PARADE!
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@dhromed said:
You could start with five doors, if you like, and have Monty open two of them and reveal they're empty, and then you make your pick from the remaining three.
Quite. So you agree, now?
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@TDWTF123 said:
@dhromed said:
You could start with five doors, if you like, and have Monty open two of them and reveal they're empty, and then you make your pick from the remaining three.
Quite. So you agree, now?And then you pick, and then monty opens another empty door, and then you switch for 2/3 chance of a prize.
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@Lorne Kates said:
If you set it up according to the conditions I've described, fine. But in fact you're just another idiot begging the question by assuming the truth of the very thing we're arguing about.In fact, I'll do you one better.
I will set up a website that simulates the MHP. I will hook it up to Paypal. I will let TDWTF play it. I will give him even payouts-- be bets $1, I will pay him $1 if he wins-- BUT ONLY if he agrees to never switch.
After all, this is a 50/50 game according to him, so even money is fair.
If he's right, at the end of, say, $10,000 worth of play, we should be even. Since it's a coin-flip game, I believe $10k of play should be enough to hit the long term, and we'll be within 1 standard dev. Neither of us will have really lost anything.
Or if I'm right, which I obviously am not-- he'll be out around $3K and an idiot will have been rightfully parted from his money.
I do like the way you've set up what you think is a no-lose bet for yourself, and then called me an idiot if I don't take it. If I'm right, you lose nothing, but if you are, I lose $3k. Does that sound like a bet anyone but an idiot would take?It is yet another demonstration that someone attempting to 'school' me doesn't actually have any abilities with maths or logic whatsoever.
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@TDWTF123 said:
Obviously, two-thirds of the time switching will win. That's simply going to follow from your starting conditions. The question is whether those starting conditions are correct, and you can't prove or disprove that experimentally.
We know our starting conditions are correct because we read the problem.
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@TDWTF123 said:
All your experiment proves is that the two-thirds answer follows from your interpretation,
Given that dhromed's interpretation is obviously the correct one, following as it does from a literal reading of a clear and unambiguous statement of the problem to be solved, I would be interested to hear from you - specifically - where the ambiguity that leaves room for your alleged viewpoint is.
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@dhromed said:
But you just agreed that there are only two doors, so why are you back on thirds again?@TDWTF123 said:
@dhromed said:
You could start with five doors, if you like, and have Monty open two of them and reveal they're empty, and then you make your pick from the remaining three.
Quite. So you agree, now?And then you pick, and then monty opens another empty door, and then you switch for 2/3 chance of a prize.
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@TDWTF123 said:
there are only two relevant doors
Which of the three doors are the relevant two doors? Obviously, this is only obvious to you.
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@flabdablet said:
@TDWTF123 said:
I suggest reading the fucking thread, then, since we've been over and over this.All your experiment proves is that the two-thirds answer follows from your interpretation,
Given that dhromed's interpretation is obviously the correct one, following as it does from a literal reading of a clear and unambiguous statement of the problem to be solved, I would be interested to hear from you - specifically - where the ambiguity that leaves room for your alleged viewpoint is.
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@dhromed said:
@TDWTF123 said:
Once more, what we're arguing about is the correct interpretation. All your experiment proves is that the two-thirds answer follows from your interpretation,
Why are you opposed to the literal interpretation?That's a good question. His thing seems to be making extra assumptions and then never letting go of them. It's why I do all my posting in Excel these days.
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@flabdablet said:
@Lorne Kates said:
AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHYour avatar is scary.
Sorry, I'll change it.
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@TDWTF123 said:
@anonymous234 said:
And yet, people who don't switch doors after picking one have exactly a 1/3 chance of of getting the prize. Where did that 3 come from?
Another question-begger. If there are only two doors, then the people who don't switch have a 1/2 chance of winning, same as those who do.Alright. I started trying to describe the experiment formally here but then realized the best way to describe it formally was with Python.
<html><body style='color:#000000; background:#ffffff; '>
import random def play_game(): #let's set up the environment doors = ['A', 'B', 'C'] winning_door = random.choice(doors) #Now let's ask the player to pick a door. #"Hmm, I have no information so I'm just going to pick a random one" player_door = random.choice(doors) #now the host opens a different door doors_that_can_be_opened = [d for d in doors if d != player_door and d != winning_door] opened_door = random.choice(doors_that_can_be_opened) #we remove it from the relevant doors doors.remove(opened_door) #now we switch, or not player_wants_to_switch = True if player_wants_to_switch: player_door = doors[0] if player_door == doors[1] else doors[1] #uglyyyyyy #did he win? if player_door == winning_door: return True else: return False
And now we run it:
>>> sum([play_game(True) for i in range(100000)])/100000.0 0.6688 >>> sum([play_game(False) for i in range(100000)])/100000.0 0.33385
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@Lorne Kates said:
Show me an Excel, you idiot.
You can't demonstrate a language trick using Excel. What the fuck is wrong with you people?
@Lorne Kates said:The burden of proof is on you
And it's yet another bit of evidence that you lot know nothing about logic. No, you can't prove a negative. You are making the claim that the third door is relevant, but you're unable to explain why or demonstrate that it is without first assuming it to be so.
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@Lorne Kates said:
Because non-idiots commonly take bets that have a downside (however unlikely) but no upside?@TDWTF123 said:
Does that sound like a bet anyone but an idiot would take?
Yes.
Is the problem here that you've confused 'idiot' and 'normally intelligent functioning adult'?
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@TDWTF123 said:
I suggest reading the fucking thread, then, since we've been over and over this.
Kindly provide a link to the comment where you already identified the specific ambiguity in the way the problem is worded that could conceivably lead to your alleged viewpoint. If you can do this, I'm prepared to concede that my reading of your points has been less than thorough. But of course you can't, because you're merely a not-very-skilful troll.
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@Lorne Kates said:
Still no good? OK, I try again.@flabdablet said:
@Lorne Kates said:
AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHYour avatar is scary.
Sorry, I'll change it.
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@TDWTF123 said:
@Lorne Kates said:
Show me an Excel, you idiot.
You can't demonstrate a language trick using Excel. What the fuck is wrong with you people?You fucking ivory tower denier. Of course you can!
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@TDWTF123 said:
You can't demonstrate a language trick using Excel.
I can. I did. You're claiming "50%" odds because of a language trick. Prove it. Otherwise you're just an idiot.
@TDWTF123 said:
What the fuck is wrong with you people?
Nothing's wrong with us. If you think everyone around you is wrong, either you have proof, or you're an idiot.
Guess which one you are.
(hint: you're an idiot)
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@Lorne Kates said:
Nothing's wrong with us.
I'd say the fact that so many of you are hooked by such an obvious troll shows that there is something wrong with you. It has been entertaining though, so well played TDWTF123 you somehow managed to hook a great many with an amazingly small hook.
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@boomzilla said:
@TDWTF123 said:
@Lorne Kates said:
Show me an Excel, you idiot.
You can't demonstrate a language trick using Excel. What the fuck is wrong with you people?You fucking ivory tower denier. Of course you can!
He argues a linguistic trick, but my example code has the full text of the problem explanation in the comments, and every word in the specification (original text of the problem) is accounted for. So, if there was a language trick, my simulation should give the results he expects.
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@joe.edwards said:
He argues a linguistic trick, but my example code has the full text of the problem explanation in the comments, and every word in the specification (original text of the problem) is accounted for. So, if there was a language trick, my simulation should give the results he expects.
And it did! The result he expects is for people to argue with him.
I think it's more fun and healthier and less illegal to take aggression out on stupid trolls like TDWTF123 than beating your dog or kids. So I'd like to thank TDWTF123 for looking out for those who, like himself, are unable to defend themselves.
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@locallunatic said:
hooked by such an obvious troll
There's a difference between being hooked and just playing along.
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Now I know why this question (and the airplane/treadmill one) are banned at straightdope.com. Christ.
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@blakeyrat said:
Now I know why this question (and the airplane/treadmill one) are banned at straightdope.com. Christ.
Only 0.999...% of the time, so there's still a small chance to ask them.
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@Lorne Kates said:
@blakeyrat said:
Now I know why this question (and the airplane/treadmill one) are banned at straightdope.com. Christ.
Only 0.999...% of the time, so there's still a small chance to ask them.
Do they ban you instantly, or is there a speed-of-light delay?
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@blakeyrat said:
Now I know why this question (and the airplane/treadmill one) are banned at straightdope.com. Christ.
Wow, they are? That's wisdom right there.
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@joe.edwards said:
Do they ban you instantly, or is there a speed-of-light delay?
It will appear to you, moving at c, as though you posted and got banned at the same time, but the mods will testify that they read your post first, and then banned you.
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@dhromed said:
@joe.edwards said:
Do they ban you instantly, or is there a speed-of-light delay?
It will appear to you, moving at c, as though you posted and got banned at the same time, but the mods will testify that they read your post first, and then banned you.
Only if you actually believe Eienstein's Folly.
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@Lorne Kates said:
Only if you actually believe Eienstein's Folly.
Look, the very first sentence of Special Relativity: The Paper is:
@Einstein said:
It is known that Maxwell’s electrodynamics as usually understood at the present time-when applied to moving bodies, leads to asymmetries which do not appear to be inherent in the phenomena.
And I have no idea what I'm trying to say by that.
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@Lorne Kates said:
@dhromed said:
@joe.edwards said:
Do they ban you instantly, or is there a speed-of-light delay?
It will appear to you, moving at c, as though you posted and got banned at the same time, but the mods will testify that they read your post first, and then banned you.
Only if you actually believe Eienstein's Folly.
I don't think anyone here is disputing that Einstein was just a pawn of the Illuminati spreading the myth of relativity. The question is whether or not this problem actually involves relativity. The speed of light is actually irrelevant. It's a simple linguistic trick which is so obvious I'm not going to bother pointing out where.
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Did somebody say counterexamples to relativity?
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@anonymous234 said:
Did somebody say counterexamples to relativity?
The very existence of that website killed my hope for humanity.
It's time to start drinking.
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@dhromed said:
@TDWTF123 said:
[quote user="dhromed"]You could start with five doors, if you like, and have Monty open two of them and reveal they're empty, and then you make your pick from the remaining three.
Quite. So you agree, now?And then you pick, and then monty opens another empty door, and then you switch for 2/3 chance of a prize.
[/quote]
Hmm... No, you'd get 3/5.
As I understand it, the expanded theory is that the group of doors, among which one is opened, holds the same probability of having the prize amongst them both before and after the opening. Therefore, the others of the group get "boosted" by dividing the share of the opened door amongst themselves equally.
This way, after 2/4 "unselected doors" were opened, the other 2 were at 2/5 chance each. Switching now would reserve a 2/5 door, the others being 1/5 and 2/5.
One more opens, and the other is 3/5.
Most gain would be had by holding after the first 2 doors are opened. The third door to be opened would then be in the same group of 4 doors, getting thus a 4/5 chance to be gained from the switch when only two doors are left in total.
Then again, I could be mistaken. Someone care to expand the python example?
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@joe.edwards said:
@anonymous234 said:
Is that a real site? It seems more like a mockery of conservatism than a site run by a legitimate conservative.Did somebody say counterexamples to relativity?
The very existence of that website killed my hope for humanity.
It's time to start drinking.
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@mott555 said:
@joe.edwards said:
@anonymous234 said:
Is that a real site? It seems more like a mockery of conservatism than a site run by a legitimate conservative.Did somebody say counterexamples to relativity?
The very existence of that website killed my hope for humanity.
It's time to start drinking.
Define legitimate. Kooks come in all flavors. To the best of my knowledge, it is meant to be serious. But then so is the New York Times.
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@mott555 said:
Is that a real site? It seems more like a mockery of conservatism than a site run by a legitimate conservative.
There is no consensus on what percentage of that site is serious and what percentage is trolling. But there is definitely some of both.
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@mott555 said:
@joe.edwards said:
@anonymous234 said:
Is that a real site? It seems more like a mockery of conservatism than a site run by a legitimate conservative.Did somebody say counterexamples to relativity?
The very existence of that website killed my hope for humanity.
It's time to start drinking.
Speaking of Poe's law? <a href="http://sciencedefeated.wordpress.com/2008/11/05/09999-1/>You ain't seen nothing, <a href="http://sciencedefeated.wordpress.com/2008/11/11/negative-numbers/>son.
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@mott555 said:
@joe.edwards said:
@anonymous234 said:
Is that a real site? It seems more like a mockery of conservatism than a site run by a legitimate conservative.Did somebody say counterexamples to relativity?
The very existence of that website killed my hope for humanity.
It's time to start drinking.
@conservapedia.com said:
16. The action-at-a-distance of quantum entanglement.[note 4]
17. The action-at-a-distance by Jesus, described in John 4:46-54, Matthew 15:28, and Matthew 27:51.
So...Jesus was a form of quantum entanglement. Interesting...
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@ookami said:
So...Jesus was a form of quantum entanglement. Interesting..
The action-at-a-distance-ness of quantum entanglement is still a contented issue. On the other hand it's plain to see that Jesus is spooky-action-at-a-distant-future.
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@mott555 said:
Is that a real site?
No. But this one is.@Monty Hall Problem said:The Monty Hall problem describes a real-life situation which contestants on Monty's popular daytime television program The Price Is Right would often find themselves to be a part of. In the scenario, a contestant is presented with three doors. One of the doors has a car behind it, and the other two conceal goats. The object of the exercise is for the contestant to pick the door with the car behind it, thus winning the car. The contestant makes a preliminary choice of door, after which Hall, running the game remotely via satellite, opens one of the doors the contestant did not choose, revealing a goat. The contestant can then discuss with the goat whether he or she should stick with the door they have already picked, or switch to the other. One of the goats will always speak the truth, the other will always lie. Although 'speak' here is a purely metaphorical term, as the goats communicate by clomping their hooves on the floor. If Monty Hall's satellite is not working correctly, and it picks a door at random instead of always picking a door with a goat, then the game is forfeit and the contestant must settle for consolation prizes such as vitamin supplements and turtle wax. Mathematicians have proven that it is possible to argue about the implications of statistics in the Monty Hall problem for a long time, ignoring the non-mathematicians present who would prefer to talk about something else.
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@flabdablet said:
@Monty Hall Problem said:
The Monty Hall problem...
This makes more practical sense than the original.
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@dhromed said:
This makes more practical sense than the original.
No more than I'd expect from such an impeccable source.
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@Monty Hall Problem said:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
If this is all you are given, you DO NOT KNOW. Because the rules, as written here, do not state that the other door MUST be opened. It only states that the other door WAS opened.
Consider the case where Monty Hall hates your guts because you ran over his dog in the parking lot. If you pick a wrong door, he's not going to offer you another door. He's just going to make you lose, instantly. But if you pick the right door, he's going to try to lure you away from that door. So, if you switch doors when given the opportunity, your odds drop to 0%. The only way to win is to get lucky on the initial choice and then stick with it.
Conversely, maybe you pulled Monty Hall's kid out of a burning building and he really wants you to win. So if you pick the correct door, he's just going to make you win instantly. But if you pick a wrong door, he's going to open the other wrong door and give you the opportunity to switch. In this scenario, switching gives you a 100% chance of winning.
You could even think of a scanario where Monty is rather indifferent. Although he was going to open a remaining door no matter what, he decided which remaining door to open based on a coin flip. It could have contained a car (leaving you with a choice between two goats) but as it happens, it contained a goat. Switching and staying would give the same odds in this scenario.
So in summary: Switching works unless Monty hates you. You can't calculate the probability of winning by switching unless you know his mindset.
(If you add the condition that Monty MUST open a door with a goat and give you a chance to switch - then yes, your odds if you switch are 2/3. But it is not the same thing to say "a door with a goat was opened" and "a door with a goat must be opened.")
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@ceasar said:
You could even think of a scanario where Monty is rather indifferent. Although he was going to open a remaining door no matter what, he decided which remaining door to open based on a coin flip. It could have contained a car (leaving you with a choice between two goats) but as it happens, it contained a goat. Switching and staying would give the same odds in this scenario.
If Monty has no choice but to open a door, then these nine ways of reaching the described scenario remain equally likely and in six of those nine cases it is to your advantage to switch. Note that when the description says that Monty "must" open a particular door, you should read that as meaning that he must do so in order to reach the described scenario. Whether he also has the option of doing something else in order to reach some other scenario is irrelevant to the problem as posed.
Only if Monty has a choice of whether to open a door do your chances depend on his mindset.
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@flabdablet said:
If Monty has no choice but to open a door, then these nine ways of reaching the described scenario remain equally likely and in six of those nine cases it is to your advantage to switch.
Even if he must open a door, you do not get the 2/3 odds if he was allowed to show you the door with the car. (And that's why people were saying it depends on the wording of the problem.) Without that restriction, any of the remaining possibilities are equally likely.
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If I open the door with the car behind it, there is a p=1 chance that I have chosen the car.
If I open the door without the car behind it, there is a p=0 chance that I have chosen the car.
P(x|x) = 1
P(x|¬x) = 0
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@Ben L. said:
If I open the door with the car behind it, there is a p=1 chance that I have chosen the car.
You don't get to open any door until after you've decided whether to switch or not, so the probabilities you've just quoted are irrelevant to the scenario described in the problem statement.If I open the door without the car behind it, there is a p=0 chance that I have chosen the car.
P(x|x) = 1
P(x|¬x) = 0
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@ceasar said:
Even if he must open a door, you do not get the 2/3 odds if he was allowed to show you the door with the car.
Yes you do, because the only way Monty's choice of door can affect your chances is if the fact of his choice was conditional upon your original choice of door. If (a) Monty is committed to opening a door you didn't choose regardless of your choice and (b) Monty has opened a door and revealed a goat, then there really are only nine ways that that can happen, all nine are equally likely, and you know for a fact that the door that you didn't choose and he didn't open is twice as likely to conceal the car as the one you initially chose.
If Monty was not pre-committed to opening a door - that is, if Monty could have chosen to end the game early and declare you an instant winner or loser, but has not done so - then you know for sure that Monty has opened the door that he did because he has chosen not to make such a declaration. In that circumstance only, Monty's attitude toward you becomes relevant to your evaluation of your chances. You and I are in heated agreement on that point.
There is a probability 1/2 variant of this game where Monty does not know which door has the car behind it, but that's not the game in the problem statement (which says explicitly that Monty does know that).
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@flabdablet said:
There is a probability 1/2 variant of this game where Monty does not know which door has the car behind it, but that's not the game in the problem statement (which says explicitly that Monty does know that).
Yes, it was specified that he knew where the car was. But it doesn't matter if he knows if he doesn't USE that knowledge, and it was NOT specified that he used that knowledge. If he knows which door has the car but instead chooses to flip a coin, that's the same as if he didn't know.
The possibilities if he flips a coin are:
1 You choose A, car is in A, Montey opens B. Stay wins.
2 You choose A, car is in A, Montey opens C. Stay wins.
3 You choose A, car is in B, Montey opens B. This did not happen since we know a goat was shown.
4 You choose A, car is in B, Montey opens C. Switch wins.
5 You choose A, car is in C, Montey opens B. Switch wins,
6 You choose A, car is in C, Montey opens C. This did not happen since we know a goat was shown.
(You could repeat for choosing B and C, but they'll be essentially the same.)
The 2/3 trick only works if Montey is forbidden from choosing 3 or 6. and must instead choose 4 or 5. If he was free to choose 3 or 6 and they merely didn't happen, that does nothing to differentiate 1 and 5 or 2 and 4 from the player's perspective.
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@ceasar said:
it doesn't matter if he knows if he doesn't USE that knowledge, and it was NOT specified that he used that knowledge.
His knowledge was mentioned in the problem statement, and is therefore a relevant factor in determining the solution. The only way that this can be so is if it does indeed force Monty to choose a losing door to open for you; if Monty's knowledge does not constrain his choice of doors, the problem has no solution for the reasons you've outlined above. If the problem has a solution at all, it is that you should switch doors. I have yet to see anybody identify an ambiguity in the wording as given that could support a reading where staying and switching are guaranteed equal chances of a win; for all his bluster, TDWTF123 certainly failed to do so.
