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Think you understand Monty Hall? Try the Tuesday boy problem.

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Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#101

Post by zerostar07 »

jfager wrote: post #89
zerostar07 wrote: post #81

Earlier quoted context omitted.

Alternatively, suppose we label the combinations of child-day BM, GM, BT, GT ... BU, GU (14 items). We have a bucket with 2 of each of these combinations, as there are 2 kids (total 28 items). The problem states that we have taken one "BT" item out of the bucket (so 27 items left), and are asking how many "B*" items are left in the bucket (13). The confusion arises because one can consider the same problem with repla…

Having 2 each of BM, GM, etc. is implicitly labeling 1BM, 2BM, 1GM, 2GM, etc. If you sampled with replacement, you have to admit the possibility that you draw the same BT twice. It's a stretch to suggest that the statement is ambiguous in this way, as it would imply that the father could have the same child twice.

Trying to figure out why it seems wrong at first: it seems my first reaction would be emotional, that the odds of having a boy should be 0.5. However it is exactly because the total probability of a boy has to be .5 that the probability of a second boy has to be less than 1/2. The fact that you can pair "boy" with any enumerable attribute that can bring the probability down to 1/3 is ... funny.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#102

Post by NickPollard »

bpm140 wrote: post #100

Wait. What? The author says there can only be one BB but that there can be both GB and BG? What he's doing here is saying that birth order functionally doesn't matter if the sibling is a boy, but it does matter if it's a girl. How is this correct? If you keep comparing apples to apple you get: Older Boy / Boy Boy / Younger Boy Older Girl / Boy Boy / Younger Girl And we're back to a 50% chance that the other child is…

Altogether now! Here's the chart:

   B   G
B BB BG

G GB GG

Satisfied?

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#103

Post by zerostar07 »

reitzensteinm wrote: post #58

I've been thinking about this for an hour, and I'm now convinced that the author is wrong. The fact that we found out about one of the children from the father means that all probabilities are not equal, even though they're treated here like they are. The difference is between the information being offered, and determined independantly. I'll do this with the boy/girl problem, for simplicities sake. If we ask a man if…

That last comparison is false. In both cases p=13/27 .

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#104

Post by suivix »

bpm140 wrote: post #100

Wait. What? The author says there can only be one BB but that there can be both GB and BG? What he's doing here is saying that birth order functionally doesn't matter if the sibling is a boy, but it does matter if it's a girl. How is this correct? If you keep comparing apples to apple you get: Older Boy / Boy Boy / Younger Boy Older Girl / Boy Boy / Younger Girl And we're back to a 50% chance that the other child is…

Since he has two children, you know first off that these are all equal chance historically:

BB BG GB GG

Now since you know the child is a boy, it eliminates the fourth option. So now we have these possibilities:

BB BG GB

In only one of these is the other child a boy, so the chance is 1/3.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#105

Post by ajanuary »

ars wrote: post #65
ajanuary wrote: post #62

Earlier quoted context omitted.

Looking at your simple example of genders using the same process the article uses to enumerate the combination of gender and days: Assuming Child 1 is a Boy: Child 2 can be: Boy, Girl Assuming Child 1 is a Girl: Child 2 can be: Boy, Girl Assuming Child 2 is a Boy: Child 1 can be: Boy, Girl Assuming Child 2 is a Girl: Child 1 can be: Boy, Girl Combining those gives us the following combinations: Child 1 | Child 2 ----…

Except you are not actually supposed to remove the duplicates!! It is a very common mistake, but it's simply incorrect, it leads to incorrect results. Also, why are you numbering the kids as child 1/2? There is no such distinction made. If you changed your list so that the fixed child is always listed first, and removed duplicates you would have 4 possibilities.

The removal of duplication and the distinguishing between GB and BG are two completely different things.

If you're not meant to remove the duplicates, then why doesn't your enumeration of two gendered children run: BB, BG, GB, GG, BB, GB, BG, GG?

I number them merely to distinguish them. The "a child has 50% chance of being a boy, and 50% chance of being a girl" applies to a single child, so you need to enumerate their states independently. To do that you need to be able to distinguish between them.

It is this fact that each child's possible states should be treated independently that means you can't combine BG and GB, not any aversion to removing duplication. Because they are independent entities BG represents one - nominally called child 1 - is a B while the other - nominally called child 2 - is a girl. GB represents one - nominally called child 1 - is a G while the other - nominally called child 2 - is a B.

If the list were changes so that the fixed child is always in the list we wouldn't be enumerating all the possibilities for each child and then combining them. That would be falling into exactly the same mistake you are keen to avoid.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#106

Post by drblast »

The crux of the issue is much better illustrated by the simplified problem where if a man tells you one of his kids is a boy, the other child is a boy in 1/3 cases. That's intuitive looking at the possibilities:

BG GB BB

Where this analysis gets confusing is when you say that if you meet the man with one of his kids at random and it's a boy, the probability changes to 1/2. The reason is in that case one of the possible outcomes was indeed GG, even if the child you met was a boy. The difference is how much knowledge you had about the situation prior to making your measurement of the outcome.

If that doesn't make your head spin even a little you're not really human.

I was able to make peace with these problems when I read about the Two Envelopes Paradox and realized that it's not really possible to pick a random integer, because by picking a random integer you're essentially limiting the domain of integers you've picked from to a finite one. If you can wrap your head around that it might help you save your sanity.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#108

Post by its_so_on »

its_so_on wrote: post #99
danparsonson wrote: post #67

Earlier quoted context omitted.

You can follow a similar process to work out the probability as for the original question. For any two children, the possibilities are: Child 1 born Mon, Tue, Wed, Thu, Fri, Sat, Sun Child 2 born Mon, Tue, Wed, Thu, Fri, Sat, Sun We know that one of the children was born on a Tuesday, but we haven't said whether it's Child 1 or Child 2 (and that lack of information is key to understanding the original problem), so ou…

This is ridiculous. Let's do some queries on the world: From all women: (some 3 billion) the ones who have exactly two living children: x matches the ones in which it is true for them to say "One of my children was born on Tuesday.". (this is true or false for every mother that comprises x. y mothers remain (can answer "true" for that question") the ones that can say "both my children were born on Tuesday": z matches…

Wait, I think I get it. (Sorry, too late to update).

This is indeed ingenious. The key is the step from x to y. The women in x who can say "one of my children was born on Tuesday" are the ones who can say "EITHER my first OR my second child was born on Tuesday"; the initial constraint (when you meet the person) is thus: "women who can say I have exactly 2 children and it is true to say EITHER my first OR my second child was born on a Tuesday"; obviously this is far more than 1/7th. Then the additional constraint "BOTH of them were" is a smaller addition than 1/7th.

It's not 1/7th the first time and 1/7th the second time, because the question wasn't "of women who can say they have exactly two children, the number who can say 'my elder child was born on a Tuesday'" and then "of these the women who can say 'my younger child was also born on Tuesday". This would indeed be 1/7th each time.

instead y is "EITHER my elder OR my younger child was born on Tuesday". This obviously results in a set that is greater than 1/7th of all women with two children, and, consequently, it is no surprise that there is a correspondingly smaller than 1/7th possibility that both were.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#109

Post by raldi »

invertd wrote: post #64

How would the solution change if the problem was stated as follows: 'You meet a man on the street and he says, “I have two children and one is a son born in March.” What is the probability that the other child is also a son?'

Or, "...one is a son born in an even-numbered month"

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#110

Post by danparsonson »

its_so_on wrote: post #108
its_so_on wrote: post #99

Earlier quoted context omitted.

This is ridiculous. Let's do some queries on the world: From all women: (some 3 billion) the ones who have exactly two living children: x matches the ones in which it is true for them to say "One of my children was born on Tuesday.". (this is true or false for every mother that comprises x. y mothers remain (can answer "true" for that question") the ones that can say "both my children were born on Tuesday": z matches…

Wait, I think I get it. (Sorry, too late to update). This is indeed ingenious. The key is the step from x to y. The women in x who can say "one of my children was born on Tuesday" are the ones who can say "EITHER my first OR my second child was born on Tuesday"; the initial constraint (when you meet the person) is thus: "women who can say I have exactly 2 children and it is true to say EITHER my first OR my second ch…

Bingo :-) I was just racking my brain for a different way to explain it but luckily you got there first!

It's a real mind-bender.

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