OT: Hurricanes Lounge XXII: Wisdom Teeth Removal edition

[Carolinas Identity*]

Someone with a higher tolerance for math explain this to me:

You're given an option of three doors. Two doors have a booby prize, one door has a car. You choose a door at random. Then, one of the doors with a booby prize is eliminated and you're given the option to choose between the two remaining doors: The one you picked originally and the one you didn't.

Apparently, mathematically-speaking, you're always supposed to switch from your originally chosen door. Because you picked that door when you had a 1-in-3 chance of being correct, and if you switch, the other door has a 1-in-2 chance of being correct. And I get that it makes sense mathematically, because your odds of being correct went from 33% to 50%.

But logically, it doesn't make any sense to me. Because once one of the doors is eliminated, I have a 50% chance of being correct regardless of whether or not I switch doors. My originally picked door either has the car or it doesn't. 50/50. I don't see why switching doors would change that.

This is true and I have been asked about this a bunch.

Look at it this way:

When you pick a door, you have a 33.3...% chance of the car and a 66.6....% of a donkey (the example I am familiar with)

That means, that between the other two doors, one is guaranteed to be a donkey because your door is either the car or the 2nd donkey and the same is true for the third door. By eliminating the one of the other two doors that is guaranteed to be a donkey, you are left with two doors, the one you picked and the third. Now, even though there are only two options left, the door you picked to start still has a 66% chance of being a donkey (1 out of 3), the elimination of the 2nd door doesn't change the constant. So by switching to the 3rd door (the one that wasn't eliminated) you actually double your chances of getting the car because since the 3rd option is eliminated, that door has a 66% chance of being the car.

Does that make sense?

 

[What the Faulk]

I think it's helpful to remember that the host is never eliminating the car in "round 1". So your 33% chance off the bat hasn't changed if you don't switch. If you do switch, think of it almost as making a new door choice.

It's sort of like Deal or No Deal. Your case has a 1/25 (?) chance of containing $1,000,000 to begin with. You go through the game until you have only your case and the one next the models, and only the amounts 0 or $1,000,000 left. You still originally had only a 1/25 chance originally in selecting the million, but you now know the other case is 50/50 to be it. Do you switch?

 

[Carolinas Identity*]

I think it's helpful to remember that the host is never eliminating the car in "round 1". So your 33% chance off the bat hasn't changed if you don't switch. If you do switch, think of it almost as making a new door choice.

This is a simpler way of thinking of it, but yes, you are correct sir.

 

[Carolinas Identity*]

We did it!!!!

 

[Blueline Bomber]

This is true and I have been asked about this a bunch.

Look at it this way:

When you pick a door, you have a 33.3...% chance of the car and a 66.6....% of a donkey (the example I am familiar with)

That means, that between the other two doors, one is guaranteed to be a donkey because your door is either the car or the 2nd donkey and the same is true for the third door. By eliminating the one of the other two doors that is guaranteed to be a donkey, you are left with two doors, the one you picked and the third. Now, even though there are only two options left, the door you picked to start still has a 66% chance of being a donkey (1 out of 3), the elimination of the 2nd door doesn't change the constant. So by switching to the 3rd door (the one that wasn't eliminated) you actually double your chances of getting the car because since the 3rd option is eliminated, that door has a 66% chance of being the car.

Does that make sense?

Not at all.

Because when it's all said and done, when you're asked to make the second choice, you can either stick with the door you picked originally or choose the other door. You know that either door could have the car behind it, so it's 50/50 on whether you choose correctly, regardless of whether you stay put or change.

Your odds have increased from 33% of being right to 50% simply by eliminating the third door. Choosing a different door doesn't change that.

 

[What the Faulk]

Say you pick door one.

Is that more helpful?

EDIT: Your door has an initial 33% chance of being right. The other two have a 66% combined chance of being right. You kind of really get to switch to the other two, because you get to see what's behind one of them, which is always a goat. The other is your new door.

 

[Blueline Bomber]

Say you pick door one.

Is that more helpful?

EDIT: Your door has an initial 33% chance of being right. The other two have a 66% combined chance of being right. You kind of really get to switch to the other two, because you get to see what's behind one of them, which is always a goat. The other is your new door.

That makes slightly more sense, but it still doesn't explain why a decision between two choices is anything but 50/50.

Once the host eliminates a door (let's say door 3), your choice becomes:

Car - Goat
or
Goat - Car

Without knowing which door holds the prize, it's 50/50.

 

[Finlandia WOAT]

If you go in with the mentality that you are going to switch no matter what, the object of the game changes from "pick the door with the car" to "pick a door with the goat". At the beginning of the game, you have a 66% chance of picking a door with a goat.

Conversely, if you go in with the mentality that you are going stay the same no matter what, the object of the game is to "pick the door with the car"; which is a 33% chance at the beginning of the game.

Taking away the door with the goat that you didn't pick is just a way for the game makers to trick people into thinking that either object has the same statistical probability, when in fact one is more significant than the other. They're making you think there are two separate decisions by informing you of something you already knew (that one of the doors you didn't pick was a goat), when in fact there is one.

 

[What the Faulk]

You're ignoring a crucial part of the problem, though: the third door. Regardless of which door you pick (and whether it's a car or goat), the host is revealing a goat behind one of the doors you didn't pick.

Think of it this way. You initially have a 33% chance of picking a car, and 66% chance of picking a goat. That's never going to change, even after the elimination of one door. If you don't switch, those are still the odds. But if you do switch post-reveal, you'll want to have picked that goat first, since that now means you'd end up with the car. And since there are two goats to one car, you double your odds.

^ EDIT: Whelp, you beat me to that explanation. I think that's probably the best way to say it, and you did it better anyway.

 

[BrindAmours Army]

Not at all.

Because when it's all said and done, when you're asked to make the second choice, you can either stick with the door you picked originally or choose the other door. You know that either door could have the car behind it, so it's 50/50 on whether you choose correctly, regardless of whether you stay put or change.

Your odds have increased from 33% of being right to 50% simply by eliminating the third door. Choosing a different door doesn't change that.

Your odds go from 33% to 66%, not to 50%.

You originally choose Door A. Door A has a 33% chance of being nothing. Ergo there is a 66% chance of the car not being behind A. The host states that the car is not behind B. So you have a 33% chance of it being behind A, a 0% chance of it being behind B, and a 66% chance of it being behind C.

Once B is eliminated, C becomes twice as likely to hide the car.

 

[Blueline Bomber]

Your odds go from 33% to 66%, not to 50%.

You originally choose Door A. Door A has a 33% chance of being nothing. Ergo there is a 66% chance of the car not being behind A. The host states that the car is not behind B. So you have a 33% chance of it being behind A, a 0% chance of it being behind B, and a 66% chance of it being behind C.

Once B is eliminated, C becomes twice as likely to hide the car.

Here's where you lost me. Because once the host states that it's not behind B, that means it's either behind A or C. Which means that there's a 50% chance it's behind A, a 0% chance it's behind B, and a 50% chance it's behind C.

I don't see how the odds change for one door and not the other once Door B is removed from the equation.

 

[What the Faulk]

Because the host is never showing you the door with the car before the opportunity to switch. He's always showing you a goat. If the car was actually behind door B, he'd show you door C instead. The two goats doors are basically one entity. It boils down to what Finlandia said above. You either want to pick the car and not switch (33% chance of winning) or pick the goat and switch (66% chance of winning). Option two is the better strategy.

 

[Blueline Bomber]

Because the host is never showing you the door with the car before the opportunity to switch. He's always showing you a goat. If the car was actually behind door B, he'd show you door C instead. The two goats doors are basically one entity. It boils down to what Finlandia said above. You either want to pick the car and not switch (33% chance of winning) or pick the goat and switch (66% chance of winning). Option two is the better strategy.

But the two goats aren't one entity. There's a car, a goat, and a goat. When you make your first choice, you've got a 33% chance of choosing the car. Then the host removes a goat from the equation.

That leaves you with a choice between a car and a goat. You don't know which door has which prize, and both doors are equally likely to contain either prize. So whether you decide to stick with your originally decided door, or decide to switch to the other door, the odds are equally as good that it contains either a goat or a car.

The 33/66% solution only makes sense if you assume that your odds of choosing the car remains the same after the host eliminates a goat. But it doesn't.

When your choice changes from 1 car/2 goats to 1 car/1 goat, your odds of choosing the car should change as well, no?

 

[Carolinas Identity*]

But the two goats aren't one entity. There's a car, a goat, and a goat. When you make your first choice, you've got a 33% chance of choosing the car. Then the host removes a goat from the equation.

That leaves you with a choice between a car and a goat. You don't know which door has which prize, and both doors are equally likely to contain either prize. So whether you decide to stick with your originally decided door, or decide to switch to the other door, the odds are equally as good that it contains either a goat or a car.

The 33/66% solution only makes sense if you assume that your odds of choosing the car remains the same after the host eliminates a goat. But it doesn't.

When your choice changes from 1 car/2 goats to 1 car/1 goat, your odds of choosing the car should change as well, no?

Okay, try looking at it this way:

Start of competition:

A) - 33% car//66% goat
B) - 33% car//66% goat
C) - 33% car//66% goat

You make a choice, say, C:

A) - 33% car//66% goat
B) - 33% car//66% goat
C) - 33% car//66% goat

Host eliminates a goat, say, A:

A) - 33% car//66% goat
B) - 33% car//66% goat
C) - 33% car//66% goat

Now we have this:

A) - 0% car//100% goat
B) - 33% car//66% goat
C) - 33% car//66% goat

Nothing about C has changed, there is still a 33% chance it's the car and a 66% chance it's the other goat, but since we know A has a 0% chance of being the car, in order for the math to add up to 100%, if C has a 33% chance of being the other goat, then there is a 66% chance B is the car. Just because you remove a variable, doesn't mean the percentages change.

Does this help at all?

 

[tomdundo]

Okay, I don't get it. You ask a question and get the correct answer a repeated number of times, but you refuse to accept it. It's based on Bayesian statistical theory, which is hard to explain in text form...so here's a video that explains the problem for your convenience:







Or, for the Khan Academy version, which does a better job...





Basically, since the host knows where the car is/isn't, you should assume that you pick wrong with your door selection. Obviously there's a 33% chance that you pick the correct door in the first place, nobody is denying that. But the host will never show you the car. He's always going to show you one of the two goats.


Option A: Car, Goat, Goat
Option B: Goat, Car, Goat
Option C: Goat, Goat, Car


Let's say that you always pick Door 1 as your first choice.
-In option A, the host can open EITHER door 2 or 3 and reveal a goat. In that case, you don't want to switch.
-In option B, the host can ONLY open door 3, since he won't open your door and won't reveal the car.
-In option C, the host can ONLY open door 2, for the same reason as above

So with those above 3 scenarios, two of the three result in you switching doors and therefore winning the car, for your 2/3 chance.

 

[Blueline Bomber]

Okay, I don't get it. You ask a question and get the correct answer a repeated number of times, but you refuse to accept it.

As I said originally, I accept it makes sense mathematically. I've seen the videos and seen the Mythbuster episode that showed its correct.

But logically, all that's happened is that you've taken a choice with three options, two of them being incorrect (meaning you had a 33% of being right), and turned it into a choice with two options, only one of them being incorrect (meaning you have a 50% of being right).

So I'm not entirely sure how that works. Since if you've ONLY got a right answer and a wrong answer, and you randomly guess, you've got a 50% chance of being right, not a 33% chance.

If I asked you what 2 + 2 was, and gave you the options of 4, 5, and 6, then eliminated 6 as an option, you've got a 50% chance of being right if you chose between 4 and 5. There were two wrong answers initially, I eliminated one of them, leaving the correct answer and one wrong answer. How is that different from this goat/car problem?

 

[tomdundo]

See my edit above. Rather than thinking about each individual scenario, think about it as a set of scenarios. It's conditional probability, not just a straight yes/no answer.

Obviously you'll have the same scenario if you pick door 2 or door 3 as your first choice, but the host will react differently.

 

[Carolinas Identity*]

As I said originally, I accept it makes sense mathematically. I've seen the videos and seen the Mythbuster episode that showed its correct.

But logically, all that's happened is that you've taken a choice with three options, two of them being incorrect (meaning you had a 33% of being right), and turned it into a choice with two options, only one of them being incorrect (meaning you have a 50% of being right).

So I'm not entirely sure how that works. Since if you've ONLY got a right answer and a wrong answer, and you randomly guess, you've got a 50% chance of being right, not a 33% chance.

If I asked you what 2 + 2 was, and gave you the options of 4, 5, and 6, then eliminated 6 as an option, you've got a 50% chance of being right if you chose between 4 and 5. There were two wrong answers initially, I eliminated one of them, leaving the correct answer and one wrong answer. How is that different from this goat/car problem?

you're not randomly guessing though, in essence you're getting advice after the fact to manipulate the odds, but nothing about the percentages change

 

[Finlandia WOAT]

If you make the goal of the game "pick a goat door at the beginning" and ignore the "second" choice; then you have a 66% chance to win if you switch.

(BTW the second quotation marks are irony quotes, there is only one choice here, not two. To switch or not to switch)

Surely you can see 66>50.

 

[tomdundo]

I think you're getting caught up in the idea of A) picking a door and B) swapping doors after a goat is revealed as two separate, unrelated events. That's not true at all. The second event is entirely dependent on the first.

 

[Carolinas Identity*]

okay, super layman terms answer:

you pick door 1, you have 33% chance of being right and 66% chance of being wrong

door 2 is a goat

you still have a 33% of being right and a 66% chance of being wrong because nothing about your door changed

 

[Blueline Bomber]

See my edit above. Rather than thinking about each individual scenario, think about it as a set of scenarios. It's conditional probability, not just a straight yes/no answer.

That makes sense, but it is 3 AM, so a lot of crazy things may make sense. I get that scenario adds up to having a 66% of being correct, but I'm still at a loss as to how a decision between two options is anything but 50/50 if you're randomly guessing.

If the host only had two doors initially and asked you to choose one, you'd have a 50% of winning the prize. But when the above scenario plays out, the choice between two doors becomes 66%?

So if there were a million doors, I'd have a 1/1000000 chance of being correct initially, but if he eliminated it down to two doors, I should switch from my initial one, because the other one has a 999999.9% of being correct?

 

[tomdundo]

That makes sense, but it is 3 AM, so a lot of crazy things may make sense. I get that scenario adds up to having a 66% of being correct, but I'm still at a loss as to how a decision between two options is anything but 50/50 if you're randomly guessing.

If the host only had two doors initially and asked you to choose one, you'd have a 50% of winning the prize. But when the above scenario plays out, the choice between two doors becomes 66%?

So if there were a million doors, I'd have a 1/1000000 chance of being correct initially, but if he eliminated it down to two doors, I should switch from my initial one, because the other one has a 999999.9% of being correct?

It's not though, that's what your mind is tricking you to think. As I said above, it's about conditional probability. What the host does depends on what you do -- the combination of events is what ultimately influences your final decision.

Here's the math if you want it, although it might confuse you more. Look under the "Direct calculation" section. https://en.wikipedia.org/wiki/Monty_Hall_problem


For the above link, the most confusing part would probably be why the P(C2|X1), P(C1|X1) and P(C3|X1) all cancel out...but that's because the probability of the car being behind any door is 1/3 at the start.

 

[tomdundo]

I'm gonna stop editing that post and walk you through it.


The probability P(H3|C1,X1) is 1/2 because he can't open door 1 since you chose it. That's situation 1 in my above explanation. He can open either door 2 or door 3 and reveal a goat.

The probability P(H3|C2,X1) is 1 because he can't open door 2 with the car in it, and he cant open door 1 because that's the door you chose. This is both option B and option C in my above scenario.

The final probability, P(H3|C3,X1) is 0 because the host can't open the door that contains the car. That ruins the entire point of the game.



It then picks a situation for you, in which you select door 1 and the host therefore opens door 3.

The probability of a car being behind any door is 1/3 at the beginning, which is what I mentioned in my edit of the above post. So those cancel out, and you're left with the final result in the Wikipedia page.



(If you don't understand those probabilities, what it means is the probability of the item before the |, given that you know the items after it. So the probability that the host opens door 3 given that he knows the car is behind door one and you pick door one...and so forth)




EDIT: You know what, let's just forget it for tonight. I can walk you through it tomorrow if you want when it's not quite so late.

 

[Carolinas Identity*]

here

i got a fun one for you guys

an apartment hallway is 10 yards (y) long

there are 3 feet (f) in a yard

write a equation that accurately portrays the relationship between y and f

have fun