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Use Bayes rule to mechanically solve probability riddles

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Re: Use Bayes rule to mechanically solve probability riddles

#31

Post by bmm6o »

FabHK wrote: post #20
kgwgk wrote: post #18

Earlier quoted context omitted.

> many formulations one finds (including the one in her column) As far as I’ve been able to find the text below would be the original question (“the host, who knows what’s behind the doors, opens”) and answer (“the host, who knows what’s behind the doors and will always avoid the one with the prize, opens”). Are you referring to that? https://web.archive.org/web/20130121183432/http://marilynvos... Suppose you’re on a…

Yes, indeed, thanks for digging it up. You see that Marilyn's answer contains the clarified correct specification ("the host, who knows what’s behind the doors and will always avoid the one with the prize..."), but the question does not fully specify it. Thus, ignoring the clarification in the answer (which furthermore pertains to a modified problem), one could interpret the question differently: the host randomly op…

There's also the cultural context of Monty Hall being a real person who had a real game show, on which he really opened doors with goats behind them. Most readers of her column would have been familiar with the mechanics of the show. And the question doesn't really make sense if there's a chance that he opens the door with a prize, there's no more hidden information in that case.

Re: Use Bayes rule to mechanically solve probability riddles

#32

Post by zaik »

AnotherGoodName wrote: post #23
kgwgk wrote: post #2

> You're told that at least one of them is a girl. > Likelihood of at least one girl What the “mechanism” requires is “likelihood of being told that at least one of them is a girl”. Use Bayes rule to correctly solve probability riddles: https://news.ycombinator.com/item?id=45056790 p(both are girls | you're told at least one is a girl) = p(you're told at least one is a girl | both are girls) * p(both are girls) / ( p…

Pretty much all of them are like this and it's actually a terrible article. For those thinking about how can the likelihood of them being a girl can change consider this; >A family has two children. You're told that at least one of them is a girl. What's the probability both are girls? "The question writer took all sets of two child families and ruled out the bb case. Then they asked the exact question above" This is…

I did not mention a lot of assumptions. For example my problem statement also does not answer the question if Monty always gives you a choice to switch or only when you picked in a certain way. There are quite a few possible variations, all of which change the answer: https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...

In the derivation of the likelihood the assumptions are clearly stated.

While my knowledge on statistics is certainly unsatisfactory, your assessment of the education I had is quite wrong. My final exam in order to obtain a Master's degree in mathematics is next week. One of the subjects I chose is Bayesian Statistics, so wish me luck.

Re: Use Bayes rule to mechanically solve probability riddles

#34

Post by kgwgk »

zaik wrote: post #9
sparsely wrote: post #7

For the last one, why does the "born on a Tuesday" information change the result? I don't see how it isn't equivalent to "born on a day", since the day of the week has no connection to the rest of the scenario. I understand why "at least one boy" does matter.

If you accept the Bayes theorem, the answer is that the likelihood of "At least one boy is born on a Tuesday" is not the same for different numbers of boys. The more boys the more likely the statement is true. Therefore this information is indicative of how many boys Mrs. Chance has.

> If you accept the Bayes theorem

That doesn’t make a lot of sense. A theorem is just a theorem. It’s proved, and in this case the proof is trivial.

The question is whether you accept that the description of the problem in terms of conditional probabilities is adequate, and then whether you accept that the values assigned to those conditional probabilities are appropriate.

Re: Use Bayes rule to mechanically solve probability riddles

#35

Post by AnotherGoodName »

zaik wrote: post #32
AnotherGoodName wrote: post #23

Earlier quoted context omitted.

Pretty much all of them are like this and it's actually a terrible article. For those thinking about how can the likelihood of them being a girl can change consider this; >A family has two children. You're told that at least one of them is a girl. What's the probability both are girls? "The question writer took all sets of two child families and ruled out the bb case. Then they asked the exact question above" This is…

I did not mention a lot of assumptions. For example my problem statement also does not answer the question if Monty always gives you a choice to switch or only when you picked in a certain way. There are quite a few possible variations, all of which change the answer: https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_... In the derivation of the likelihood the assumptions are clearly stated. While my knowle…

I do wish you luck. Fwiw I'd always work forward from the question and talk to the ambiguities present especially when people may come across this and not realize. Stating the Bayesian probabilities of one possible interpretation and having that as your way of stating assumptions is a bit like working backwards from one possible answer.

As an example of how presenting this way can hurt; The link you had for the sisters paradox doesn't talk to the ambiguities of a question that is well known to have no answer - https://en.wikipedia.org/wiki/Boy_or_girl_paradox . This information spreads and people see it restate it. This leads to a false belief that the sisters paradox has no ambiguities and that the answer is clearly 1/3 (the wikipedia page on this has the correct statement that it's ambiguous and the answer can be 1/3 or 1/2 depending on one of two reasonable interpretations).

I think a far better article would talk to ambiguities of each of these. Not subtly stating assumptions. I'll also point out that these types of ambiguous questions are commonly used in DS interviews (I've worked in big tech for many years now). The expected response of a strong candidate is a discussion on the ambiguities. You'll usually be prompted "is there any other interpretation of this question that leads to a different result?". If people read this as their guide rather than the more detailed wikipedia articles etc they may be misled which is why i'm strongly negative on this article as written. I'd hope no one reads and doesn't realize the assumptions here since they are critical.

Re: Use Bayes rule to mechanically solve probability riddles

#36

Post by JeffJor »

amluto wrote: post #16
sparsely wrote: post #7

For the last one, why does the "born on a Tuesday" information change the result? I don't see how it isn't equivalent to "born on a day", since the day of the week has no connection to the rest of the scenario. I understand why "at least one boy" does matter.

I think this particular question illustrates a major oversimplification in the entire premise of the webpage. If you have a probability problem that isn't well-specified, no amount of "mechanical" magic, Bayesian or otherwise, will give you a fully correct answer, since you are missing relevant details . Let's consider this particular question: "Mrs. Chance has two children of different ages. At least one of them is…

Mr. Bertrand has (exactly - this needs to be included) two children (not twins, which is not quite the same as different ages). A gender, and a day of the week, that apply to at least one of his children have been written inside a sealed envelope. What is the probability that both children have that gender?

In this problem, we have no gender- or day-specific information. So the answer can only be the probability that he has two of the same gender. Which is 1/2.

Now open the envelope. If the answer changes to P based on what you see written, it has to change to the same P regardless of what you see written. Which means you didn't need to unseal the envelope; the answer was P before, not 1/2.

This is what Joseph Bertrand identified as his Box Paradox in 1889. That word was used to describe an actual contradiction, not a non-intuitive result. It disproves any answer except P=1/2. FOR ANY OF THESE PROBLEMS.

In fact, it is the same reason why the Monty Hall Problem's answer is what it is. Many "explanations" will claim that your original probability can't change, but never justify it. This is the justification - if it changes when one door is opened, it must change the same way when either door is opened.

Re: Use Bayes rule to mechanically solve probability riddles

#37

Post by 0xfaded »

glitchc wrote: post #33
0xfaded wrote: post #28

Just a tidbit for remembering bayes rule: P(A|B)P(B) = P(A,B) = P(B|A)P(A) The familiar forms P(A|B) = P(A,B)/P(B) = P(B|A)P(A)/P(B) immediately follow

Comma is not a defined operator. Do you mean intersection?

It's a joint distribution :)

https://en.m.wikipedia.org/wiki/Joint_probability_distributi...

Re: Use Bayes rule to mechanically solve probability riddles

#38

Post by SAI_Peregrinus »

FabHK wrote: post #25
kgwgk wrote: post #21

Earlier quoted context omitted.

Did any of those readers of her column, including PhD and mathematicians, that declared her solution wrong do so because they objected to the clarification? (Does it make sense to declare the answer wrong ignoring the answer?)

I think they didn't notice the significance of that clarification (which, again, was given in the context of a modified problem). Reading the exchanges again, though, I see that many respondents were not only mistaken, but also impolite or sexist in accusing Marylin of being wrong. If you specify the problem carefully, anyone with some training in probability should get the answer right. But clearly back then it was…

The original problem is the one you're thinking of as modified, since the original problem was about a real-world game show, and the rules of the real-world show in question included Monty Hall (the host) opening a door that the player hadn't picked and which did not contain the car. The problem assumed familiarity with the game show, the clarification is for people who didn't watch the show.

Re: Use Bayes rule to mechanically solve probability riddles

#39

Post by bobbylarrybobby »

0xfaded wrote: post #28

Just a tidbit for remembering bayes rule: P(A|B)P(B) = P(A,B) = P(B|A)P(A) The familiar forms P(A|B) = P(A,B)/P(B) = P(B|A)P(A)/P(B) immediately follow

In words, for those who need help with the first line: the chance that A and B both happen is the chance that A happens, and (given that A happened) B happens. (Or vice versa.)

Re: Use Bayes rule to mechanically solve probability riddles

#40

Post by griffzhowl »

FabHK wrote: post #15
Shorel wrote: post #5

This Monty Hall problem was asked to Marilyn vos Savant, a woman with an extremely high IQ, who solved it correctly, and many readers of her column, including PhD and mathematicians, declared her solution wrong. Then careful analysis proved her correct. https://en.wikipedia.org/wiki/Monty_Hall_problem#Savant_and_...

To be fair, I think this problem needs to be formulated very carefully and specifically [0], because the "correct" answer is predicated on that, and many formulations one finds (including the one in her column) are not that. [0] It has to be specified that a) Monty knows what's behind which door, and b) he will on purpose always open a door such that there's a goat behind it.

> [0] It has to be specified that a) Monty knows what's behind which door, and b) he will on purpose always open a door such that there's a goat behind it.

You don't need condition (a) here. It's enough to just stipulate that a door with a goat behind it will be opened, however that comes about

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