Maths

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Re: Maths

Post by Ashenai » Tue Apr 13, 2021 7:48 pm

Jeb Bush 2012 wrote: Tue Apr 13, 2021 7:47 pm
Ashenai wrote: Tue Apr 13, 2021 7:46 pm The question is not well defined, so the answer is gonna be contingent on what you think "probability of an event from her perspective" means.

It's quite annoying to define it correctly, though.
well I mean "how should we define probability" is the point of the question, it's not ill-defined in, say, the way the typical telling of the monty haul problem is
I think it's quite similar! Both cases involve unclear priors.

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Re: Maths

Post by pterrus » Tue Apr 13, 2021 7:48 pm

It is kind of funny that if she wakes up on Tuesday she should say 1/3, but then if 2 seconds later the experimenter tells her the experiment is over she should revise to 1/2.

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Re: Maths

Post by Jeb Bush 2012 » Tue Apr 13, 2021 7:52 pm

Ashenai wrote: Tue Apr 13, 2021 7:48 pm I think it's quite similar! Both cases involve unclear priors.
I don't think that's right - the experimental procedure here is very carefully defined, and the problem is how you want to define probability

the ambiguity of the monty haul problem, on the other hand, comes exclusively from the fact that the experimental procedure is underspecified

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Re: Maths

Post by Skeletor » Tue Apr 13, 2021 7:54 pm

She should give whatever answers she wants, and then after the experimenter corrects her, force the amnesia medicine on him, go back to sleep, and when he wakes her up the next day she should give his answer back to him
wow, [you]. that all sounds terrible. i hope it gets better for you

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Re: Maths

Post by Doug » Tue Apr 13, 2021 7:56 pm

Do you guys have strong opinions about Grandfather's Axe/Ship of Theseus

Or how about Unexpected Hanging
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Re: Maths

Post by Jeb Bush 2012 » Tue Apr 13, 2021 7:57 pm

I'm strongly against all forms of hanging but I suppose if I am to be hanged I would prefer it to be unexpected

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Re: Maths

Post by pterrus » Tue Apr 13, 2021 7:58 pm

Jeb Bush 2012 wrote: Tue Apr 13, 2021 7:52 pm the ambiguity of the monty haul problem, on the other hand, comes exclusively from the fact that the experimental procedure is underspecified
How is it underspecified and where is the ambiguity? My understanding of that problem is that peoples' intuition is usually wrong, but that the math is pretty unambiguous.

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Re: Maths

Post by Khaos » Tue Apr 13, 2021 8:00 pm

i think the ambiguity is that it's not clear that the dude will always open an empty door for you, not a random door

i have a good way of explaining the monty hall problem to people in an intuitive way

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Re: Maths

Post by Jeb Bush 2012 » Tue Apr 13, 2021 8:01 pm

pterrus wrote: Tue Apr 13, 2021 7:58 pm How is it underspecified and where is the ambiguity? My understanding of that problem is that peoples' intuition is usually wrong, but that the math is pretty unambiguous.
the underspecified part is why the guy opened the door he did. the "right" answer is right if you assume that his policy is to always open a door without a prize behind it, but usually the problem is stated as just "he opens a door without a prize behind it". but if he opened a door at random and it just happened to not have a prize behind it, you should be indifferent between switching or not switching

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Re: Maths

Post by Doug » Tue Apr 13, 2021 8:03 pm

"He then opens a losing (empty/goat) door" is a variable instruction, which makes people go wrong when trying to evaluate the probabilities

It represents an entire tree of actions
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Re: Maths

Post by Jeb Bush 2012 » Tue Apr 13, 2021 8:03 pm

and tbc this is not just retellers screwing it up, the original statement of the problem has the same ambiguity
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?

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Re: Maths

Post by Rylinks » Tue Apr 13, 2021 8:06 pm

i don't see how the sleeping beauty thing is a problem

P(heads) is 0.5 and P(Heads|Sleeping beauty is awake) is 2/3

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Re: Maths

Post by pterrus » Tue Apr 13, 2021 8:44 pm

Jeb Bush 2012 wrote: Tue Apr 13, 2021 8:01 pm
pterrus wrote: Tue Apr 13, 2021 7:58 pm How is it underspecified and where is the ambiguity? My understanding of that problem is that peoples' intuition is usually wrong, but that the math is pretty unambiguous.
the underspecified part is why the guy opened the door he did. the "right" answer is right if you assume that his policy is to always open a door without a prize behind it, but usually the problem is stated as just "he opens a door without a prize behind it". but if he opened a door at random and it just happened to not have a prize behind it, you should be indifferent between switching or not switching
Ah I see. That is sort of ambiguous. I don't think it's much of a leap to assume the game never just randomly gives you the car, but it wouldn't have hurt to be explicit about this in the description.

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Re: Maths

Post by Jeb Bush 2012 » Tue Apr 13, 2021 8:51 pm

pterrus wrote: Tue Apr 13, 2021 8:44 pm Ah I see. That is sort of ambiguous. I don't think it's much of a leap to assume the game never just randomly gives you the car, but it wouldn't have hurt to be explicit about this in the description.
I mean, seems like a pretty weird game to start with, who knows how it works

but consider another possibility: the host wants you to lose, but also is contractually forbidden from actually lying. so he thinks, huh, everyone "knows" the correct answer to this problem, so what I'll do is, if they get the right door, I'll open another one and then offer them the chance to switch. otherwise I won't offer them the choice at all

this is also totally consistent with the problem as described!

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Re: Maths

Post by pterrus » Tue Apr 13, 2021 9:13 pm

Jeb Bush 2012 wrote: Tue Apr 13, 2021 8:51 pm
pterrus wrote: Tue Apr 13, 2021 8:44 pm Ah I see. That is sort of ambiguous. I don't think it's much of a leap to assume the game never just randomly gives you the car, but it wouldn't have hurt to be explicit about this in the description.
I mean, seems like a pretty weird game to start with, who knows how it works

but consider another possibility: the host wants you to lose, but also is contractually forbidden from actually lying. so he thinks, huh, everyone "knows" the correct answer to this problem, so what I'll do is, if they get the right door, I'll open another one and then offer them the chance to switch. otherwise I won't offer them the choice at all

this is also totally consistent with the problem as described!
Actually yeah, what the fuck. This is the more common sense interpretation of the problem if you don't know it's supposed to be a math question.

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Re: Maths

Post by Crunchums » Wed Apr 14, 2021 1:39 am

Skeletor wrote: Tue Apr 13, 2021 7:26 pm If you think of probability as being some aspect of the coin, you'll say 1/2, if you think of it as being a statement about your (sleeping beauty's) best possible guess, then it's 1/3
this seems correct
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Re: Maths

Post by Crunchums » Wed Apr 14, 2021 1:42 am

Jeb Bush 2012 wrote: Tue Apr 13, 2021 7:47 pm monty haul problem
heh
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Re: Maths

Post by Jeb Bush 2012 » Wed Apr 14, 2021 1:55 am

another way in which that problem is ill-posed - it's specified that one door has a car behind it and the other has goats, but it doesn't say how many goats, how expensive the car is, what kind of goat, etc.

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Re: Maths

Post by Jeb Bush 2012 » Wed Apr 14, 2021 1:57 am

google suggests goats are worth 100-800 dollars each so if it's like a cheap used car you could easily fit enough goats behind the door to make it worth your while

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Re: Maths

Post by Jeb Bush 2012 » Wed Apr 14, 2021 1:59 am

idea: game show where you're shown a series of car/goat herd pairs, and you have to guess which one is the most valuable

provisional title is "you've goat a deal"

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Re: Maths

Post by Doug » Wed Apr 14, 2021 3:08 am

rofl
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Re: Maths

Post by pterrus » Wed Apr 14, 2021 10:47 am

We need more farm animal based shows. Last night we were watching a documentary on the Sichuan-Tibet highway and there was a 5 minute aside about this family that herds yaks on the side of the road and sells yak butter and when they cut away I was upset. Those people deserve their own show. Real Yak Herders of Tibet.

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Re: Maths

Post by Doug » Thu May 06, 2021 12:18 am

There are 121 commons in a given Magic set

Let's say a draft has 10 commons in each pack. There are 24 packs in a given draft.

If we use only 1 of each common, that's not enough to create the draft, because the draft needs 240 commons.

If we use 2 of each common, that's enough to make the draft with only 2 cards left over. The odds that they are 2 copies of the same card are 1/121? So we're pretty likely to have every common in the draft at least once.

What if we use 3 of each common, such that there are plenty left over -- that is, we have 363 commons, 3 of each, and we take 240 of those -- what is the probability for a given common that none of it are in the draft?

What if we use 4 of each common?

What if we use 1 million of each common
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Re: Maths

Post by Crunchums » Thu May 06, 2021 1:11 am

Doug wrote:If we use 2 of each common, that's enough to make the draft with only 2 cards left over. The odds that they are 2 copies of the same card are 1/121?
i'm confused; are you talking about the odds of 2 copies of the same card one pack? where does this 1/121 number come from?
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Re: Maths

Post by Rylinks » Thu May 06, 2021 1:13 am

it's the odds a common does not appear in the draft at all, and since there are 2 cards left over a card doesn't appear if the leftovers are a duplicate

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Re: Maths

Post by Jeb Bush 2012 » Thu May 06, 2021 1:23 am

it's 1/241 though; when you pick the second card to be left over there are 241 cards you haven't picked yet and only one of them matches the one you've already picked

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Re: Maths

Post by Jeb Bush 2012 » Thu May 06, 2021 1:26 am

I don't think there's a nice closed form for this problem in general, in the limit as the number of copies of each common goes to infinity it's the coupon collector's problem

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Re: Maths

Post by Jeb Bush 2012 » Thu May 06, 2021 2:00 am

okay, but, we can apply some lower bounds to the problem, and we get that even if there are only 3 of each card (the situation gets worse the more copies of each there are), you have a >99% chance of missing at least one card

let's think about what the chance is of missing any given card. without loss of generality, that card is squire. you're going to pick 240 cards out of 363. after picking k of them, if you haven't picked squire yet, there are 360 - k non-squires out of 363 - k cards. so your chance of never picking squire are prod_{k=0}^239 (360 - k)/(363 - k). this turns out to be ~3.83%. not too bad!

but there are 121 cards, so how likely is it that you miss any of them? well, this is tricky to calculate exactly, but we do know that the "you get at least one copy of card x" events are negatively associated - if you get at least one copy of a card, it's less likely you get at least one copy of any given other card. so you can get an upper bound on the chance that you get at least one of each card by just looking at what the chance would be if these events were independent

this is (1 - 0.0383)^121, so about 0.89%. so yeah, once there are at least three of each card to choose from, you're not gonna get every card

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Re: Maths

Post by Doug » Thu May 06, 2021 2:07 am

Jeb Bush 2012 wrote: Thu May 06, 2021 1:23 am it's 1/241 though; when you pick the second card to be left over there are 241 cards you haven't picked yet and only one of them matches the one you've already picked
Oh derf right
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Re: Maths

Post by Crunchums » Thu May 06, 2021 3:03 am

Doug wrote:What if we use 3 of each common, such that there are plenty left over -- that is, we have 363 commons, 3 of each, and we take 240 of those -- what is the probability for a given common that none of it are in the draft?
isn't it hypergeometric? i.e. (360/363)*(359/362)*...*(121/124)
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