I think it's quite similar! Both cases involve unclear priors.Jeb Bush 2012 wrote: ↑Tue Apr 13, 2021 7:47 pmwell 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
Maths
What are birds
Re: Maths
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Re: Maths
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
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
Re: Maths
Do you guys have strong opinions about Grandfather's Axe/Ship of Theseus
Or how about Unexpected Hanging
Or how about Unexpected Hanging
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Re: Maths
I'm strongly against all forms of hanging but I suppose if I am to be hanged I would prefer it to be unexpected
Re: Maths
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.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
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Re: Maths
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
Re: Maths
"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
It represents an entire tree of actions
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Re: Maths
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?
Re: Maths
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.Jeb Bush 2012 wrote: ↑Tue Apr 13, 2021 8:01 pmthe 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
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!
Re: Maths
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.Jeb Bush 2012 wrote: ↑Tue Apr 13, 2021 8:51 pmI 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
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
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
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"
provisional title is "you've goat a deal"
Re: Maths
rofl
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Re: Maths
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.
Re: Maths
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
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
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?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?
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Re: Maths
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
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
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
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
Re: Maths
Oh derf rightJeb 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
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Re: Maths
isn't it hypergeometric? i.e. (360/363)*(359/362)*...*(121/124)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?
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