Re: Monty Hall, we have a PROBLEM
This whole thread hurts my head.
The previous statement is false.
This whole thread hurts my head.
The previous statement is false.
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Monty Hall, we have a PROBLEM
- Thread starter owslachief
- Start date
owslachief
occupe toi de tes oignons
Re: Monty Hall, we have a PROBLEM
If you think of it as N people trying to dodge N-1 persons' birthday's it's apparent that N would be pretty low (like 23) to make it a safe bet.
If you think of it as one person trying to dodge another person's birthday, you'd tend to to make that safe-bet number high.
If you think of it as N people trying to dodge N-1 persons' birthday's it's apparent that N would be pretty low (like 23) to make it a safe bet
.
bigblue_dl
Armed
Re: Monty Hall, we have a PROBLEM
Why not ask a question you know the answer to, and go in the door guarded by the one that gives the correct answer? There were no limits of what the question has to be, so ask: Who won Super Bowl I, or Who was the 1st president of the USA? or What is capital city of Ohio?
owslachief
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Re: Monty Hall, we have a PROBLEM
4. The question must be one whose answer is unknown to you beforehand (except regarding the facts given in the other conditions listed).
Thanks for bringing that up; I think we all missed that. There needs to be one more condition (thanks to my 13-year-old niece who said "Just ask one of the guardians if Elvis is dead."):
4. The question must be one whose answer is unknown to you beforehand (except regarding the facts given in the other conditions listed).
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Re: Monty Hall, we have a PROBLEM
Or, again, just assign the guardians to random doors. The Elvis question then tells you which is the liar, but doesn't tell you which door to pick. It's just a much cleaner way to state the problem.
Twitch Boy
Defacing this tagline is a MAX foul
Re: Monty Hall, we have a PROBLEM
The way I've always heard is to ask "If I asked your partner which is the correct door, what would he say?" Either you asked the truth teller and he'll tell the truth about the liar lying, or you asked the liar and he'll lie about the truth teller telling the truth. Both lead to the same outcome: the WRONG door. Take whichever door they don't indicate.
Alternatively, you could ask "If I asked YOU if this is the correct door, what would YOU say?" then take the door that IS indicated. Either you asked the truth teller and he tells the truth about telling the truth, or you asked the liar and he will lie about lying, telling the truth in the process. Kind of a copout for "one question" though if you ask me.
The way I've always heard is to ask "If I asked your partner which is the correct door, what would he say?" Either you asked the truth teller and he'll tell the truth about the liar lying, or you asked the liar and he'll lie about the truth teller telling the truth. Both lead to the same outcome: the WRONG door. Take whichever door they don't indicate.
Alternatively, you could ask "If I asked YOU if this is the correct door, what would YOU say?" then take the door that IS indicated. Either you asked the truth teller and he tells the truth about telling the truth, or you asked the liar and he will lie about lying, telling the truth in the process. Kind of a copout for "one question" though if you ask me.
XYZ
Smoking in the selection room.
Yeah, in the examples I've seen of this problem it's been a given that the two are just standing in a room with two doors as opposed to each being in front of a door. Which are the guidelines I was referring to when I said a question that leads to the same answer from each is the goal, just to clarify.
owslachief
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Re: Monty Hall, we have a PROBLEM
* These are Andrew's possible sons-in-law
+ These are Bernard's possible daughters-in-law
% These are Bernard's son's possible sisters-in-law
# These are Claude's possible daughters-in-law
& These are Donald's possible sons-in-law
~ These are Donald's daughter's possible brothers-in-law
@ Possible fathers of Donald's son-in-law
^ Possible sons-in-law of the father of Donald's son-in-law
= Possible mothers-in-law of the son-in-law of the father of Donald's son-in-law
o Possible fathers of Bernard's daughter-in-law
x Possible daughters-in-law of the father of Bernard's daughter-in-law
- Possible mothers-in-law of the daughter-in-law of the father of Bernard's daughter-in-law
As I did this, I was able to start eliminating some relationship possibilities.
When I got to the end, it reduced to a mother-in-law/son-in-law relation between Eugene's "wife" and Bernard's son.
So Bernard's son married Eugene's daughter.
Thanks FreshFish for this time-consuming problem. I like probably many others beat my head against the wall until I realized that the last clue really is very important. Here is why, and what I did -> The last clue eliminated half the possible child-in-law relations right off. I then saw this as an exercise in writing out every possible relation given and assigning symbols to all 20 people involved (father, mother, son, daughter) and proceeded through all the clue like this:
* These are Andrew's possible sons-in-law
+ These are Bernard's possible daughters-in-law
% These are Bernard's son's possible sisters-in-law
# These are Claude's possible daughters-in-law
& These are Donald's possible sons-in-law
~ These are Donald's daughter's possible brothers-in-law
@ Possible fathers of Donald's son-in-law
^ Possible sons-in-law of the father of Donald's son-in-law
= Possible mothers-in-law of the son-in-law of the father of Donald's son-in-law
o Possible fathers of Bernard's daughter-in-law
x Possible daughters-in-law of the father of Bernard's daughter-in-law
- Possible mothers-in-law of the daughter-in-law of the father of Bernard's daughter-in-law
As I did this, I was able to start eliminating some relationship possibilities.
When I got to the end, it reduced to a mother-in-law/son-in-law relation between Eugene's "wife" and Bernard's son.
So Bernard's son married Eugene's daughter.
owslachief
occupe toi de tes oignons
Re: Monty Hall, we have a PROBLEM
P.S. There's probably a really good algebraic way to solve that brain-buster, but it was murder when I went down that path.
P.S. There's probably a really good algebraic way to solve that brain-buster, but it was murder when I went down that path
.Re: Monty Hall, we have a PROBLEM
You are in the hallway outside an apartment. There are three light switches on the wall next to the door. Each switch controls a lamp in the entryway on the other side of the door (you can't see the lights through or under or around the door). Your task is to identify which switch controls which lamp. You can arrange each of the switches to off or on as you like. You can only open the door once and cannot change the switch arrangements once you decide to open the door (you do get to walk inside after you open the door).
how do you figure out which switch controls which lamp?
You are in the hallway outside an apartment. There are three light switches on the wall next to the door. Each switch controls a lamp in the entryway on the other side of the door (you can't see the lights through or under or around the door). Your task is to identify which switch controls which lamp. You can arrange each of the switches to off or on as you like. You can only open the door once and cannot change the switch arrangements once you decide to open the door (you do get to walk inside after you open the door).
how do you figure out which switch controls which lamp?
unofan
Well-known member
Presuming the switches are identical for all three (thus if all three switches are in the same position, theyre all either on or off) and presuming we're still using incandescent lights: you start with all three switches in one position and then flip two switches and then wait 5 or 10 minutes. You then flip one of those back and immediately enter the room.
If two are on, the dark one is the one you switched once. The one you can still touch is the one you switched twice, and the hot one is the one you didn't switch at all.
If two are off, the one on is the one you switched once. The hot one is the one you switched twice, and the cold one is the one you didn't switch at all.
johnnypohlfan
New member
Re: Monty Hall, we have a PROBLEM
Something that I just can't wrap my head around with the original problem posted in this thread and the solution given.
When you first pick a door, your chances of it being right are one out of three.
When Monty shows you what's behind one of the other two doors, that action occurs AFTER you already made your choice.
The chances of your choice being right do not change as a result of something that happens subsequently.
Therefore, your chances of being right, if you stick with your original choice, are still one out of three.
Since there is now only one other door available, the chances of THAT door being right must be 1 minus 1/3, or two out of three.
Why is it that the odds for one door cannot change due to subsequent events, but the odds for another door can change. I've read enough about this problem to be convinced that this solution has merit. But I just don't understand how different mathematical rules apply to different doors.
Something that I just can't wrap my head around with the original problem posted in this thread and the solution given.
When you first pick a door, your chances of it being right are one out of three.
When Monty shows you what's behind one of the other two doors, that action occurs AFTER you already made your choice.
The chances of your choice being right do not change as a result of something that happens subsequently.
Therefore, your chances of being right, if you stick with your original choice, are still one out of three.
Since there is now only one other door available, the chances of THAT door being right must be 1 minus 1/3, or two out of three.
Why is it that the odds for one door cannot change due to subsequent events, but the odds for another door can change. I've read enough about this problem to be convinced that this solution has merit. But I just don't understand how different mathematical rules apply to different doors.
Twitch Boy
Defacing this tagline is a MAX foul
Re: Monty Hall, we have a PROBLEM
Think of your door as one package, and BOTH the doors you didn't pick together as Monty's package.
Odds of the car being in your package = 1/3.
Odds of the car being in Monty's package = 2/3.
Now, Monty knows where the goats are. Since there is one car and two goats, Monty's package of two doors must have at least one goat, and he knows where it is. We also know he will 100% show a goat. So really his showing a goat has not changed any probabilities because we know he has one and he's 100% going to show one.
Now if Monty DIDN'T know where the goats are, and/or could possibly (accidentally or on purpose) show the car, this whole problem falls apart.
Think of your door as one package, and BOTH the doors you didn't pick together as Monty's package.
Odds of the car being in your package = 1/3.
Odds of the car being in Monty's package = 2/3.
Now, Monty knows where the goats are. Since there is one car and two goats, Monty's package of two doors must have at least one goat, and he knows where it is. We also know he will 100% show a goat. So really his showing a goat has not changed any probabilities because we know he has one and he's 100% going to show one.
Now if Monty DIDN'T know where the goats are, and/or could possibly (accidentally or on purpose) show the car, this whole problem falls apart.
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Re: Monty Hall, we have a PROBLEM
The point is that your probability of being right initially *doesn't* change - it remains at 1 in a million. The probability that you got it right doesn't "magically" go up to 50-50 after the fact just because the host opened those other doors.
It's the order that counts. I had a hard time with this one the first time I saw it, until I thought about extending it - when the odds are all 1/2s and 1/3's, those are "pretty close" anyway, so it's hard to get an intuitive feel. But what if there were a million doors? Then, the probability that your door is correct is 1 in a million. The host then opens 999,998 doors, showing you that there's nothing behind them, leaving just two doors, yours and one other. Now, if you initially made that 1 in a million guess, then your door is correct. But if you got it wrong (as you will 999,999 out of a million times) then the prize is behind the other door. Therefore, your door has a 1 in a million chance of being right, and the other door has 999,999 in a million chance - in other words, its probability of being "right" is equal to the probability that your initial guess was "wrong."
The point is that your probability of being right initially *doesn't* change - it remains at 1 in a million. The probability that you got it right doesn't "magically" go up to 50-50 after the fact just because the host opened those other doors.
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Re: Monty Hall, we have a PROBLEM
Now, if he picks a door at random and DOES show the car, well, then I'd say you should DEFINITELY pick a different door!
I disagree. Even if he's guessing at random, if he picks a door and shows a goat, then the remaining door (not the one you initially picked and not the one he opened) still has a 2/3 chance of being correct. For it to be correct, you just had to get the initial door wrong, and you had a 2/3 chance of doing that.
Now, if he picks a door at random and DOES show the car, well, then I'd say you should DEFINITELY pick a different door!
unofan
Well-known member
I disagree. Even if he's guessing at random, if he picks a door and shows a goat, then the remaining door (not the one you initially picked and not the one he opened) still has a 2/3 chance of being correct. For it to be correct, you just had to get the initial door wrong, and you had a 2/3 chance of doing that.
Now, if he picks a door at random and DOES show the car, well, then I'd say you should DEFINITELY pick a different door!
But 1/3rd of the time he'd open the car, which leaves your overall likelihood at 1/3rd.
Go back to your millions example. In order to even make it to the final two, you have to survive 999,998 chances of him opening the door to the car.
If he really doesn't know where the car is, your overall odds do not change: 1/3rd of the time it'll be your door, 1/3rd it'll be the unopened door, 1/3rd it'll be the opened door.
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Twitch Boy
Defacing this tagline is a MAX foul
Re: Monty Hall, we have a PROBLEM
Mathematically, you're correct at the point a goat is shown. At the start, if the possibility is there that he could show the car, then we don't know it's 2/3 in your favor to switch YET. The possibility exists that he could open the car, leaving you a goat to switch to (illogical as that may be, it's mathematically possible, plus we have to assume that you can't switch to the open door, etc.)
My head hurts.
I disagree. Even if he's guessing at random, if he picks a door and shows a goat, then the remaining door (not the one you initially picked and not the one he opened) still has a 2/3 chance of being correct. For it to be correct, you just had to get the initial door wrong, and you had a 2/3 chance of doing that.
Now, if he picks a door at random and DOES show the car, well, then I'd say you should DEFINITELY pick a different door!
Mathematically, you're correct at the point a goat is shown. At the start, if the possibility is there that he could show the car, then we don't know it's 2/3 in your favor to switch YET. The possibility exists that he could open the car, leaving you a goat to switch to (illogical as that may be, it's mathematically possible, plus we have to assume that you can't switch to the open door, etc.)
My head hurts.
Re: Monty Hall, we have a PROBLEM
No. If he opens the car then your likelihood is *0*, not 1/3. If he opens a goat, then your overall likelihood remains at 1/3, so the remaining door must be 2/3. You're still better off switching if he opens a goat - even if he opens the goat based on chance. For that remaining door to have the car, all you had to do was make the "wrong" initial guess, and you had a 2/3 chance of doing so at the time you made your choice.
Re: Monty Hall, we have a PROBLEM
Ah - now that's a good point, probably why unofan and I are talking past each other. It makes a huge difference whether the host just randomly selects a door or if he actually OPENS the door before offering you the chance to switch. If you select, he selects, and then you have the chance to switch without knowing whether he selected the car, then it doesn't matter whether you switch. But if he opens his randomly selected door first and reveals a goat and THEN offers you a chance to switch, make the switch every time.
unofan
Well-known member
But 1/3 of the time you never get to switch, cause he opens the car. Which, as you say, reduces your chances to zero for that iteration.
If you survive that, you are correct that you should switch. But your overall odds of winning are only 1/3rd because you wont survive that everytime, unlike the original version of the problem where your overall odds are 2/3rds so long as you always switch.
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