Game Show
You are on a game show in which there are 3 doors. Behind one door is a pot of gold, but the other doors contain rocks. You choose door #1.
The host of the game show pulls a lever which will randomly open one of the two doors you didn't choose. To your relief, door #3 was opened and turned out to have rocks!
The host offers you a decision: stick with door #1, or switch to door #2. To maximize the probability that you get the gold, what should you do?
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3 solutions
Discussions for this problem are now closed
But seeing how that the door randomly DID open to the one with the rocks then doesn't that turn it into the traditional Monty Hall Proble?
No -- it's very different to open a door randomly and get rocks than it is to open a door that is guaranteed to be rocks.
Do you see why? (You can write out the 3 cases for where the gold is located, and then work out the probabilities.)
But isn't this question talking about this specific situation? where the door opened was a rock and thus being the same as the Monty Hall Problem?
@Brendan Moore – No -- in the Monty Hall problem, the host opens a door which he/she knows has rocks (goats). In this problem, an unselected door (#2/#3) is opened at random by the lever. This is a fundamentally different action as far as the probabilities are concerned.
Unlike what some comments have suggested, when the door opened reveals rocks, this does not reduce the problem to the Monty Hall problem. The specifics of the action that led to the door being opened (e.g., randomly or guaranteeing rocks) directly affect the probabilities.
The idea of gaining information from an outcome via the a priori distribution of the possibilities for that outcome is fundamental to conditional probability (and probability in general).
As a similar example of why this is this case/to try to illuminate these ideas further, consider this example:
- There are 3 doors: one with $1, one with $10, and one with $100. You pick a door. There are then two variants of the process:
(Variant A) The host uses the level which opens one of the two unselected doors, then gives you the choice to switch.
(Variant B) The host knows what's behind the doors, and opens the unselected door containing the most amount of money, and then gives you the choice to switch.
In (A), we gain no additional information since the door is opened randomly; this is analogous to the problem statement here. We are indifferent to sticking/switching, no matter what is revealed. You can work out the probabilities directly by considering the 3 cases of what was behind the door we initial picked.
In (B), because the host is making a non-random action (as in the traditional monty hall problem), we can actually gain information . In particular, if the host reveals $10, then we know we have selected the door with $100 (since otherwise the host would have revealed $100), so we should certainly "stick". If the host reveals $100, then we should be indifferent to sticking/switching since he would reveal $100 whether we had initially chosen the door with $1 or $10. Thus, in this variant, we are indifferent to switching/sticking if $100 is revealed, but if $10 is revealed, it is in fact in our interest to stick.
@Eli Ross – I guess the other way to think about it is this:
1) You try this 1000 times.
2) It's going to be behind the door you pick 333 times.
3) Out of the randomly exposed other doors, it will show up 333 times.
4) Therefore, it's going to be behind the unexposed, unselected door the remaining 333 times.
5) Thus original = 333, unexposed/unselected = 333.
Cool... learned something today.
I think you're right Shivam, but I am tempted to write a script that: 1) Randomly places something in 1 of 3 containers. 2) Then randomly selects a container. 3) Randomly exposes one of the other two containers. 4) If the object is exposed, then to discard this result. 5) If the object is not exposed, to simply count the number of times it was in the first container vs. the unexposed/unselected container.
... brb
Isn't it also true though that when you select your door, there is a 1/3 chance that you've selected the door with gold behind it and a 2/3 chance that the gold is behind one of the other doors. When the door with rocks is revealed, there's still a 2/3 chance that you've selected the wrong door, so you double the probability of selecting the correct door by switching.
A: event gold is behind selected door
B: event gold is behind revealed door
C event gold is behind last door
before rocks revealed:
P(A) = 1 - P(B) - P(C) = 1-1/3-1/3 = 1/3
after rocks revealed:
P(B) = 0 since we know rocks behind door B
P(A) = 1/3 since we selected before we knew rocks were behind door B
P(C) = 1 - P(A) - P(B) = 1 - 1/3 - 0 = 2/3
Switching to door C gives you the highest chance of winning
Am I the only one who didn't read the whole lengthy question and thought it to be Monty Hall prob n now regretting to see the word 'Randomly' there?
The moment we get to know there are rocks behind door B, you said the probability of gold being in door B is 0. This is absolutely correct. But in the next step, you have kept probability of gold being in door A as 1/3, which is wrong. The moment P(B) became 0, P(A) and P(C) would become 0.5 each. How will P(A) still remain to be 1/3, through which u concluded P(C) is 2/3?
Consider another example. There is an unbiased die. The probability of getting 1 on the die is 1/6. Now I give you another die. This die is biased such that number 6 does not fall. Now this means that the probability of you getting a 1 is 1/5 (since the die will never show a 6). Say I give you another die. This die is biased in such a way that numbers 6 and 5 do not fall. Now this means that the probability of you getting a 1 is 1/4 (since the die will never show a 6 or a 5). This is the concept of equal probability theory. This simply means, when the probability of occurrence of an event is (1/x) and one of the probability of one of the x events becomes 0, automatically the probability of occurrence of the said event would become [1/(x-1)].
To conclude, the moment you got to know that door B contained rocks, your probability increased from 1/3 to 1/2. With each door getting rocks, your probability changes from 1/x to [1/(x-1)].
but each door has probabilty of 1/3 but we choosed door 1 and host opened with no information door 3 so door 3 + door 2 probabilty is 2/3 but after opening rock door 3 probability got 0 but door 2 got 2/3 ... now isn't better to choose door 2?
Consider an example. There is an unbiased die. The probability of getting 1 on the die is 1/6. Now I give you another die. This die is biased such that number 6 does not fall. Now this means that the probability of you getting a 1 is 1/5 (since the die will never show a 6). Say I give you another die. This die is biased in such a way that numbers 6 and 5 do not fall. Now this means that the probability of you getting a 1 is 1/4 (since the die will never show a 6 or a 5). This is the concept of equal probability theory. This simply means, when the probability of occurrence of an event is (1/x) and one of the probability of one of the x events becomes 0, automatically the probability of occurrence of the said event would become [1/(x-1)].
To conclude, the moment you got to know that door B contained rocks, your probability increased from 1/3 to 1/2. With each door getting rocks, your probability changes from 1/x to [1/(x-1)].
Thanks @Shubhank Sbk ... now i got the concept :)
@Ahsan Azhar – Your welcome! :)
In order to apply this theory to this problem, you need to assume that the events you are considering are all equally likely at every stage, which is what you are trying to decide! You shouldn't assume your answer and then use this to prove your answer.
@Joel Toms – Im not! This is in fact the right thing to apply and what gives the answer I have contented.
Say there are 100 doors, gold behind one door and rocks behind the other 99. U chose a door at RANDOM. The host RANDOMLY opens an unselected door. U find rocks. U sigh a relief because the probability of u finding gold in the opened door is 0, and that INCREASES your chance of getting gold. Host again opens an unselected door at RANDOM. U again find rocks. U sigh a relief again because it INCREASED your chance of winning gold again. Similarly, every time a door is eliminated by finding rocks, it actually increases your chance of getting gold. The moment any door reveals gold, your probability of winning gold automatically becomes 0. Until then it just keeps increasing. This is what is represented by my formula.
@Shubhank Sbk – You need to be very careful when you say that you have "increased your chances of getting gold". Perhaps be more specific, i.e. you have "increased the probability that the gold is behind your door".
In any case, you have to justify your equal-probability conclusion based on the problem, not just by counting the number of possible outcomes. It is very easy to think of examples where there are two outcomes with non-equal probability: for example, it will/will not rain on a given day; I will/will not roll 5 or 6 on a fair die; I will pick a red counter from an unevenly filled bag.
It's better to switch, because you can not trust the host. What if the game was developed to "randomly" show always a rock? You never know, and in this case you win by switching. If it happens to be really random, then it is the same chance to stay or to switch, so you don't lose by switching. All together it is worth switching :) And you make upset the people who think if you switch in a game show then you are a bad person, not loyal, untrustworthy etc. Only making those people upset worth the switching :)
RANDOM DOES NOT EXIST !
I disagree. The same principle carries over from the monty hall problem. The chance you choose rock is 2/3 and gold is 1/3. If the gold is revealed, then it doesn't matter what you choose. But since it is a rock, there is a possibility that it is a gold behind door 1. So from our perspective, it is either we chose the one with gold behind or not. Since it is more likely to choose the one with rocks behind hence it is preferable to switch the door. You can run this game in a computer simulation, and you would see that whenever you switch the door when the host reveal the rock, your odds of choosing the gold increase.
well that is incorrect. because this is exactly the Monty hall problem. why? because it only talks of 1 event. it only happenED once, in the past. so it has/had 100% chance of getting the rocks, since it happened already. this does not ask you, 'if repeated, what is the best choice'. it asks you 'this is the situation that has just happened, now, what do you do?'.
of course this is not maths, but a little logic and language nitpicking. just happened(?) to be bored right now. :P
- 2nd para of question states that...........
-
The host of the game show pull a lever which will randomly open one of the two doors u did not choose.........
this means that host knows that door 2 & door 3 both having rocks there fore no problem to open any door 2 Or 3 randomly....... - if host pull a lever randomly & open one door out of 3 & then Door 3 get opened & shows rock then answer is : "Doesn't Matter"
- there fore as per given question answer is " Stick with one"
"The host of the game show pull a lever which will randomly open one of the two doors u did not choose......... this means that host knows that door 2 & door 3 both having rocks there fore no problem to open any door 2 Or 3 randomly......."
It is incorrect to make such a statement. If you had chosen door 1, then as per the statement, the host would pull a lever which would randomly open either door 2 or door 3. If you had chosen door 2, the host would pull a lever which would randomly open either door 1 or door 3. If you had chosen door 3, the host would pull a lever which would randomly open either door 1 or door 2. It does not mean that the host knew doors 2 and 3 contained rocks AND THEREFORE he made the selection of the doors to be opened as random. This has nothing to do with the host knowing which door contained rocks. It means that the system is designed in such a way that the lever opens a random door which you did not select (whether or not it contains gold). Therefore there was a chance that door 3 contained gold. This just means the probability became zero when you got to know it contained rocks.
Going to the next step, the probability would still be equal (ie. 1/2) and therefore the answer "Does not matter" would hold correct.
It turns out to be 50/100 percent chance that you will get the door with gold, so the luck will lead you.
Search up Monty Hall problem on YouTube.Its not 50/50
Actually, this whole problem is wrong and so id your answer. If you chose door one, you have a 33.3% chance of getting gold. If he shows you that door number three is rocks, then you still have 3 parts except one has been moved. There is now a 66.7% chance of gold behind door two. Pick door two. This is easier to understand on a larger scale. Imagine you have 100 doors. You pick door one. He opens 98 doors, so yours and door 38 are left. There is now a 99% chance that your gold is behind door 38, because the others have been ruled out, but the odds are still there. Would you switch or not?
Let the doors be labelled Gold, Rocks 1 and Rocks 2. There are six equally probable possibilities for this problem, before you or the host choose doors.
A. You choose Gold; Host chooses Rocks 1
B. You choose Gold; Host chooses Rocks 2
C. You choose Rocks 1; Host chooses Gold
D. You choose Rocks 1; Host chooses Rocks 2
E. You choose Rocks 2; Host chooses Gold
F. You choose Rocks 2; Host chooses Rocks 1
P ( A ) = P ( B ) = ⋯ = P ( F ) = 6 1 . These possibilities are all equally likely because the choice at each stage is random. So when the host reveals Rocks, the ONLY thing you know is that you are NOT in option C or E. The other possibilities (A,B,D,F) are still equally likely.
In two cases you should Stick; in two cases you need to Switch. So it makes no difference to your odds whether you Stick or Switch.
In the Monty Hall Problem, the options are as follows:
A. You choose Gold; Host chooses Rocks 1
B. You choose Gold; Host chooses Rocks 2
C. [not possible]
D. You choose Rocks 1; Host chooses Rocks 2
E. [not possible]
F. You choose Rocks 2; Host chooses Rocks 1
P ( A ) = P ( B ) = 6 1 ; P ( D ) = P ( F ) = 3 1 . So you can see that the problem is different when the host chooses randomly.
That one was lame. If the odds are 50/50 it is kinda obvious.
It is very important to note that the lever opens one of the other two doors randomly, rather than making sure to reveal rocks. (In the traditional Monty Hall problem , the host guarantees that you will see rocks (goats), but here that is not the case.)
We want to find the probability that door 1 contains the gold (G) given that the door that was randomly opened contained rocks (R). Via Bayes' theorem :
P ( G ∣ R ) = P ( R ) P ( G ) ⋅ P ( R ∣ G ) = gold in 1 and rocks revealed 3 1 ⋅ 1 + gold in 2 and rocks revealed 3 1 ⋅ 2 1 + gold in 3 and rocks revealed 3 1 ⋅ 2 1 3 1 ⋅ 1 = 2 1 ,
so it doesn't matter if you switch or not.
Remark: In the traditional Monty Hall problem, P ( R ) = 1 , which is why you get P ( G ∣ R ) = 3 1 . The key here, as with any conditional probability, is information gain . Here, we actually gain information about what is likely to be behind door 1 because of the randomness element. Intuitively, we are more likely to see rocks revealed when door 1 contains the gold, so probability shifts to door 1 when we see rocks. (Whereas in the traditional Monty Hall problem, we are always going to see rocks, so it doesn't add information about door 1.)