Three Doors

[Scotch78]

There are three doors (A, B and C) in front of you. One, and only one, contains a prize. You choose a door, say A, and an empty door, say C, is revealed. Some girl is insisting to me that she took a stats course that told her B is now a 2-1 favorite over A. I'm telling her she's fool of it, but she keeps insisting her textbook proved it. Of course, she doesn't have the proof anymore. Anyone?

Scott

[mantasm]

SHe's talking about the monty hall problem, try googling that

[Scotch78]

Ok, so from what I found, the answer really does appear to be 33-67. So, there must be a fallacy somewhere in the 50-50 argument. Has it been identified yet?

Scott

[Arnfinn Madsen]

[ QUOTE ]
Ok, so from what I found, the answer really does appear to be 33-67. So, there must be a fallacy somewhere in the 50-50 argument. Has it been identified yet?

Scott

[/ QUOTE ]

2/3 of all instances you choose a door which does not contain the price. In those 2/3-instances the door not opened has a 100% of containing the price. In the 1/3-instances that you choose the correct door, the other door not opened has 0% chance of containing the price. Thus she is right (2/3 vs 1/3 is 2 vs 1).

Tell her you have never met such a smart girl and invite her for dinner /images/graemlins/cool.gif.

[kyro]

Welcome to the Probability section. I can see this is your first time here.

[mostsmooth]

[ QUOTE ]
Welcome to the Probability section. I can see this is your first time here.

[/ QUOTE ]
welcome to earth would have fit better. i didnt think there was a person alive that hasnt heard this one. should we let him in on the 3 guys and the bellboy and the missing $1? /images/graemlins/cool.gif

[BruceZ]

[ QUOTE ]
There are three doors (A, B and C) in front of you. One, and only one, contains a prize. You choose a door, say A, and an empty door, say C, is revealed. Some girl is insisting to me that she took a stats course that told her B is now a 2-1 favorite over A. I'm telling her she's fool of it, but she keeps insisting her textbook proved it. Of course, she doesn't have the proof anymore. Anyone?

Scott

[/ QUOTE ]

The way you phrased the problem, the answer could be either 1/2 or 2/3. It depends on whether or not the empty door was chosen randomly from all 3 doors so that the prize door could have been revealed, or if only an empty door could have been chosen. The later is the way the game is supposed to be played, and then the answer is 2/3.

[kyro]

[ QUOTE ]
[ QUOTE ]
Welcome to the Probability section. I can see this is your first time here.

[/ QUOTE ]
welcome to earth would have fit better. i didnt think there was a person alive that hasnt heard this one. should we let him in on the 3 guys and the bellboy and the missing $1? /images/graemlins/cool.gif

[/ QUOTE ]

Only if he tells us the odds of two people flopping a flush draw.

[Scotch78]

[ QUOTE ]
Only if he tells us the odds of two people flopping a flush draw.

[/ QUOTE ]

Assuming there are two people dealt two of the same suit as each other . . . with four of the suit seen and 48 unseen cards, I'm going to say it's ((9/48 + 9/47) * 8/46) - (9/48 * 8/47 * 7/46) = 6.1%.

And my guess for two people sharing four of the same suit is 12/51 * ((11/50 * 10/49) + ((13-m)/(52-n) + ((13-m1)/(52-n2) + . . .) where the m series represents the number of said suit previously dealt and the n series represents the number of total cards dealt, with ((n/2) - 1) = x terms in the parenthetical where x equals the number of players. I'm sure there's a formulaic way to write all that, but I'll let the mathematicians in here deal with that; I'm just a philosopher.

Now, what's this other brain teaser?

Scott

[kyro]

Touche...

3 guys stay at a hotel room. It comes to $30. The guys give the manager $10 each. Later, the manager realizes he overcharged the 3 men and sends his bellboy to the room to return $5. The bellboy, (who is a dishonest fellow), only gives $3 back to the men and pockets the other $2.

So the men each paid $9 for the room, or $27 altogher. The bellboy kept $2. $27 + $2 = $29.

Where'd the missing dollar go?

[capone0]

[ QUOTE ]
Touche...

3 guys stay at a hotel room. It comes to $30. The guys give the manager $10 each. Later, the manager realizes he overcharged the 3 men and sends his bellboy to the room to return $5. The bellboy, (who is a dishonest fellow), only gives $3 back to the men and pockets the other $2.

So the men each paid $9 for the room, or $27 altogher. The bellboy kept $2. $27 + $2 = $29.

Where'd the missing dollar go?

[/ QUOTE ]

So if he keeps 2 dollars. That means each guy pays 28/3 for the room instead of paying 25/3 for the room which was the intention of the manager. The thing is, there is no missing dollar. They started with 30 bucks. So they are all paying 10 dollars each. 28/3 does not equal nine dollars each. I know thats the tricky part. They are really paying 28/3 each, not 9, 9.3333333 or whatever.

[Scotch78]

Is there some etiquette rule in the probability forum that I'm unaware of and violated? Because unless I've acted like a newb so far, there's no need to insult me with kindergarten "problems". Answer: <font color="white"> The men paid $27. The hotel received $25 and the bellboy received $2. </font>

Scott

[LetYouDown]

You have. This question has been asked/answered several times. The search function is wonderful. I wouldn't be surprised if it's been asked in the past week.

The answer to the 3 doors question is intuitively obvious and frankly, less thought provoking than the missing dollar question...even though that's pretty straightforward.

[mostsmooth]

[ QUOTE ]
there's no need to insult me with kindergarten "problems".

Scott

[/ QUOTE ]
you started it

[kyro]

I was joking around.

You asked a question that gets asked at least once a month. mostsmooth referenced another problem that finds its way around these forum often, and I merely brought up another one. You asked me to propose the missing dollar riddle, so I did.

[danzasmack]

The missing dollar is behind door C.

next question?

[kyro]

[ QUOTE ]
The missing dollar is behind door C.

next question?

[/ QUOTE ]

/images/graemlins/grin.gif

[LetYouDown]

No, door C is empty. There's a 2/3 probability that it's behind B and a 1/3 probability it's behind A. Definitely switch.

[Scotch78]

[ QUOTE ]
I was joking around.

[/ QUOTE ]

I was too, but we all know humor can get lost online. I do disagree with the person who said the bellboy riddle is more challenging than the three doors problem though. One was incredibly easy for me and the other is still messing with me. I hate repeat questions in my home forum (HUSH) just as much as you guys probably hate this one, and if I had any idea that such a non-poker related thing was so common here (or that the name of it was Monty Hall), I wouldn't have posted without searching first. My apologies folks.

Scott

[kyro]

Don't sweat it too much. If you're still confused about the 3 doors problem, I'd suggest searching it. There are some great laid-out solutions for it. Good luck.

[Scotch78]

I'm pretty sure I understand why the answer is 2-1 for switching, what's tripping me up is the fact that there must now be a fallacy somewhere in the 1-1 argument and I can't find it.

Scott

[moomoocow]

Slight variation to move towards generalizing the 3 doors problem.

Instead of 3 doors - now 5 doors, 2 of them have prizes hidden. You choose one door. It is not opened. The host of the gameshow randomly chooses another door, opens it and shows you a prize. You cannot chosoe that door any longer -

You are given the option to switch your door choice (to any of the 3 other doors remaining). Do you do so (why/why not)?

I haven't worked through the answer yet but thoughts woudl be much appreciated.

Cheers,

Moo!

[moomoocow]

This is absolutely right - I actually had to run a spreadsheet to convince myself that this was the case. I ran a simulation of the 3 door case 250 times- where Hero always picks door 1, door 2 is always opened, and if door 2 is empty, then Hero has a chance to switch.

I get numbers hovering aroudn 50/50 for switching and sticking to door 1 strategies given door 2 is empty.

Moo!

[PairTheBoard]

Maybe you should lay out exactly what the 1-1 argument is. Also, clearly describe the procedure whereby door C got opened.

PairTheBoard

[Scotch78]

[ QUOTE ]
clearly describe the procedure whereby door C got opened.


[/ QUOTE ]

This part is what helps me understand why the probability is 2-1.

[ QUOTE ]
Maybe you should lay out exactly what the 1-1 argument is

[/ QUOTE ]

That's what I'm having trouble doing. Philosophy taught me how to critique an argument once it is made, but I don't know enough statistics to formulate this argument.

Scott

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
clearly describe the procedure whereby door C got opened.


[/ QUOTE ]

This part is what helps me understand why the probability is 2-1.

[ QUOTE ]
Maybe you should lay out exactly what the 1-1 argument is

[/ QUOTE ]

That's what I'm having trouble doing. Philosophy taught me how to critique an argument once it is made, but I don't know enough statistics to formulate this argument.


Scott

[/ QUOTE ]


I think the 1-1 argument goes something like this. Doors A and B were equally likely to have the prize before Door C was opened. Opening door C has no effect on whether the prize is in A or B. Therefore A and B remain equally likely and with door C eliminated are now both 50-50 to have the prize.

This argument is actually correct if the procedure for opening door C went like this. I choose door A. I'm thinking about whether to switch to door B. So I say, let's open door C and see what's in it. If the prize is in C then A and B remain equally likely, both with 0% chance. If the prize is not in C then A and B remain equally likely, both with 50% chance.

Of course that's not Monty's procedure. See what happens as you try to apply the 1-1 argument within Monty's procedure.

PairTheBoard

[Corleone]

[ QUOTE ]
The way you phrased the problem, the answer could be either 1/2 or 2/3. It depends on whether or not the empty door was chosen randomly from all 3 doors so that the prize door could have been revealed, or if only an empty door could have been chosen. The later is the way the game is supposed to be played, and then the answer is 2/3.

[/ QUOTE ] That did it for me...

I would also like to add that I'm obviously a noob, and that I find problems like this very interesting. Keep 'em coming!
/images/graemlins/smile.gif

[MontyHall]

Thank you for such an original problem that has never been posted here before.

[Jazza]

if you liked the three door riddle, here is one to ponder (i posted this in OOT a while ago):

you know a woman has 2 children, you ask her if she has at least one girl, she says yes, what is the chance the other child is a girl?

and a harder one:

assume that when naming a girl, parents will name the girl Lisa with probability 1/1000 (even if they already have a daughter named Lisa)
you know a woman has two children, you ask her if she has at least one girl named Lisa, she says yes, what are the chances the other child is a girl?

[moomoocow]

Is the answer 1/3 in the first problem?

and approximately 1/2 in the second answer?

[Jazza]

yeah, the first one is a fairly well know riddle, it is 1/3

but for the second one i would like an exact answer

[moomoocow]

For the second one - gievn 2 children - combinations are
BB - chance of Lisa = 0
BG - chance of Lisa = 1/1000
GB - chance of Lisa = 1/1000
GG - chance of Lisa = 1 - chance of no Lisa = (1 - (999/1000)^2) = 1.999/1000 approx. = 2/1000

each of these has 25% probability. So

p(other child being daughter|one is "Lisa")
= (1.999/1000)/(1/1000+1/1000+1.999/1000)
= 49.9874968%
or apprx. = 1/2

[TomCollins]

[ QUOTE ]
if you liked the three door riddle, here is one to ponder (i posted this in OOT a while ago):

you know a woman has 2 children, you ask her if she has at least one girl, she says yes, what is the chance the other child is a girl?

and a harder one:

assume that when naming a girl, parents will name the girl Lisa with probability 1/1000 (even if they already have a daughter named Lisa)
you know a woman has two children, you ask her if she has at least one girl named Lisa, she says yes, what are the chances the other child is a girl?

[/ QUOTE ]

A) 1/3

B) (1999/4M) / (1999/4M + 1/2000) ~= 49.987%

[Guernica4000]

It was done on "Numb3rs" on CBS

[Jazza]

looks good to me

[somapopper]

This whole thing depends on Monty choosing a door at random when you pick the door the car is behind, correct?

Say, the car is behind door number 2, Monty is determined to pick door number 3 if he can so when you pick door number 1 or 2 he will open door number 3 when you pick door number 3 he will open door number 1.

Wouldn't this make it a 50/50 proposition for the contestant to pick either door?

edit: nevermind, 2/3rds of the time I pick the wrong door, when I pick the wrong door the other door is the right one, it doesn't make one whit of difference what Monty does. I get it but I don't feel it.

[jjacky]

[ QUOTE ]
The answer to the 3 doors question is intuitively obvious and frankly, less thought provoking than the missing dollar question...even though that's pretty straightforward.

[/ QUOTE ]

you underestimate the difficulties, many people have with monty hall problem. afaik there was a lengthy discussion in the new york times about it. with many people who insisted the probability was 1/2 and many missleading arguments. i really don't see why we shouln't discuss this topic every month.

[MikeL05]

Wow. Once again, let's see who can try to post the least helpful response. I'm glad I'm not the only one who gets [censored] for not being as smart as you guys.

For the door problem... the best explanation I have heard was to imagine that instead of one door, there were 100 doors but still only one prize. So you pick one door... and so there are 99 other doors left. Before your door gets opened, the host of the show opens up 98 of the 99 other doors to reveal no prize there. Now you've got your door, and the option to switch to the the 1-in-99 door. Now it seems a lot better to switch, right?

[mosta]

[ QUOTE ]
[ QUOTE ]
Ok, so from what I found, the answer really does appear to be 33-67. So, there must be a fallacy somewhere in the 50-50 argument. Has it been identified yet?

Scott

[/ QUOTE ]

2/3 of all instances you choose a door which does not contain the price. In those 2/3-instances the door not opened has a 100% of containing the price. In the 1/3-instances that you choose the correct door, the other door not opened has 0% chance of containing the price. Thus she is right (2/3 vs 1/3 is 2 vs 1).

Tell her you have never met such a smart girl and invite her for dinner /images/graemlins/cool.gif.

[/ QUOTE ]

you can do a probability table, but I like the way it is put above. to spell it out a little more: if monty will always open an empty door, then you are left with one empty and one non-empty. therefore the switching turns all winning picks into losing picks and vice versa, and inverts the odds of the original process. all your original 2/3's chance losers become 2/3's chance winners. just focus on how switching when a loser is removed always reverses the original choice.

I've gone back on forth on whether it is 50-50 when the door is opened at random. now I'm not sure why it would be different (but at times I convinced myeself it was). seems like you would have to model the entire process including the expectation based on the chance the winner was exposed.

edit: I guess it's a little more of an abstract point: if hte door opening is random, is not based on your initial selection and the known whereabouts of the prize, then the content of your choice does not enter the system (it does not inform the choice of doors to open). your initial choice has no meaning or relevance. so the new situation is not a continuation, but is rather an entirely new situation. Monty would say, "okay, now I guess we're playing the two-door version of the game." and you would be playing a 50-50 game. to make the same point another way, you can't improve your odds with the following reasoning: it's probably in one of the other two, so if I switch (without any doors opening) I'll do better. again, in this case your original choice does not add any information to or affect the system (however many times you change your mind before settling on one of the 3). okay I convinced myself again. but I'm still not quite sure how to do two different probablility tables (which is what these problems always have to boil down to).

[mosta]

to go over it a little more (for my own sake at least): if the door opening is random (like a meteor striking the door) and is not related to your pick (Monty's opening a door does relate to your pick) then it is just the same as if a door were hit by a meteor the night before anyone showed up to have the game. if everyone shows up in the morning and one door is smashed and only two are left, then you are playing a two-door 50-50 game. if the meteor strikes just after you have committed to you pick, well the meteor isn't paying attention to you and your "pick" has no effect or relevance--it's the same as when it hit at night. your 3-door game disappears and you now have a 2-door game. the 3-door game never got played. with monty opening a door, there is a logical relation between your first pick and the door opening.

[TomCollins]

The problem with this is that Monty will open the correct door some of the time, which makes the game completely different.