michael99000
michael99000
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January 14th, 2014 at 12:39:42 AM permalink
In other words, if I tell you I have a son and a pregnant wife, the odds of that second kid being a boy are 50/50.

An hour later the kid is born. I tell you I have a son and another kid at home. Now the odds that other kid is a boy is 33% ?

Doesn't make sense, the only thing that changed is the physical location of the second kid .
Jeepster
Jeepster
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January 14th, 2014 at 12:48:29 AM permalink
Quote: michael99000



Let's change the question slightly and say...

You currently have one child and your wife is home pregnant with you're next child. The child I see you with is a boy. What's the chances that the unborn child is also a boy?



Because you are now putting a time line on the events, making the first born a boy.
That only leaves two possibilities. BG or BB, a 50/50 chance of the second being a boy.
When the order of events is not known there are 3 possibilities. BB, BG or GB, in two of these cases the other child is a girl, thus only a 1/3 chance of the other being a boy.
A photon without any luggage checks into a hotel, he's travelling light.
24Bingo
24Bingo
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January 14th, 2014 at 1:01:15 AM permalink
Quote: michael99000

In other words, if I tell you I have a son and a pregnant wife, the odds of that second kid being a boy are 50/50.

An hour later the kid is born. I tell you I have a son and another kid at home. Now the odds that other kid is a boy is 33% ?

Doesn't make sense, the only thing that changed is the physical location of the second kid .



An hour later you tell someone else that you have two children, one of them a son. Not differentiating between them in any way, the odds that you have two sons rise from 25% to 33%. But tell that someone else your younger child is a son, it goes up to 50%.

But here's the thing - distinguish them in any way unrelated to gender, and it goes up to 50%. The child you've seen is a son; it goes up to 50%. The child earlier in the alphabet is a son; it goes up to 50%. One child has a birthmark and one doesn't, and the one without is a son; it goes up to 50%.

This is why I said "the westward coin." It could have been the faster-falling coin, or the shinier coin, or anything else, but it's just the same as saying "the first coin came up heads." What matters isn't chronology, but the fact that the two are distinct.

Let's think about it with the old familiar Monty Hall problem. A common way to illustrate that the change in probability isn't as ridiculous as it first appears is to go from three doors to ten, opening eight. Anyone can see then you should switch. But if the doors had been opened one at a time, with the host not knowing whether the car was behind each one, then it doesn't matter, because it was so improbable that you should have gotten there.

So let's put a similar scaling to this problem. "I've flipped a coin ten times, and got at least nine heads. With no other information, what's the chance I got ten?" 1/11. "You saw me flip heads nine times. I flipped the coin one more time, at some point buried in the nine flips, and you don't know the chronology, but which flips you saw were unrelated to their outcome. Which way was that hidden flip more likely to have come up?" They're equal.

Unless, of course, you'd never be seen in public with your daughter.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
Jeepster
Jeepster
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January 14th, 2014 at 1:07:24 AM permalink
You're opening a can of worms with the Monty Hall problem.
I'm not going near that one, most people will not accept the correct answer.
A photon without any luggage checks into a hotel, he's travelling light.
Canyonero
Canyonero
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January 14th, 2014 at 1:18:10 AM permalink
Let me jump into the fray here:

The following is assuming that you are not a Taliban, i.e. you might take your girl outside as well, unbiased. So we start with the following options:

B1B2 G1B3 B4G2 G3G4
The chances of meeting you are 12.5% for each of the children possible.
Now, by observation, we can rule out all the girls, that leaves us with :

B1 B2 B3 B4
So there is a 50% chances it is either boy 1or boy 2 form the two boy combination– you have two boys.
Or it might be a boy from either BG combination – 50% chance here as well.

The chance is 1/2.
Jeepster
Jeepster
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January 14th, 2014 at 1:28:34 AM permalink
Quote: Canyonero

Let me jump into the fray here:

The following is assuming that you are not a Taliban, i.e. you might take your girl outside as well, unbiased. So we start with the following options:

B1B2 G1B3 B4G2 G3G4
The chances of meeting you are 12.5% for each of the children possible.
Now, by observation, we can rule out all the girls, that leaves us with :

B1 B2 B3 B4
So there is a 50% chances it is either boy 1or boy 2 form the two boy combination– you have two boys.
Or it might be a boy from either BG combination – 50% chance here as well.

The chance is 1/2.



Sorry, I really have no idea of what you are saying.
A photon without any luggage checks into a hotel, he's travelling light.
gameterror
gameterror
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January 14th, 2014 at 1:45:10 AM permalink
Quote: Jeepster

Because you are now putting a time line on the events, making the first born a boy.
That only leaves two possibilities. BG or BB, a 50/50 chance of the second being a boy.
When the order of events is not know there are 3 possibilities. BB, BG or GB, in two of these cases the other child is a girl, thus only a 1/3 chance of the other being a boy.



ok, let's do all possibilities in a tree:

0
B G
BB BG GB GG

In total there are 4 possible outcomes of having 2 children. Now all we know that one of same is a boy. This eliminates only the solution GG. Leaving us with 3 possiblities (BB, BG, GB) which are equally likely. So the probability of the other also being a boy is 1/3.

edit:
I guess the problem that Canyonero has (and i also had for a short moment just now) is that he "creates" another solution where the order of the BB solution matters. Creating two variants B1B2 and B2B1 which doesn't exist. See table above. There is only one BB possibility and it doesn't matter that we don't know if we see the old or young boy of the boy-boy combo.
Things have never been so swell I have never failed to fail
Canyonero
Canyonero
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January 14th, 2014 at 2:55:31 AM permalink
Quote: gameterror


edit:
I guess the problem that Canyonero has (and i also had for a short moment just now) is that he "creates" another solution where the order of the BB solution matters. Creating two variants B1B2 and B2B1 which doesn't exist. See table above. There is only one BB possibility and it doesn't matter that we don't know if we see the old or young boy of the boy-boy combo.



No, there is only BB BG GB GG, the number I only put there as a point of reference (to little effect it seems).

In short, for a BB parent the chances of meeting them with a boy is 100%, while for either of the GB combinations it is only 50%. That is why it is not 1/3 but 1/2.
gameterror
gameterror
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January 14th, 2014 at 3:08:37 AM permalink
Quote: Canyonero

No, there is only BB BG GB GG, the number I only put there as a point of reference (to little effect it seems).


ok, then i don't understand how you come to a different solution. 4 variants. each variant of birth has the same probability. now the fact that we see at least one boy eliminates the GG variant and only this one. so we are left with the 3 others BB, BG, GB. leaving the second kid we don't see as a B, G, G ...so 1/3.

Quote: Canyonero


In short, for a BB parent the chances of meeting them with a boy is 100%, while for either of the GB combinations it is only 50%. That is why it is not 1/3 but 1/2.



the fact that we would only see a boy in a GB or BG combination 50% of the time is no longer relevant...as we have already SEEN him.
Things have never been so swell I have never failed to fail
AZDuffman
AZDuffman
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January 14th, 2014 at 3:09:10 AM permalink
Quote: michael99000

Here's why I'm having trouble understanding why the chances the other child is a boy arent 50%.

Let's change the question slightly and say...

You currently have one child and your wife is home pregnant with you're next child. The child I see you with is a boy. What's the chances that the unborn child is also a boy?

According to the question, the world rate for sex of babies is 50/50, so the answer to my question must be the chances that the other unborn child will be a boy is 50%

So, for the question posed by Jeepster, the only difference is the child has already been born. Why does that fact alone make it 17% less likely that the child is a boy. ???

The only difference between the two questions is the physical location of child #2, inside the mother or outside . Seems that shouldn't change the odds it's a boy




You mostly beat me to this. Here is another way of putting it:

There is a "no zero" roulette game, the last time the wheel was spun it landed on red. WThat are the chances it will land on red the next spin?

The chances of the other child being a boy should be 50%. The sex of the child with the parent at the time should not matter.
All animals are equal, but some are more equal than others

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