Quick Logic Problem

KingCondanomation

KingCondanomation

New member
Say that Jane and Joan (who are unrelated) each has two children. We know that at least one of Jane's children is a boy and that Joan's oldest child is a boy. What are the odds that each of the women has 2 boys?
 
The chances for "each" would be the same since they'd have to be figured separately.

It is a 50% chance that each would have two boys.
 
Regular logic, or Danologic?

If it's the latter, there was a lady I used to work with at McDonald's who had 2 boys, so the chances are 100%.
 
KingCondanomation said:
No
Click to expand...
Then you asked the question incorrectly. You did not want the odds figured for "each" you wanted them figured together.

We know each woman has one boy, and they each have two children. There is a 50% chance that the child we don't know is a boy for each woman. Taken together, Thorn would be correct. .5 X .5 = .25 or 25%. The odds are 1 in 4 when taken together, they are 1 in 2 when taken separately.
 
Onceler said:
Regular logic, or Danologic?

If it's the latter, there was a lady I used to work with at McDonald's who had 2 boys, so the chances are 100%.
Click to expand...
No that's the Danocdotal story remember? Try not to get your strawmen mixed up when you're looking for an easy out to a debate or problem.
 
Damocles said:
Then you asked the question incorrectly. You did not want the odds figured for "each" you wanted them figured together.

We know each woman has one boy, and they each have two children. There is a 50% chance that the child we don't know is a boy for each woman. Taken together, Thorn would be correct. .5 X .5 = .25 or 25%. The odds are 1 in 4 when taken together, they are 1 in 2 when taken separately.
Click to expand...

I must admit that I am a little lost here as well. If we are calculating odds based on the given info then it seems thorn should be right with 25%.

If you are calculating the odds of two women each having two boys, then it is a different equation.

I've looked at the wording of your question and there seems to be no deception, so I can only assume that you are asking the question incorrectly.
 
Damocles said:
Then you asked the question incorrectly. You did not want the odds figured for "each" you wanted them figured together.

We know each woman has one boy, and they each have two children. There is a 50% chance that the child we don't know is a boy for each woman. Taken together, Thorn would be correct. .5 X .5 = .25 or 25%. The odds are 1 in 4 when taken together, they are 1 in 2 when taken separately.
Click to expand...
No, I know what I asked.

Here's a hint:
When you said .5 X .5 = .25 or 25% you are not incorrect.
 
Depends(pun ) on whether the dude wore 'Boxers' or 'Briefs'...temperature controls the outcome...then again maybe I was asleep during this class...naw this is a fact!
 
Epicurus said:
6.25%

If not I give up.
Click to expand...

Not because it's been established that each already has one boy. So we're calculating the probability only that the second child of each is a boy. The knowledge that each already has one boy is actually irrelevant and is one of those nasty little tricks that such questions often include just to mess with you.
 
You must log in or register to reply here.
Back
Top