Brain Teasers, Riddles, etc...
Still nobody got my second one correct. I gave a hint that everyone was omitting the human element.
Think. You're on a game show. Regardless of your source and how trustworthy it is, WHY would the host give you a chance to switch doors unless you chose the right one? It's a win situation for the game host if you get a donkey.
-Albatross
Think. You're on a game show. Regardless of your source and how trustworthy it is, WHY would the host give you a chance to switch doors unless you chose the right one? It's a win situation for the game host if you get a donkey.
-Albatross
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About the game show one:
Duoas was correct, Grey Wolf and Zykker are incorrect. Statistically, the correct answer is 50/50. Which door you picked the first time has absolutely no impact on the odds of picking the right door the second time. Ultimately, you're being asked to pick from one of two doors. The revealed door is statistically insignificant.
Here are all the possible outcomes:
3 doors, A, B, and C
in the form 'first selection, revealed, correct answer'. IE: 'BCA' means you selected door B first, door C was reveled as the wrong answer, and door A had the car.
Rows followed by a * mean you get the correct answer if you switch:
ABA
ABC *
ACA
ACB *
BAB
BAC *
BCA *
BCB
CAB *
CAC
CBA *
CBC
There are 6/12 situations where you will get the car by switching doors
and 6/12 situations where you will get the car by keeping the original door you picked.
IE: 50/50
Introductory statistics: Don't get distracted by irrelevent information. If you flip a coin 49 times in a row and it comes up heads every time... the 50th time you flip STILL has a 50/50 chance of being heads. It doesn't matter what the other coin flips were.
Duoas was correct, Grey Wolf and Zykker are incorrect. Statistically, the correct answer is 50/50. Which door you picked the first time has absolutely no impact on the odds of picking the right door the second time. Ultimately, you're being asked to pick from one of two doors. The revealed door is statistically insignificant.
Here are all the possible outcomes:
3 doors, A, B, and C
in the form 'first selection, revealed, correct answer'. IE: 'BCA' means you selected door B first, door C was reveled as the wrong answer, and door A had the car.
Rows followed by a * mean you get the correct answer if you switch:
ABA
ABC *
ACA
ACB *
BAB
BAC *
BCA *
BCB
CAB *
CAC
CBA *
CBC
There are 6/12 situations where you will get the car by switching doors
and 6/12 situations where you will get the car by keeping the original door you picked.
IE: 50/50
Introductory statistics: Don't get distracted by irrelevent information. If you flip a coin 49 times in a row and it comes up heads every time... the 50th time you flip STILL has a 50/50 chance of being heads. It doesn't matter what the other coin flips were.
@Albatross
Thats just obnoxious. But anyway, You were told before you got on the show that you would be given the chance to switch your door. So it wouldnt matter if you had chosen the correct door, because the host would have asked anyway.
Thats just obnoxious. But anyway, You were told before you got on the show that you would be given the chance to switch your door. So it wouldnt matter if you had chosen the correct door, because the host would have asked anyway.
@Zykker:
I never said by whom. C'mon, have a little imagination!
@Disch:
Probability tree, please.
-Albatross
I never said by whom. C'mon, have a little imagination!
@Disch:
Probability tree, please.
-Albatross
| @Disch: Probability tree, please. |
Is my list of all possible permutations really not good enough? Can't you just use that to construct your own tree(s)?
I don't have time to make a tree now because I have to go to work, but I might during my break.
closed account (z05DSL3A)
Disch, you are wrong.
No, you are still being asked to pick from three doors, but you know the probability of one of those doors is now zero.
You are effectively being asked to group the doors in two groups. You pick one door to go into group one the other two doors go into group two.
So the winning door has a 1:3 chance of being in group 1 and a 2:3 chance of being in group 2.
Now if you open one of the doors in group 2, revealing that it is not the winning door, you have not changed the chances of the winning door being put into that group. so the remaining door still has 2:3 chance of being the winning door.
| Ultimately, you're being asked to pick from one of two doors. |
No, you are still being asked to pick from three doors, but you know the probability of one of those doors is now zero.
You are effectively being asked to group the doors in two groups. You pick one door to go into group one the other two doors go into group two.
So the winning door has a 1:3 chance of being in group 1 and a 2:3 chance of being in group 2.
Now if you open one of the doors in group 2, revealing that it is not the winning door, you have not changed the chances of the winning door being put into that group. so the remaining door still has 2:3 chance of being the winning door.
@Zykker:
*Clap* *clap* *clap*
@Disch:
If you can draw a mathematically sane tree showing that the probability is really 50/50, then you'd be pretty smart. That said, your list of possible combinations is not mathematically sane, because you forgot to list the probability of each happening.
EDIT: Wow... I knew this would draw a lot of posts, but didn't expect this much discussion.
-Albatross
*Clap* *clap* *clap*
@Disch:
If you can draw a mathematically sane tree showing that the probability is really 50/50, then you'd be pretty smart. That said, your list of possible combinations is not mathematically sane, because you forgot to list the probability of each happening.
EDIT: Wow... I knew this would draw a lot of posts, but didn't expect this much discussion.
-Albatross
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| No, you are still being asked to pick from three doors, but you know the probability of one of those doors is now zero. |
But you can't pick the revealed door, so you're limited to two doors.
The fact of the matter is, once one of the doors is revealed, that means the car have to be behind one of the TWO other doors. One of the two = 50% chance. Which door you picked initially has absolutly no effect on which door the car is behind.
| So the winning door has a 1:3 chance of being in group 1 and a 2:3 chance of being in group 2. |
This is crap. Run a simulation. I really wish I wasn't at work right now or I'd write one myself.
If door A is revealed, then the car must be behind either door B or C. What difference does it make which of those doors you picked initially? How could picking door C initially possibly make the odds of the car being behind door any B more likely? The car is either behind door B or it's not. Your initial choice doesn't magically improve your odds somehow... that's nonsense.
You guys are letting unrelated information cloud the simple fact that you're choosing between two doors.
Past events don't matter. Flip a coin 49 times, etc. What only matters is the most recent decision. The most recent decision is you choosing between 2 doors.
| If you can draw a mathematically sane tree showing that the probability is really 50/50, then you'd be pretty smart. |
I'm no expert on probability trees, but the logic here isn't all that complicated.
I really thought my COMPLETE list of all possible outcomes would make this pretty open and shut. I mean there it is ... right there, just look at it.
All possible situations, and all possible outcomes. And based on all situations and outcomes, half of the outcomes result in you getting the car.
I don't know how I can explain it any better, but I know for a fact I'm right.
I'm with Disch, I don't know why Albatross is arguing tbh. Disch's explanation seems fairly sound to me.
Edit: As Disch said; you can't pick the first door twice so you're only choosing one of two doors. You don't know which of the two has the car. It could be one of two. One divided by two is 0.5, or a half, or 50%. You therefore have a 50% chance of choosing the car (those are excellent odds, too; where can I sign up for this game show?).
Edit 2: Here's your blasted probability tree: http://i40.tinypic.com/rjqp2r.png
Edit: As Disch said; you can't pick the first door twice so you're only choosing one of two doors. You don't know which of the two has the car. It could be one of two. One divided by two is 0.5, or a half, or 50%. You therefore have a 50% chance of choosing the car (those are excellent odds, too; where can I sign up for this game show?).
Edit 2: Here's your blasted probability tree: http://i40.tinypic.com/rjqp2r.png
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closed account (z05DSL3A)
Disch,
I haven't read fully read your post yet, but for now I will say that the Monty Hall problem is a well known veridical paradox, Google it. It does not seem correct but it is better to switch (and it is a 2:3 probability).
The main thing you seem to be missing is the choice is made when there are three doors, not two...
I haven't read fully read your post yet, but for now I will say that the Monty Hall problem is a well known veridical paradox, Google it. It does not seem correct but it is better to switch (and it is a 2:3 probability).
The main thing you seem to be missing is the choice is made when there are three doors, not two...
I agree with Ditch, too. The probability is 0.5.
If you picked the winning door, then both other doors have donkeys behind them, and if you didn't, there's still one more door with a donkey behind it, so if the host knows the solution, he can always pick a door with a donkey behind it. The first door you pick has no influence on the final outcome. After one donkey is revealed, you only have two choices left, since you can't rationally choose the donkey.
If you picked the winning door, then both other doors have donkeys behind them, and if you didn't, there's still one more door with a donkey behind it, so if the host knows the solution, he can always pick a door with a donkey behind it. The first door you pick has no influence on the final outcome. After one donkey is revealed, you only have two choices left, since you can't rationally choose the donkey.
The probability of you getting the car on the first go is 1/3; but once you've picked the donkey the probability becomes 1/2 because there are only two doors left to pick, one with the donkey, and one without; thus, the probability of getting the car is 0.5.
Oh, I think I see the problem.
The question is
In other words, what's the probability that the car is behind the door you didn't choose, which is indeed not even.
Ditch, chrisname, and I have been calculating the probability a player has of winning if he randomly chooses to stay or switch. That probability is even.
The question is
| What is the probability that changing the door will get you the car? |
Ditch, chrisname, and I have been calculating the probability a player has of winning if he randomly chooses to stay or switch. That probability is even.
closed account (z05DSL3A)
| ...the probability becomes 1/2 because there are only two doors left to pick... |
but you are NOT picking again, you are deciding on whether or not to change. Your initial choice was made at 1:3 this will not change, therefore the other door is 2:3
| In other words, what's the probability that the car is behind the door you didn't choose, which is indeed not even. Ditch, chrisname, and I have been calculating the probability a player has of winning if he randomly chooses to stay or switch. That probability is even. |
No, it's the same problem, just phrased differently.
If you picked door A and door C is revealed.. then:
You either pick door A again (keep the door)
or
You pick door B (change the door)
It's still 50/50.
| WHY would the host give you a chance to switch doors unless you chose the right one? |
Probability here is 50:50
| No, it's the same problem, just phrased differently. |
No matter which door you initially pick, you have a 1/3 chance of having chosen the winning door, which means the other two doors combined have a 2/3 chance. After one donkey is revealed, your door still has 1/3 chance, but since now one door has a 0 chance, the other door now has a 2/3 chance. This is a case where the probability of choosing (assuming random behavior on the player's part) the winning door is not equal to the probability of a door being the winning door.
A player that consistently chooses to switch, has a 2/3 chance of winning.
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closed account (z05DSL3A)
http://en.wikipedia.org/wiki/Monty_Hall_problem
http://barryispuzzled.com/zmonty
http://www.cut-the-knot.org/peter.shtml
...
Yes you have two doors left, but they do not have an equal probability of being the winning door.
http://barryispuzzled.com/zmonty
http://www.cut-the-knot.org/peter.shtml
...
| If you picked door A and door C is revealed.. then: You either pick door A again (keep the door) or You pick door B (change the door) It's still 50/50. |
Yes you have two doors left, but they do not have an equal probability of being the winning door.
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