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Mike Honcho

Non-political Wednesday diversion-Brainteasers

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John  and Jane are a married couple  They have two kids, one of them is a girl. Assume safely that the probability of each gender is 1/2.
What is the probability that the other kid is also a girl?

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Normal answer: 0%

2019 answer: Genders are fluid, people who seek to assign constructs of gender based on the body organs present at birth are enforcers of patriarchal oppression.

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22 minutes ago, Mike Honcho said:

John  and Jane are a married couple  They have two kids, one of them is a girl. Assume safely that the probability of each gender is 1/2.
What is the probability that the other kid is also a girl?

.25%

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13 minutes ago, Voltaire said:

Normal answer: 0%

2019 answer: Genders are fluid, people who seek to assign constructs of gender based on the body organs present at birth are enforcers of patriarchal oppression.

I'm guessing this is the answer from the Chinese government.    

Keep guessing guys.

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3 minutes ago, Mike Honcho said:

I'm guessing this is the answer from the Chinese government.    

Keep guessing guys.

The first is the answer : 0%.

The second is modernity's LBGTQPZ2 answer.

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0%.  they have two kids, one of them is a girl.   The other isn't a girl then.

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33.3% because you have these possibilities:  GB, BG, GG, BB

Eliminate BB cuz you know it's not that.  That leaves 1 in 3.

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34 minutes ago, Mookz said:

33.3% because you have these possibilities:  GB, BG, GG, BB

Eliminate BB cuz you know it's not that.  That leaves 1 in 3.

And Mookz is the winner!   

He can supply a new one or I'll put up another in a little bit.

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2 minutes ago, Mike Honcho said:

And Mookz is the winner!   

He can supply a new one or I'll put up another in a little bit. 

Okay, here's one for teh geek bored:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

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52 minutes ago, Mookz said:

33.3% because you have these possibilities:  GB, BG, GG, BB

Eliminate BB cuz you know it's not that.  That leaves 1 in 3.

this can not be right

 

you know the sex of 1 child F

The question is what are the odds of child 2 being F and that is a 50 50 chance regardless of what Child 1 was

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56 minutes ago, Mookz said:

33.3% because you have these possibilities:  GB, BG, GG, BB

Eliminate BB cuz you know it's not that.  That leaves 1 in 3.

BB is not a viable option and GB/BG are the same.

It's either GG or GB................. 

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25 minutes ago, Mookz said:

Okay, here's one for teh geek bored:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Always switch.

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1 hour ago, patweisers44 said:

0%.  they have two kids, one of them is a girl.   The other isn't a girl then.

Forest for the trees..... :cheers:

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36 minutes ago, Mike Honcho said:

And Mookz is the winner!   

He can supply a new one or I'll put up another in a little bit.

It’s 50%. I typed a response to this post but deleted it because I didn’t like the way I worded it. Others have covered it. 

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16 minutes ago, Filthy Fernadez said:

BB is not a viable option and GB/BG are the same.

It's either GG or GB................. 

Seems right to me.  Even though the BG and GB cases are the same case, they would happen twice as often as a GG combo.

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Here's one.  

You've been diagnosed with a rare disease.  It only occurs in 1 in 10k people.  The test for the disease gives a true positive 99.9% of the time.  What are your chances of having the disease?

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Just now, Mike Honcho said:

 

Boy/Girl   Girl/Boy are the same probability given that we weren't factoring in age (boy first then girl  or girl first then boy).

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1 minute ago, nobody said:

Here's one.  

You've been diagnosed with a rare disease.  It only occurs in 1 in 10k people.  The test for the disease gives a true positive 99.9% of the time.  What are your chances of having the disease?

If you've already been diagnosed - I'd say 100%.

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Just now, Cruzer said:

If you've already been diagnosed - I'd say 100%.

You can say you've "been told by a doctor" if you prefer that to diagnosed.

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50 minutes ago, Mookz said:

Okay, here's one for teh geek bored:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Monty Hall says switch.  

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I admit I kind of backed into my answer.  :D

It's not intuitive.  But the way I justify it in my mind is that if it were 50%, then out in the real world you would expect to find just as many couples with GG as with B and G, and that just don't seem right. 

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18 minutes ago, nobody said:

Here's one.  

You've been diagnosed with a rare disease.  It only occurs in 1 in 10k people.  The test for the disease gives a true positive 99.9% of the time.  What are your chances of having the disease?

The answer involves Bayes Theorem but I don't feel like figuring that out.  Basically we know the accuracy if you have the disease; the question is the probability you have the disease given the result.

Also put me in the 50% camp for the OP, at least how it was worded.

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Multiple Events

Independent and Dependent Events

Suppose now we consider the probability of 2 events happening. For example, we might throw 2 dice and consider the probability that both are 6's.

We call two events independent if the outcome of one of the events doesn't affect the outcome of another. For example, if we throw two dice, the probability of getting a 6 on the second die is the same, no matter what we get with the first one- it's still 1/6.

On the other hand, suppose we have a bag containing 2 red and 2 blue balls. If we pick 2 balls out of the bag, the probability that the second is blue depends upon what the colour of the first ball picked was. If the first ball was blue, there will be 1 blue and 2 red balls in the bag when we pick the second ball. So the probability of getting a blue is 1/3. However, if the first ball was red, there will be 1 red and 2 blue balls left so the probability the second ball is blue is 2/3. When the probability of one event depends on another, the events are dependent.

https://revisionmaths.com/gcse-maths-revision/statistics-handling-data/probability

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20 minutes ago, Strike said:

Multiple Events

Independent and Dependent Events

Suppose now we consider the probability of 2 events happening. For example, we might throw 2 dice and consider the probability that both are 6's.

We call two events independent if the outcome of one of the events doesn't affect the outcome of another. For example, if we throw two dice, the probability of getting a 6 on the second die is the same, no matter what we get with the first one- it's still 1/6.

On the other hand, suppose we have a bag containing 2 red and 2 blue balls. If we pick 2 balls out of the bag, the probability that the second is blue depends upon what the colour of the first ball picked was. If the first ball was blue, there will be 1 blue and 2 red balls in the bag when we pick the second ball. So the probability of getting a blue is 1/3. However, if the first ball was red, there will be 1 red and 2 blue balls left so the probability the second ball is blue is 2/3. When the probability of one event depends on another, the events are dependent.

https://revisionmaths.com/gcse-maths-revision/statistics-handling-data/probability

This is Strikes way of saying that I am right, this must be a very very painful day for him.   :)

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The video that Honcho posted is good.  The people saying 50% are assuming it's the second case that he goes through, that we already had the girl, and now we're having another baby.  But in reality the girl could have been born first or second, which means the boy could also be born either first or second, not just second.

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4 minutes ago, Mike Honcho said:

This is Strikes way of saying that I am right, this must be a very very painful day for him.   :)

Wow, I always knew reading comprehension was a challenge for you, but this seems pretty straightforward.  Even Forrest Gump could understand it.

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2 minutes ago, Mookz said:

The video that Honcho posted is good.  The people saying 50% are assuming it's the second case that he goes through, that we already had the girl, and now we're having another baby.  But in reality the girl could have been born first or second, which means the boy could also be born either first or second, not just second.

It doesn't matter what baby he was talking about in the OP.  And I didn't watch the video so not going to comment on it.  But having a baby is completely independent of any other baby.  The order doesn't matter.  The probability of each baby is the same.  We're being told that this probability is 1/2.  Therefore, it's 1/2 for the first baby, the second baby, and the 1000th baby.

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2 minutes ago, Strike said:

Wow, I always knew reading comprehension was a challenge for you, but this seems pretty straightforward.  Even Forrest Gump could understand it.

The most ironic post ever made at the geek club.  You might want to watch the video I posted what you posted shows why the answer is 33%---like I said.   

So the probability of getting a blue is 1/3.

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The question is not worded what is the probability of a couple having a GG

 

The question is what is the probability of the unknown child being a girl. Knowing child 1 has 0 impact on the probability of the unknown child

 

im still saying its 50%

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