Quick Logic Problem

KingCondanomation said:
MottleyDude got it, the problem is right here and explained, they even explain the exact mistake Damo made (and possibly you).
Please read it and see for yourself:
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
Click to expand...
No, I have explained the mistake they make.

The Given is set, regardless of birth position.

John will always be John and cannot be both John and James.

Only if both children were variable would the solution be 1/3.

There would be a 1/3 probability that she would have no boys at all.

They misapply the variable.
 
Damocles said:
If both coins are variable (both unknown) you have three possibilities.

heads-tails
heads-heads
tails-tails

When one is given, you are figuring the probability of only one coin and it becomes 50%.

Given - Heads
Given - Tails

The problem Dano is having is understanding the principal of "The Given".
Click to expand...


Each toss is independent of all others, and whatever has occurred previously or subsequently has no influence at all on the outcome of a single toss. Each toss has the probability of 0.5 for each side.

Similarly, in this question, one boy in each family has been established and has no bearing on the gender of the other child, whether that child is older or younger. The probability for each family is 0.5 that the other child is one or the other gender. EOS.
 
KingCondanomation said:
Damo, the problem is at the beginning, you do not have 3 possibilities, you have 4 possibilities:

heads-tails
heads-heads
tails-tails
tails-heads

There is only ONE way to get heads-heads, you flip the first coin and get heads and then you flip the 2nd coin and get heads. That's it, no other way.

To get a combination of heads and tails, you can either:
Flip the first coin and get heads and flip the 2nd and get tails
OR
Flip the first coin and get tails and flip the 2nd and get heads

Tell me you can agree with that. All the evidence produced backs what I said up.
Click to expand...


tails-heads is the SAME as heads-tails, it's not a different outcome.
 
KingCondanomation said:
Damo, the problem is at the beginning, you do not have 3 possibilities, you have 4 possibilities:

heads-tails
heads-heads
tails-tails
tails-heads

There is only ONE way to get heads-heads, you flip the first coin and get heads and then you flip the 2nd coin and get heads. That's it, no other way.

To get a combination of heads and tails, you can either:
Flip the first coin and get heads and flip the 2nd and get tails
OR
Flip the first coin and get tails and flip the 2nd and get heads

Tell me you can agree with that. All the evidence produced backs what I said up.
Click to expand...
tails-heads is equal to heads-tails.

There would be a 1/3 probability (if both coins were variable) that you would have one heads and one tails. (This is the principle they are misapplying in the Dano scenario, once you have a Given of "heads" for one, only the one coin has a probability factor and that factor will always be 50% regardless of how often you flipped it previously.)

There would be 1/3 probability that you would have two heads, and 1/3 again that you would have two tails.
 
AssHatZombie said:
King. Are we NOT assuming that when we flip a coin we have a 1 in 2 chance of getting a tail?
Click to expand...
Of course. But we are flipping 2 coins.

So if I flip 2 coins, I will either get:
1st heads, 2nd heads
1st heads, 2nd tails
1st tails, 2nd heads
1st tails, 2nd tails

Now say I tell you a fact that at least one of my coins is heads. Well then you can only cross off the tails-tails option, which leaves you with 3 different possible combinations left.
1 of those 3 combinations is heads-heads.
 
KingCondanomation said:
Of course. But we are flipping 2 coins.

So if I flip 2 coins, I will either get:
1st heads, 2nd heads
1st heads, 2nd tails
1st tails, 2nd heads
1st tails, 2nd tails

Now say I tell you a fact that at least one of my coins is heads. Well then you can only cross off the tails-tails option, which leaves you with 3 different possible combinations left.
1 of those 3 combinations is heads-heads.
Click to expand...
The problem you have is one result is "Given", thus only making one coin variable.

Given - Heads
Given - Tails

If you want to be superfluous you can add the irrelevant second tier to the chart.

Heads - Given
Tails - Given

Yet it will still always figure to be 50%.

If you have no given result you have three possible results.

Heads - Tails
Tails - Tails
Heads - Heads.

Then your probability will be 1/3 for each.

In your scenario you had a given, one is a boy.

There are only two possible results with an equal probability.
 
Damocles said:
tails-heads is equal to heads-tails.

There would be a 1/3 probability (if both coins were variable) that you would have one heads and one tails. (This is the principle they are misapplying in the Dano scenario, once you have a Given of "heads" for one, only the one coin has a probability factor and that factor will always be 50% regardless of how often you flipped it previously.)

There would be 1/3 probability that you would have two heads, and 1/3 again that you would have two tails.
Click to expand...
You are wrong. Use basic math. If I flip ONE coin, there are 2 different possible outcomes. If I flip 2 coins, there are 4 different possible outcomes.
2 to the power of 2
If I flip 3 coins, there are 8 different possible outcomes (2 to the power of 3) and on an on.
 
KingCondanomation said:
You are wrong. Use basic math. If I flip ONE coin, there are 2 different possible outcomes. If I flip 2 coins, there are 4 different possible outcomes.
2 to the power of 2
If I flip 3 coins, there are 8 different possible outcomes (2 to the power of 3) and on an on.
Click to expand...
Only if you maintain that one heads and one tails is "different" than one heads and one tails...
 
KingCondanomation said:
Of course. But we are flipping 2 coins.

So if I flip 2 coins, I will either get:
1st heads, 2nd heads
1st heads, 2nd tails
1st tails, 2nd heads
1st tails, 2nd tails

Now say I tell you a fact that at least one of my coins is heads. Well then you can only cross off the tails-tails option, which leaves you with 3 different possible combinations left.
1 of those 3 combinations is heads-heads.
Click to expand...

But one is known.
 
Another source:
"It gets more complicated if you have 2 coins. To make it easier to see, pretend you have a nickel and a dime. What is the probability that you get 2 heads or P(2h)? We know that this means we must have a head for the nickel and a head for the dime, but to finish the problem we need to know all the possibilities, so we will make a list:

nickel dime from the table we can see that there are four different possibilities,
but only one way to get 2 heads
head head
head tail probability of two heads = P(2h) = 1 way out of 4 = ¼
tail head
tail tail
"
http://www.capitan.k12.nm.us/teachers/shearerk/probability.html

It is inescapable that there are 4 possibilities and if the only given is that ONE of the coins must be heads, then all you can elininate is the tails-tails combination.
 
KingCondanomation said:
Another source:
"It gets more complicated if you have 2 coins. To make it easier to see, pretend you have a nickel and a dime. What is the probability that you get 2 heads or P(2h)? We know that this means we must have a head for the nickel and a head for the dime, but to finish the problem we need to know all the possibilities, so we will make a list:

nickel dime from the table we can see that there are four different possibilities,
but only one way to get 2 heads
head head
head tail probability of two heads = P(2h) = 1 way out of 4 = ¼
tail head
tail tail
"
http://www.capitan.k12.nm.us/teachers/shearerk/probability.html

It is inescapable that there are 4 possibilities and if the only given is that ONE of the coins must be heads, then all you can elininate is the tails-tails combination.
Click to expand...
Sigh.

One of the coins is a given.

Therefore only the other coin is variable. Since only one coin is a variable you have only four possible solutions two of which are redundant.

Heads - Given
Tails - Given
Given - Heads
Given - Tails

The problem you have in just "crossing this out" is dismissing the principal of the given quantity.
 
Damocles said:
Sigh.

One of the coins is a given.

Therefore only the other coin is variable. Since only one coin is a variable you have only four possible solutions two of which are redundant.

Heads - Given
Tails - Given
Given - Heads
Given - Tails

The problem you have in just "crossing this out" is dismissing the principal of the given quantity.
Click to expand...
I can't explain it anymore, so I'll give up.
But before I do, you are saying, I am wrong, the smartest woman in the world is wrong, the algebra teacher is wrong, the results of the survey of 18000 PEOPLE are wrong and the well known and documented probability problem that is so well known and examined that I showed you a link to it on Wikipedia is also wrong.
Is that correct?
 
KingCondanomation said:
Maybe if you guys realized that both children's sexes are known, but just known by the parents, the outsider is only told one is a boy.
Click to expand...
It doesn't matter if "both sexes are known by the parent" we are figuring the probability of the unknown quantity.

One is a given. I tried to help you out by "naming" that Given. Well, now his name is given.

Given - Boy
Given - Girl
Boy - Given
Girl - Given

Are the only possible scenarios. The only ones. BECAUSE you have Given, you have limited the variable to only one child and only two possibilities for outcome.
 
You must log in or register to reply here.
Back
Top