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When intuition and math probably look wrong

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Re: When intuition and math probably look wrong

#21

Post by bad_user »

RiderOfGiraffes wrote: post #15
bad_user wrote: post #12

Earlier quoted context omitted.

> Let's toss two coins until at least one shows a head. By your reasoning the odds of them both being heads is 1/2. It's not. Try it. You haven't read the article then ... the problem as stated in the article is that you know one coin is going to be a head, so what's the probability of the other one also being a head? Of course ... the events aren't connected ... the second coin toss doesn't depend in any way on the…

Here's the problem as Gary originally stated it, and as the article quotes it: > I have two children, one of whom is a son born > on a Tuesday. What is the probability that I > have two boys? You say: the problem as stated in the article is that you know one coin is going to be a head, No. The point of the article is that you don't know how or why you are given this information. Suppose I toss two coins until I get o…

> The probability that they are both heads is 1/3.

The probability they are both heads is 1/2.

You're trying to formulate this as ....

  P(head AND head | one is a head) =
  P(head AND head) / P(one is a head) =
  1/4 * 4/3 = 4/12 = 1/3
But this is wrong ... you know that you have a head ... which makes ...

  P(one is a head) = 1 && P(head_a AND head_b) = P(head_a) = P(head_b)
1/3 would be the probability only on the first try (instead of stopping when you've got a head).

Re: When intuition and math probably look wrong

#22

Post by bad_user »

RiderOfGiraffes wrote: post #18
bad_user wrote: post #6

Earlier quoted context omitted.

> But since the boy could be either the younger or the older child, the analysis is more subtle. Devlin started by listing the children’s sexes in the order of their birth Personally, when I read that I could spot the error. For lazy people (tl;dr types) ... order doesn't matter as you're not given any info about that order. So given 2 children, there are only 3 possibilities ... boy, boy girl, girl boy, girl So if y…

Actually, I'm going to stop now: http://xkcd.com/386/ It's pretty clear I won't convince you. Let me leave you with these questions: If I toss two coins until at least one shows a Head, what's the probability that both are Heads? If I roll two dice until at least one shows a 6, what's the probability that both are 6's? If I spin two roulette wheels until at least one shows a Red-23, what's the probability that both a…

It's been fun :-)

> If I toss two coins until at least one shows a Head, what's the probability that both are Heads?

I answered in my last reply ...

  P(head_a AND head_b | head_b) = 1/3
The problem is that head_a and head_b are independent events, and that P(head_a) = 1 (you made sure of that).

   P(head_a AND head_b | head_a) = 
   P(head_a AND head_b) / P(head_a) =
   P(head_a AND head_b) =
   P(head_a) = 1/2
QED :)

Re: When intuition and math probably look wrong

#23

Post by RiderOfGiraffes »

bad_user wrote: post #22
RiderOfGiraffes wrote: post #18

Earlier quoted context omitted.

Actually, I'm going to stop now: http://xkcd.com/386/ It's pretty clear I won't convince you. Let me leave you with these questions: If I toss two coins until at least one shows a Head, what's the probability that both are Heads? If I roll two dice until at least one shows a 6, what's the probability that both are 6's? If I spin two roulette wheels until at least one shows a Red-23, what's the probability that both a…

It's been fun :-) > If I toss two coins until at least one shows a Head, what's the probability that both are Heads? I answered in my last reply ... P(head_a AND head_b | head_b) = 1/3 The problem is that head_a and head_b are independent events, and that P(head_a) = 1 (you made sure of that). P(head_a AND head_b | head_a) = P(head_a AND head_b) / P(head_a) = P(head_a AND head_b) = P(head_a) = 1/2 QED :)

I have no idea what problem you think you're solving, but it's not the one I posed. I never said a specific coin was heads, I just said at least one of them was heads.

Given that I make money on these sorts of questions (doing it for real in a semi-military context where it's important to be right, and you get tested against reality) I feel pretty confident that I know what I'm talking about. You're clearly talking about something completely different, and I really don't understand what you're saying.

And that's the last I'll say.

Re: When intuition and math probably look wrong

#24

Post by tlammens »

bad_user wrote: post #22
RiderOfGiraffes wrote: post #18

Earlier quoted context omitted.

Actually, I'm going to stop now: http://xkcd.com/386/ It's pretty clear I won't convince you. Let me leave you with these questions: If I toss two coins until at least one shows a Head, what's the probability that both are Heads? If I roll two dice until at least one shows a 6, what's the probability that both are 6's? If I spin two roulette wheels until at least one shows a Red-23, what's the probability that both a…

It's been fun :-) > If I toss two coins until at least one shows a Head, what's the probability that both are Heads? I answered in my last reply ... P(head_a AND head_b | head_b) = 1/3 The problem is that head_a and head_b are independent events, and that P(head_a) = 1 (you made sure of that). P(head_a AND head_b | head_a) = P(head_a AND head_b) / P(head_a) = P(head_a AND head_b) = P(head_a) = 1/2 QED :)

did you just prove that 1 == 1/2?

Re: When intuition and math probably look wrong

#25

Post by bad_user »

RiderOfGiraffes wrote: post #23
bad_user wrote: post #22

Earlier quoted context omitted.

It's been fun :-) > If I toss two coins until at least one shows a Head, what's the probability that both are Heads? I answered in my last reply ... P(head_a AND head_b | head_b) = 1/3 The problem is that head_a and head_b are independent events, and that P(head_a) = 1 (you made sure of that). P(head_a AND head_b | head_a) = P(head_a AND head_b) / P(head_a) = P(head_a AND head_b) = P(head_a) = 1/2 QED :)

I have no idea what problem you think you're solving, but it's not the one I posed. I never said a specific coin was heads, I just said at least one of them was heads. Given that I make money on these sorts of questions (doing it for real in a semi-military context where it's important to be right, and you get tested against reality) I feel pretty confident that I know what I'm talking about. You're clearly talking a…

Look, don't take this the wrong way ... probabilities are easy to get wrong ... I'm only having this conversation with you because my math skills are rusty.

> I never said a specific coin was heads

OK, so let's make it mathematically correct (let's say we're painting them) ... making an effort here :)

  P( blue_head AND red_head | blue_head OR red_head)
  P(blue_head OR red_head) = 1  
      you said that you're retrying until this happens
      normally this would be 3/4
Making the problem ...

  P( blue AND red | blue OR red) 
  = P( blue OR red | blue AND red ) * P( blue AND red ) / P(blue OR red)
      (applied bayes)
  = 1 * P(blue) * P(red) / 1
But ...

   P(blue OR red) = P(blue) + P(red) - P(blue) * P(red) = 1 =>
   P(blue) * P(red) = 1 - P(blue) - P(red) = 1 - 1/4 - 1/4 = 1/2
Show me the error.

[EDITED] ... modified the stuff as I've totally fucked up the previous version :) ... as I said, I'm rusty

Re: When intuition and math probably look wrong

#26

Post by tlammens »

bad_user wrote: post #25
RiderOfGiraffes wrote: post #23

Earlier quoted context omitted.

I have no idea what problem you think you're solving, but it's not the one I posed. I never said a specific coin was heads, I just said at least one of them was heads. Given that I make money on these sorts of questions (doing it for real in a semi-military context where it's important to be right, and you get tested against reality) I feel pretty confident that I know what I'm talking about. You're clearly talking a…

Look, don't take this the wrong way ... probabilities are easy to get wrong ... I'm only having this conversation with you because my math skills are rusty. > I never said a specific coin was heads OK, so let's make it mathematically correct (let's say we're painting them) ... making an effort here :) P( blue_head AND red_head | blue_head OR red_head) P(blue_head OR red_head) = 1 you said that you're retrying until t…

multiplication and addition are not the same thing 1/4 * 1/4 = 1/16

Re: When intuition and math probably look wrong

#27

Post by bad_user »

tlammens wrote: post #26
bad_user wrote: post #25

Earlier quoted context omitted.

Look, don't take this the wrong way ... probabilities are easy to get wrong ... I'm only having this conversation with you because my math skills are rusty. > I never said a specific coin was heads OK, so let's make it mathematically correct (let's say we're painting them) ... making an effort here :) P( blue_head AND red_head | blue_head OR red_head) P(blue_head OR red_head) = 1 you said that you're retrying until t…

multiplication and addition are not the same thing 1/4 * 1/4 = 1/16

Oh shit ... I'm tired :)

Re: When intuition and math probably look wrong

#28

Post by RiderOfGiraffes »

bad_user wrote: post #25
RiderOfGiraffes wrote: post #23

Earlier quoted context omitted.

I have no idea what problem you think you're solving, but it's not the one I posed. I never said a specific coin was heads, I just said at least one of them was heads. Given that I make money on these sorts of questions (doing it for real in a semi-military context where it's important to be right, and you get tested against reality) I feel pretty confident that I know what I'm talking about. You're clearly talking a…

Look, don't take this the wrong way ... probabilities are easy to get wrong ... I'm only having this conversation with you because my math skills are rusty. > I never said a specific coin was heads OK, so let's make it mathematically correct (let's say we're painting them) ... making an effort here :) P( blue_head AND red_head | blue_head OR red_head) P(blue_head OR red_head) = 1 you said that you're retrying until t…

Your first error is here:

  P(BH or RH) = 1
This is supposed to be the probability without conditionals, in which case P(BH or RH) = 3/4. Using the Bayesian formula is what is taking into account the fact that at least one head has shown.

Your second error is here:

    P( blue_head ) * P( red_head ) (independent events)
    = 1/4 * 1/4
That's probably a typo, because you probably know that P(BH)=1/2.

Your third error is here:

    1/4 * 1/4 = 1/2
1/4 * 1/4 = 1/16

Even if you correct this to the 1/2 * 1/2 that you intended, you now get that P(BH and RH)=1/4. Surely you realise that this must be wrong. I've already told you that there's at least one head, so you can't get the same answer as when I've given you no information.

Here's the correct Bayesian sum:

    P(RH and BH | RH or BH)
  = P(RH and BH and (RH or BH)) / P(RH or BH)
  = P(RH and BH) / (3/4)
  = P(RH) * P(BH) / (3/4)
  = 1/2 * 1/2 / (3/4)
  = 1/4 / (3/4)
  = 1/3
The fact that I'm using the conditional probability is what uses the information I've given you, so you can't use that again and claim P(RH or BH)=1.

Show me the error.

Here, do an experiment. Toss two coins of different denominations. Consider those cases where there's at least one head. How often are they both heads? Are you a programmer? Run the simulation - show me the code.

Here, let me help:

  #!/usr/bin/python

  import random

  def toss():
    if random.random()

Re: When intuition and math probably look wrong

#29

Post by RiderOfGiraffes »

bad_user wrote: post #25
RiderOfGiraffes wrote: post #23

Earlier quoted context omitted.

I have no idea what problem you think you're solving, but it's not the one I posed. I never said a specific coin was heads, I just said at least one of them was heads. Given that I make money on these sorts of questions (doing it for real in a semi-military context where it's important to be right, and you get tested against reality) I feel pretty confident that I know what I'm talking about. You're clearly talking a…

Look, don't take this the wrong way ... probabilities are easy to get wrong ... I'm only having this conversation with you because my math skills are rusty. > I never said a specific coin was heads OK, so let's make it mathematically correct (let's say we're painting them) ... making an effort here :) P( blue_head AND red_head | blue_head OR red_head) P(blue_head OR red_head) = 1 you said that you're retrying until t…

Now replying to the modified version. My response to your first version is here: http://news.ycombinator.com/item?id=1473088

Anyway ...

Don't take this the wrong way ... probabilities are easy to get wrong ... I'm only having the conversation with you because you're getting it completely wrong, and yet you appear to want to learn.

You say:

  P(blue_head OR red_head) = 1  
      you said that you're retrying until this happens
      normally this would be 3/4
This is an incorrect application of Bayes' Theorem. The whole point of taking P(X|Y) isn't that P(Y) can be then taken as being 1. P(Y) remains the probability of Y. The point is that when we then only consider those events where Y occurs, then we have conditioned our probabilities, and the formula gives us what we want. Specifically, relative to Y, the probability of X changes.

Using "BH" for "Blue coin shows Head" and "RH" for "Red coin shows Head" we have:

    P(BH and RH | BH or RH)
  = P(BH and RH and (BH or RH)) / P(BH or RH)
  = P(BH and RH) / P(BH or RH)
  = 0.25 / 0.75
  = 1/3
You then go on to say:

  P(BH OR RH) = P(BH) + P(RH) - P(BH) * P(RH) = 1
Taking just the second part of this:

  P(BH) + P(RH) - P(BH) * P(RH) = 1
and adding P(BH) * P(RH) to both sides we get:

  P(BH) + P(RH) = 1 + P(BH) * P(RH)
Since by symmetry P(RH)=P(BH), and letting x=P(BH)=P(RH) this simplifies to

  x + x = 1 + x^2
The only solution to that is x=1, so P(BH)=P(RH)=1

Your math is clearly screwed at this point.

So I've given you the correct interpretation, I've shown you where your calculations are wrong, and I've explained your incorrect use of Bayes' Formula.

Twice.

Re: When intuition and math probably look wrong

#30

Post by dododo »

bad_user wrote: post #9
dododo wrote: post #4

isn't this the same kind of "intuitive" reasoning that fails at simpson's paradox? http://en.wikipedia.org/wiki/Simpsons_paradox conditioning upon more events can lead to a higher probability. p(boy = 2 | boy >= 1) = 1, tuesday) (or more precisely, conditioning upon more events can yield a distribution with less entropy)

You know, the problem with that conditional probability is that the sex of the second child is in no way conditioned by the sex of the first child, so ... p(boy = 2 | boy >= 1) = p(any child = boy) And this was the original problem that led them to the 33% probability.

in the sample space i was intending, that would not be the case. i was imagining boy as a random variable that counts instances in an order tuple of genders (the underlying sample space).

you're right, without this explicit construction, it's problematic.

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