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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.

#111

Post by onemoreact »

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…

The Monty hall problem is based on the idea that you will always do the same thing in response to my choice aka he can always pick an open door. It's feels add to think about it in terms of what's already happens but seems more reasonable to say it in terms of something that will happen.

So if you say I will flip 2 coins and if I get zero heads I will flip again. So, if the first coin is a head second one either a head or a tail, but if it's a tail you know the second one is a head or I would have flipped again. Thus 3 options one of which is HH.

Assuming you used the same approach with the Tuesday boy problem, aka the first one can be BMTWTFSS or GMTWTFSS and the second one can be BMTWTFSS, GMTWTFSS but if I don't get a BT from the first or second try's I will pick again. Thus BT + BMTWTFSS or GMTWTFSS, OR BMTWTFSS or GMTWTFSS + BT minus a BT,BT which would otherwise be counted twice. Thus it's 14 + 13 options with 7 + 6 being BB. Which works out to 13/27.

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

#112

Post by bpm140 »

suivix wrote: post #104
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.

This explanation makes a ton of sense. I get it now. Thank you.

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

#113

Post by twelvechairs »

incremental wrote: post #97
twelvechairs wrote: post #59

Earlier quoted context omitted.

Okay. To respond to both of you - I have put this in programming just to check, and I think you are taking a different inference from the question than I am. Under your inference, the man wouldnt have mentioned anything unless he had at least one male child. (in which case you can say the GG scenario is gone, but GB BG and BB are equally probable) Under my inference, the man just told me the sex of one child at rando…

"Under your inference, the man wouldnt have mentioned anything unless he had at least one male child. (in which case you can say the GG scenario is gone, but GB BG and BB are equally probable)" No. I'm just assuming he has two children, and randomly mentions something about one of them. The GG scenario is only eliminated after he makes his statement, because we then know he has at least one boy.

Ah, then I do think you have an error of logic.

Put it this way. before he says it, we have GG, GB, BG, BB

after he says "I have a child that is [MALE OR FEMALE]" we have (where the capital letter is the child whose sex has been mentioned, and the lowercase letter is the other child):

Gg, gG, Gb, gB, Bg, bG, Bb, bB

So if he has said the sex is male, then we have four combinations left:

gB, Bg, Bb, bB.

Understand that we go to more scenarios (8) based on which child is mentioned, before we go to fewer. Actually the order of the children is something you can and should ignore, however as you are holding on to it, I show it this way....

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

#115

Post by [deleted] »

nadam wrote: post #21

As the currently top voted comment does not get it, I try to intuitively explain the paradox. No, it is not ambiguity of language. It says formally: I have two children. There exists a child of mine who is (boy and born on tuesday). And yes, the probability of the other child being a boy is 13/27. To understand this, try a more extreme case: When a child is born we generate a random number: rnd(1billion) Now the man…

[deleted]

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

#116

Post by pkteison »

ashishb4u wrote: post #91

I have a son born on xDay. What is the probability that my next kid will be a son? I dont get Monty Hall :(

The Tuesday Birthday problem is related to you not having info about the order leading to extra possibilities that you don't intuitively consider. Like in the simple example, if one son is a boy, probability of other being a boy is 1/3rd because the info you have the sets available could be BB, BG or GB. But if the younger son is a boy, then the GB possibility is removed, so it's 1/2 on the older son - BB or BG.

In your case, your younger kid is a son born on xDay. If you assume you will have another kid, chances on being a son are still 50%. It only gets interesting if either kid could have been a son.

In Monty Hall, the info that you don't intuitively consider is that Monty is providing additional information. You think of it as 50/50 because Monty ruled out a goat and car has to be behind one of two remaining doors, but actually the way it works is you chose a door that 1/3rd had a car, leaving 2/3rds chance of car on the other two doors. Then Monty eliminated one of those two doors, still leaving 2/3rds chance of car on the remaining door. So you should switch to that door. In Monty Hall, first you divide the set into a 1/3rd chance group of 1 door and a 2/3rd chance group of 2 doors, then Monty makes the second group a 2/3rd chance group of just 1 door.

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

#117

Post by ajanuary »

danparsonson wrote: post #88
ars wrote: post #53

Earlier quoted context omitted.

No, you do not remove the duplication. If you have two kids there are 4 possibilities, not 3: BG, BG, BB, GG BG seems to be the same as GB, except that it's not. And it's not in this case either.

I think I've spotted where your misunderstanding is. BG is only not the same as GB if there is some other information available - which was born first, what their names are, hair colour, etc., because then you'd be saying something like Boy born first, Girl born second Girl born first, Boy born second and those are two distinct possibilities. The point is that they are only distinct if you have this extra information…

No, they are correct in saying that GB and BG are distinct, even with no other information.

It is not order that is important, but considering each child as a distinct entity.

The 50% chance of being a boy and 50% chance of being a girl applies to a single independent child. When enumerating the possible combinations we need to first enumerate the possibilities for each child, and then combine these two enumerations into our overall enumeration.

To help us distinguish between the two children, let's call one Sam and the other Alex. There's no ordering over them, it's just to help us tell which one we're talking about.

Sam can be: Boy, Girl

Alex can be: Boy, Girl

Combining those gives us:

Sam is a Boy, Alex is a Boy

Sam is a Boy, Alex is a Girl

Sam is a Girl, Alex is a Boy

Sam is a Girl, Alex is a Girl

Or, to use a shorter notation, BB, BG, GB, GG.

Using oldest and youngest is just another handy way of distinguishing between the two children. Even if they were nameless, faceless children with no distinguishing characteristics other than gender we still need to consider then separately, each as their own entity with their own enumerations of possible genders.

Where the parent commenter is wrong is saying that (correctly) considering GB and BG as distinct combinations is the same as considering "Oldest Boy born on Tuesday and Youngest Boy born on Tuesday" distinct from "Youngest Boy born on Tuesday and Oldest Boy born on Tuesday". They are not. When you stop thinking about ordering and instead think about the enumerated states of each entity (child) involved it becomes clear both are saying the same information. They are not distinct, but the same combination phrased differently.

The parent commenter is basically saying BB should be distinct from BB just because the first one talks about Sam first and the second one talks about Alex first. This is clearly wrong.

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

#118

Post by suivix »

bpm140 wrote: post #112
suivix wrote: post #104

Earlier quoted context omitted.

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.

This explanation makes a ton of sense. I get it now. Thank you.

Also I don't think the exact day of birth being Tuesday makes one bit of difference to the gender. For example, if you were to say the known boy was born crying, and the chance of this is 70%, it doesn't affect the gender of the other children. It's just useless trivia.

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

#119

Post by zerostar07 »

nadam wrote: post #21

As the currently top voted comment does not get it, I try to intuitively explain the paradox. No, it is not ambiguity of language. It says formally: I have two children. There exists a child of mine who is (boy and born on tuesday). And yes, the probability of the other child being a boy is 13/27. To understand this, try a more extreme case: When a child is born we generate a random number: rnd(1billion) Now the man…

  Randomly pick one of my children. He is a boy. What is the probability of the other being a boy? 
  Yes, that would be 1/2.
To be clear, that probability would be 1/3. But I believe i understand where you are getting at. You have to phrase the question in the form "pick one of my 2 children, he is a boy with property x=x0, what s the prob that the other is a boy?" .

The probability is prob = ((N/2)-1)/(N-1) where N=2 * 2 * (number of different property values for x). For N>>1, prob tends to 1/2

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

#120

Post by _dps »

If you generalize "born on Tuesday" to a generic attribute, with probability p (equal for boys and girls) it becomes easier to see what's going on. For brevity, let's say people with the attribute are positive and people without are negative.

There are seven (ordered) possibilities involving at least one positive boy, and we'll subdivide them into two subgroups: B+B- B-B+ B+G- G-B+ / B+B+ B+G+ G+B+. The important thing to note is that within each group the outcomes are equiprobable.

If p is near 1, then the first group has almost zero probability and we have ~1/3 chance of two boys. If p is near 0 then the second group has almost zero probability and we're left with ~2/4 chance of two boys.

Intuitively, if p is near 1 then the statement "I have at least one positive boy" is almost equivalent to "Both my children are positive, and I have at least one boy" (group 2, 33% chance). If p is near 0, then the statement is almost equivalent to "I have exactly one positive boy" (group 1, 50% chance).

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