not equivalent of opening two because when he is offered the chance to switch, he isn't allowed to look through his first chosen door, then decide whether or not to swap, he is only allowed to swap. So either way, he can only see through one door.
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@Ashlin Martin – If he is allowed to look through his first chosen door there would be no problem in deciding to stay or switch. He would win 100% of the time. What are you saying??
yeah, you're wrong.
Yer wrong.
After opening the goat door, you need to chose one of two doors. This is 50% chance whatever choice you make.
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The remaining door has a 66.666..% chance of being the door to the room containing a car.
Well the Question is based on variable change the questioner should watch the movie "21".................
you copied this from wikipedia
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I did, but there is no rule stopping me from that
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Source should be mentioned.
It does not indicate in the narrative that Monty must reveal a door with a goat as is implied by the answer. Without that fact, one could deduce that if the choice was made and the car door was selected, then Monty opening a goat door would result in 50/50 odds. However, if a goat door was selected initially, Monty could simply choose to not open a door and take the goat door as the choice which would make D the better choice. I think that if C is the result, then there should be something in the narrative indicating that Monty must open a door.
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What does this imply?: then Monty opens one of the other doors which he knows contains a goat.
"The contestant chooses one door and then Monty opens one of the other doors revealing a goat."
The odds changed form 25/75 to 50/50 but the door he chose was unable opens why change?
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It changed from 33.33/66.66 to 50/50.
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But the word "should" is wrong. There's a chance he got right the first attempt.
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@David Cohen – well no, because if you take into account the 3 possible out comes you will see that if you make it a rule to always swap doors if given the chance then statistically you would win the car 2 out of three times.
No they don't
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Thy change from 33 to 66, due to the fact the other goat is shown, he has 33% chance choosing the car first and hence if he swaps boxes then he looses so 33% chance of goat if he swaps. Similarly he has 66% chance of choosing a goat at first, so when he swaps, he chooses the car. So 66% chance of getting the car when he swaps. And hence the odds have changed
The problem with the diagram of this "solution" is that the first frame contains two scenarios, not one. It's 2/4 not 2/3...
We can all agree that initially the odds are 1/3 to get the car if we forget about Monty, simple.
But Monty has his heart set on opening a door so now we have a new experiment. There are actually FOUR possible outcomes, two of which are depicted in first frame of the "solution", the other two which are depicted in the second and third frames. So the odds are 2/4 if you choose to switch and 2/4 if you decline.
Possible scenarios
Pick door 1 - Monty opens 2 - switch to 3 - Goat
Pick door 1 - Monty opens 3 - switch to 2 - Goat
Pick door 2 - Monty must open 3 - switch to 2 - Car
Pick door 3 - Monty must open 2 - switch to 1 - Car
Obviously this is a 50/50 game when he opens a door and you might as well stick with your gut or assess if Monty is trying to mess with you. Personally, I don't trust him.
You have three doors at the start to guess from - 1/3 odds you're right. Monty shows you you're not wrong, in fact there's a 50% chance you're right and going to win the car! But still zero reason to change your guess now because no matter what door you pick first he can always show you a stinkin' goat and you'll always be at 50/50 odds of winning when he does.
What you 'should' do for the answer is not trust all the solutions on this app..
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the first frame has two sub-scenarios, but they both are under the umbrella and probability of the first frame itself. Each frame represents a 1 in 3 choice (choosing one of the three doors). the two sub-scenarios in the first frame, then, are splitting that 1 in 3 odds, meaning each of them has a 1 in 6 chance of happening. Meanwhile in the other frames, there is only one sub-scenario, so the result inherits the full 1 in 3 odds.
Weighted Possible Scenarios
(1/6) Pick door 1 - Monty opens 2 - switch to 3 - Goat (1/6) Pick door 1 - Monty opens 3 - switch to 2 - Goat (1/3) Pick door 2 - Monty must open 3 - switch to 2 - Car (1/3) Pick door 3 - Monty must open 2 - switch to 1 - Car
Which means if you switch, you have a (1/6+1/6) = 1/3 chance of getting a Goat, and a (1/3+1/3) = 2/3 chance of getting the Car.
The odds are always 50\50 since you know he will take away one of the wrong doors. Simple. Odds never change.
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Strategy A: Don't switch.
Pick car = win.
pick goat = lose.
pick goat = lose.
Odds of winning by not switching = 1/3.
Strategy B: Switch.
pick car = lose.
pick goat = win.
pick goat = win.
Odds of winning by switching = 2/3.
Are you trying to prank those who don't understand with this poor show of knowledge?
The answers were in the wrong order, so I ended up choosing B instead of C. Why did you do this?!
The question does not say whether the goats are in adjacent rooms or not. What if the car is in the middle room ?
Nice explanation. Really needed it.
The answer "Yes, one should switch" is indubitably correct. However, in the question option "C" then adds "because the odds have changed", which is incorrect. One should switch because the probably of winning by switching is 2/3. That probability never changes regardless of Monty opening a door to reveal a goat. It is misguided to think that the odds change to 50:50. Marlyn Von Savant's book, The Power of Logical Thinking" has a good explanation of why the 50:50 thinking is ill-conceived.
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Who said that the odds changed to 50:50? I said that the odds changed from 1:2 to 2:1 if you switched.
In the original situation his probability of being correct is 1/3 , odds are 1:2. After the door opening switching would yield a probability of being correct of 2/3, odds are 2:1, therefore I stand by my contention that the odds have changed.
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Thank you for taking the time to reply. Mathematical education and debate can only be a good thing. Regarding standing by you position, I commend to you the book I've mentioned. You'll note that many learned mathematicians stood by their position when Marlyn Von Savant published her solution. They were, nevertheless, incorrect. See the response of Justin Malme above, which nicely demonstrates the position. The odds do not change, which is why the original TV programme is so clever. It seduces one into the misapprehension that the odds must have improved in one's favour, when then have not.
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@Harry Hutchinson – Are you saying that picking 2 doors have the same odds of winning as picking 1?
The key of this problem is actually Monty himself. You see, he doesn't just show you a random door. He shows you a goat. There are two cases:
Case 1:
You initially pick a goat with probability 2 / 3 . Because Monty can't show you your own choice, and he (obviously) can't show the car, he MUST show you the other goat. Thus by switching, you guarantee that you will get the car.
Case 2:
You initially pick the car with probability 1 / 3 . In this case, by switching you will obviously get the remaining goat that Monty hasn't shown you.
In other words, as long as you've initially picked the goat, you WILL get the car by switching. Thus your chances double by switching. :D
@Paramjit Singh @Guiseppi Butel @Milly Choochoo Here is a legit solution. Guiseppi's is incorrect.
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Yeah, so with the original 3 1 probability of choosing the car, switching will raise the probability to 3 2 .
Suppose there are 2 0 , 0 0 0 doors for you to choose from, and, just like the original game, you choose one and then Monty opens every door containing a goat except one door and the one you originally picked. If you pick door number 1 , clearly with a 2 0 , 0 0 0 1 chance of getting the car, and then Monty opens every door (showing goats) except door number 1 3 , 4 4 2 , you better damn well switch. The chances then, between your original door number 1 and number 1 3 , 4 4 2 are clearly not 2 1 . There is actually now a 2 0 , 0 0 0 1 9 , 9 9 9 probability that the door you didn't originally pick has the car behind it.
Translating this back down to the original 3 door problem, the probability after he shows the goat is 3 2 . It's not that much better than 2 1 , and this situation probably isn't as ideally uniformly random as this 'probability' shows, but it is what it is (for this problem).
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I like this explanation. The other one I like is this:
There is originally 1/3 chance that he picked the right door. The fact that the host has shown him the goat doesn't change the fact that the probability of the car being behind one of the other two doors is 2/3. Therefore he should change.
Hmm hadn't really thought about that. Cool!
I've said this exact response a million times this problem.
Thanks! I kinda got it, but was on the fence as to whether I believed it. This makes it obvious
Where am I incorrect, Finn?
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You claim that because the probability is 1 / 2 once he's revealed a goat that you should change, when the odds are (seemingly) clearly equal.
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@Finn Hulse – the odds are (seemingly) clearly equal.
What's your comment if I claim the odds are 2 to 1 if he changes.
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@Guiseppi Butel – That's correct.
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@Finn Hulse – What's correct, Finn?
the odds are (seemingly) clearly equal
or the odds are 2 to 1 ?
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@Guiseppi Butel – Think of it this way; If your strategy is to not switch, you must pick the door with the car to win. You have a 1/3 chance of doing it. If your strategy is to switch doors, you must pick a door that has a goat behind it to win (since you will then switch to the door with the car). You have a 2/3 chance of picking a goat.
@Guiseppi Butel – Yeah Guiseppi, if you meant that the moment he shows you one of the goats the probability that the car is behind the other one that you didn't pick is now 2 1 , while the probability that the car is behind the one that you did pick is lower than that, you must be wrong.
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@Milly Choochoo
–
title
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@Guiseppi Butel – This picture makes absolutely no sense at all. How are the odds of D double of B and also of C when the odds of D, B, and C are all different values?
@Guiseppi Butel – This is why we don't use 'odds' in mathematics. We use probability.
UNLESS you had chosen the car. Then your odds are changing when you choose again. Although better odds still don't guarantee a victory.
Your odds in your initial decision were 1 in 3. Regardless of your next decision (to choose a new door or keep the one you initially chose) you have 1 in 2 odds.
None of the options are fully correct answers.
For those who are uncomfortable with probabilities, let's explain this in terms of "many cases" instead.
Suppose we play the game 999 times, with the car being placed behind each of the doors A, B, and C, 333 times. To keep things simple, let's assume we always pick door A initially.
Case 1: We never switch
We will win the car 333 times, when it's placed behind door A.
Case 2: We always switch.
If the car is behind door B, Monty will open C, and we will win. Likewise, if the car is behind door C, Monty will open B and we will win again. If the car is behind door A, we will lose, however. Still, this makes for an impressive total of 666 cars!
You win twice as many times if you switch. Right on!
I don't know what you're all on about, I'd prefer the goat
LOL, mbeeeee ( goat sound )
Lol. For the amount you could probably sell the car for, you could buy a bunch of goats.
You have a 1/3 chance of picking the door with the car. That means there is a 2/3 chance that the car is behind one of the doors you DID NOT pick. Eliminating one of those doors does not change the odds. Once one door is opened, the remaining door still has a 2/3 chance of containing the car.
Still convinced the odds are 50/50? Look at it this way; instead of 3 doors say there is 1,000 doors. 1 door has a car, the other 999 have goats. You have a 1/1,000 chance of picking the door with a car. You pick a door and 998 doors are opened. Would you switch to the remaining door?
Initially, he has a 1/3 chance of being right and a 2/3 chance of being wrong. Once the goat is revealed, the first door still has the 1/3 chance of being right, and the other two doors combined share a 2/3 chance of being right. Since he knows that the door that was revealed is a goat, that door has a 0 chance of being the car and the other door has the 2/3 chance on its own which is why your probability goes up if you switch
The overall probability is what needs to increase, not just a single step. At first, one's chances are 1/3, but that doesn't matter if the next move is going to change the overall probability. When he shows you the goat, the probability for that step is 50/50. That step isn't the whole problem though. The initial guess has to be included, as well as what is chosen when asked to switch. If the original answer is kept, you would of had to of guessed it correctly the first time to get it right, so there is no overall increase in the probability from the original 1/3. If one switches, one would have had to get it wrong the first time to get it right. The probability of you guessing wrong the first time is 2/3, so that's your new probability by switching.
The best way to convince someone of the answer to is to draw up a table. It's a pain to describe the table in detail, but I'm sure you can figure it out pretty easily, you want to cover all scenarios for the guess/reveal/switch combinations. Then run the numbers and see that switching benefits him 2/3 of the time.
Now, this is where I slightly disagree with the answer. I say his odds have not changed. But this lacks specificity, as does the answer, so I assumed the traditional explanation when answering. The traditional interpretation being that his odds of winning have changed. This is, unfortunately, a complicated and subtle answer, and fewer people understand the solution, even if convinced by the table.
So, the best way to understand the problem. Consider the odds that he is wrong. To begin with, no matter the choice made, there is a 2/3 chance that he's wrong. When Monty opens one of the remaining doors, the original guess has not in any way been altered. It fits better with our understanding of choice to claim then that the odds on that first guess can't have changed. And in fact, the odds that he was wrong haven't changed at all. 2/3 of the time his first guess was wrong, now that there's only one other door, he should change his answer, because he was probably wrong the first time.
The moment he opens the first door the odds have changed
Yes, yes he did
No they do not. The fact that one door is opened, does not change the fact that there is a 2/3 chance the car is behind a door you didn't pick. Instead of 3 doors, imagine there are 1,000,000 doors...
This is a hotly debated puzzle and while I knew the answer they were looking for I reject it. Your original guess had a 1/3 chance of being correct and if you choose a new door (once prompted) you will have a 50/50 chance of being correct. However, if you re-affirm your original guess - you NOW have a 50/50 chance of being correct with the same guess. Once he reveals the goat the odds change on your original guess. The whole topic is madness.
Almost nobody got it right - and the official answer is wrong.
The contestant had 1/3 chances to pick a car. He made his initial choice, which remained unrevealed.
Then Monty increased his chances to 50/50 by opening another door, revealing a goat behind it. So, at this point the contestant's initial choice has already increased from 33% to 50% chances for winning the car! Then, why to switch if the chances would just remain the same (50%)?
This whole thing looks completely wrong. In my understanding, both A and B would be correct answers. But a better answer would be rewriting D to "Maybe, but the odds have changed."
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So obviously when you pick a door at the beginning you have a 1/3 chance. Assume you pick door A. Door A has a 1/3 chance so obviously Door B and C have a combined chance of 2/3 of being correct. Now the host opens Door C to show a goat then offers a switch. Well it's in your favor to switch to Door B. Remember Doors B and C have a combined chance of 2/3 and we just eliminated C. That means Door B gets all of the 2/3 chance because Door C now has a 0% chance. Always switch.
So what is your answer to the problem? A, B, or D?
Strategy A: Don't switch.
Pick car = win.
pick goat = lose.
pick goat = lose.
Strategy B: Switch.
pick car = lose.
pick goat = win.
pick goat = win.
Switching doors gives you a 2/3 chance of winning.