08 May 04
OK lets begin
You are on a games show,
there are 3 doors
behind one door is a car and behind the other doors are nothing
You choose a door at random
the gameshow host opens one of the doors which you have not picked and reveals it to be empty.
He says you may change your choice to un opened door or stay with your original choice.
QUESTION
should you stay with your choice, change door or dosent it matter?
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09 May 04
Originally posted by jimmyb270You have a 1 in 3 chance of the car being behind the door you chose
I remember this question from before, but I never got the explanation...anyone know?
You have a 2/3 chance of the car being behind one of the other two doors.
Assuming the presenter knows where the car is, he has now shown you WHICH of the other two doors the car is behind, if it is behind one of them. You can now switch doors and open the door left by the presenter and its as if you opened both of the other two doors - i.e. you have a 2/3 chance of getting the car.
Mind you, a clever presenter would not be so predictable. If he doesn't always open one of the other doors (on every run of the show) he could chose to weight his openings so he always opens another door if you have the right door, and opens another door 1/x of the times when you have the wrong door. What value of x will a clever presenter use and what chance of getting the car will you have now, with best play? (assume you know the value of x when you are on the show)
10 May 04
1 edit
Originally posted by iamatigerThe contestant now has 4 strategies, with associated probabilities of winning:
You have a 1 in 3 chance of the car being behind the door you chose
You have a 2/3 chance of the car being behind one of the other two doors.
Assuming the presenter knows where the car is, he has now shown you WHICH of the other two ...[text shortened]... st play? (assume you know the value of x when you are on the show)
Always stick: 1/3
Switch only if host opens a door: 2/3x
Stick only if host opens a door: 1/3 +2/3*(1-1/x)*1/2 = 2/3 - 1/3x
Always switch: 2/3(1/x + (1-1/x)*1/2) = 1/3 + 1/3x
Assuming optimal play by the contestant, and that the contestant knows whether the potential door-opening moment has passed or not, x=2 and the chance of winning is 1/2. If the presenter is even more unpredictable, and chooses a random time T to open the door (if at all) basing his decision on the contestant's current choice at that moment, and also accepting the contestant's 'final answer' at a predetermined but random time U after time T, things get harder for the contestant, even if he knows how T and U are distributed. Anyone have any idea what the optimal strategies are?
11 May 04
I've been thinking abou this, these answers imply something that is not implicit in the question. That is, that the host is opening a door he knows to be empty
It doesn't actually say that in the question, it could be one of those 'Let's see if you're right moments', in which case it makes no difference whether you change your door or not.
12 May 04
Originally posted by jimmyb270I am not sure I agree. I think that what is important is that the revealed door has nothing behind it, not what the host's knowledge is.
I've been thinking abou this, these answers imply something that is not implicit in the question. That is, that the host is opening a door he knows to be empty
It doesn't actually say that in the question, it could be one of those 'Let's see if you're right moments', in which case it makes no difference whether you change your door or not.
If he is without knowledge, he may open a door that has the prize behind it, in which case we are all done. But let's assume that (by chance instead of by knowledge) he opens a door that has nothing behind it.
In that case, you STILL have gained the added information as described above by iamatiger. In other words, you still effectively control two doors by switching and only one by staying.
14 May 04
Originally posted by Brother EdwinI'm sorry, but I'm not quite following the explanations of this riddle. I have a few questions:
...the gameshow host opens one of the doors which you have NOT picked and reveals it to be empty. He says you may change your choice to un opened door or stay with your original choice.
1) How is the wording SUPPOSED to read on the last line? I've been assuming that it's meant to be, "He says you may change your choice to AN UN-OPENED door or stay with your original choice." Is this the case? (side note: Isn't 'an un-opened door' and 'your original choice' essentially the same thing??)
2) Also, why has there been so much discussion about what the presenter knows/desires/plans? Isn't one of the 'rules' of riddling that all the information you need to solve the riddle has to be in the question itself? So, if the question doesn't refer to any inention/design on the part of the host, why are we talking about it?
3) If you (the contestant) KNOW that there are two 'empty' doors and one 'car' door, and the host has just shown you that one of the doors you DID NOT PICK is 'empty,' wouldn't that mean that, of the remaining 2 doors (one of which is the one you picked), one is 'empty' and one is 'car'? Therefore, wouldn't your chances of having the car behind the door you picked be 1/2? If you switched your choice to the other un-opened door, it would remain 1/2, would it not? So, isn't the answer, "It doesn't matter"?
Thanks; I am typically very bad at riddles, and this is probably one of those times, but this question seems very straightforward to me...
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14 May 04
1 edit
Originally posted by ddebenedActually, it does make quite a difference if the host doesn't know where the prize is:
I am not sure I agree. I think that what is important is that the revealed door has nothing behind it, not what the host's knowledge is.
If he is without knowledge, he may open a door that has the prize behind it, in which case we are ...[text shortened]... ffectively control two doors by switching and only one by staying.
(assuming doors a, b, c - prize is behind a)
We can populate the following table of possibilities:
1st pick, Host opens, Car revealed by host?, correct to switch?
a,b,no ,no
a,c,no ,no
b,a,yes,
b,c,no ,yes
c,a,yes,
c,b,no ,yes
Assuming the car was not revealed by the host, we know that either the 1st, 2nd, 4th or 6th possibility has happened - in exactly 2 of the 4 it is correct to switch and 2 of the 4 it is incorrect - so now you may as well stay with your first choice as it makes no difference at all whether you switch or not.
This thread will now degenerate into a rather long argument about whether my two posts on this topic are both correct. 😀
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21 May 04
5 edits
Originally posted by swisscheeseWhich bit stumped you? Here's the table with attempted more correct spacing (hard to do with the pesky blank space contraction here)
please explain in eng.
1st pick....Host opens....Car revealed by host?....correct to switch?
...door a......door b........................no.........................no.........
...door a......door c........................no..........................no.........
...door b......door a.......................yes.....................irrelevant....
...door b......door c........................no..........................yes........
...door c......door a.......................yes.....................irrelevant....
...door c......door b........................no..........................yes........
The point is that now the host doesn't know where the car is he might reveal the car by accident - which gives us the rows marked irrelevant in the table, since you would have no chance to switch if he did that. Because you know he hasn't revealed the car this increases the chance that your first choice was right.
Think of it as a bag with 99 black balls in and one red ball. With your eyes closed you reach in and take a ball and the host hides it without showing you its colour. You know you are very unlikely to have randomly picked the red ball.
If the host now looks in the bag and carefully removes 98 black balls, you know the ball remaining in the bag is very likely to be red as you were unlikely to randomly get the red ball at the start. In fact the chance is 1/100 that your first choice was right.
But if instead he randomly flings 98 balls out of the bag and they are all black then the fact that he didn't fling out the red ball by chance makes it more likely that red ball was the one you chose at the start. In effect the host has randomly selected a ball to leave in the bag and you randomly selected a ball to take out at the start. You know that against the odds one of those two balls turned out to be the red one so the chance that your choice is red is 1/2
21 May 04
Originally posted by iamatiger1st pick....Host opens....Car revealed by host?....correct to switch?
Which bit stumped you? Here's the table with attempted more correct spacing (hard to do with the pesky blank space contraction here)
1st pick....Host opens....Car revealed by host?....correct to switch?
1..door a......door b........................no.........................no.........
2..door a......door c........................no................. ...[text shortened]... one of those two balls turned out to be the red one so the chance that your choice is red is 1/2
1..door a......door b........................no.........................no.........
2..door a......door c........................no..........................no.........
3..door b......door a.......................yes.....................irrelevant....
4..door b......door c........................no..........................yes........
5..door c......door a.......................yes.....................irrelevant....
6..door c......door b........................no..........................yes........
I think it is agreed that this table represents the correct data up until the last column.
The question comes down to: What is the difference between scenarios 1, 2 and 4,6? I see no functional difference.
I have chosen one of the doors.
A door is opened.
It reveals no car.
My chosen door could still either have a car behind it or not.
The chances of me actually finding the car have gone up from 33% to 50%. Which is nice but randomly picking one of the remaining doors or not is not going to change a dang thing.
How would it?
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21 May 04
Originally posted by gregofthewebSorry - I stipulated in my message with the poorly formatted table that that the prize was behind door a. That's why it turns out to be correct to switch when you switch to a! I was just trying to show that the number of times it is correct to switch are equal to the number of times when it is incorrect to switch.
1st pick....Host opens....Car revealed by host?....correct to switch?
1..door a......door b........................no.........................no.........
2..door a......door c........................no..........................no.........
3..door b......door a.......................yes.....................irrelevant....
4..door b......door c....... ...[text shortened]... picking one of the remaining doors or not is not going to change a dang thing.
How would it?