Think the Airplane on a treadmill was tough?

The fact that the problem involves the day of the week as a whole different combination is just retarded.
 
However I disagree and think it is still 50/50. Very tough though.
 
How exactly does this relate to the Airplane on a treadmill? This is a probability question.

And no, the airplane on a treadmill wasn't tough.
 
It's a hereditary thing. You know there is already one boy, so the chance of there being two boys in a row is less than having just one boy.
 
It's interpretation.

If your wife gives birth to a boy and then falls pregnant again, the likelihood of the second child being a boy is 50/50 (medical factors excluded). Probability has no memory.

The link posted above takes "Boy - Girl" and "Girl - Boy" to be separate combinations, which is true if you live in ******ry world. So it takes;

Boy - Girl
Boy - Boy
Girl - Boy
Girl - Girl

To all be possibilities, but the possibility of Boy - Girl and Girl - Boy is shared. This is particularly important because the question doesn't state whether the boy is the first child or the second, therefore the boy is interchangeable.

There are 2 possible combinations here, BG - BB. The chance is 50/50.

Anyone spot any flaws in my attempted logic?
 
Acepilotf14 said:
It's a hereditary thing. You know there is already one boy, so the chance of there being two boys in a row is less than having just one boy.
Click to expand...

Has nothing to do with hereditary... It's a simple probability thing.

Possible outcomes of having 2 children:

A boy then a girl
A boy then a boy
A girl then a boy
A girl then a girl

Last one is obviously crossed out, so that leaves the first 3 options, 2 of which involve having a girl. So the odds are 1/3 of the other being a boy. Assuming for each birth that the sex is 50/50.

However, this would be different depending on the time the question is asked. If it said that someone had a baby boy, and was going to have another baby, what are the odds it will be another boy. Then it would be 1/2, because that would cross out the "a girl then a boy" option.

The link posted above takes "Boy - Girl" and "Girl - Boy" to be separate combinations, which is true if you live in ******ry world. So it takes;
Click to expand...
Um, why? You're wrong. The different orders are in fact 2 separate outcomes, and must be treated as such.
 
It has nothing to do with the problem though, the problem could be about ANYTHING, not just babies.

Say a coin toss. I tossed 2 coins, etc.
 
THE POSSIBILITY OF EVERYTHING IS 50/50 BECAUSE IT EITHER WILL OR IT WON'T HAPPEN

derp derp derp derp

...

No but seriously, I anxiously await the Mythbusters special on this problem.
 
Mythbusters? Jesus christ, anyone with basic education in probability should know the answer to this after working it out.

This is 10x simpler than that 3 doors problem.
 
Vegeta897 said:
Um, why? You're wrong. The different orders are in fact 2 separate outcomes, and must be treated as such.
Click to expand...

Yes, but since you aren't told which child your boy is (child 1 or 2) then he is interchangeable is how I took it.

Anyway, I have now think that the answer is 1/3, because B/G is twice as likely to occur (BG + GB).

I'm lost but i think it's 1/3. I'll never be able to bend my mind such that the boy being born on a Tuesday has any kind of relevance at all.
 
FLICK the three door problem. I still can't accept that shit.
 
I'm afraid I can't see how it's not 50/50. Viewing the second child as an individual event after the first child, you have a 50% chance that it's a boy.
 
kazsymonds said:
I'll never be able to bend my mind such that the boy being born on a Tuesday has any kind of relevance at all.
Click to expand...
I'm pretty sure it isn't relevant.

Sheepo said:
I'm afraid I can't see how it's not 50/50. Viewing the second child as an individual event after the first child, you have a 50% chance that it's a boy.
Click to expand...

Read my post, if you haven't already. The thing is you can't view the second child as an individual event after, because it already happened and you have to look at all the possible sex outcomes of having 2 children and weigh each option equally. Since there are 4 total possible outcomes and the girl/girl is obviously ruled out, that leaves 3 options and only 1 of those 3 is two boys.
 
I've read it but I still don't understand how you can consider a coin flip which already happened relevant to the probability the next one will have the same outcome. I completely understand where you're coming from granted that neither of the children have been born.
 
If you isolate and remove the order, its 50/50

The answer of the question is in context of the COMBINATION of two children, rather than the number of either sexes of the children
 
I completely misunderstood the problem. I thought that it stated the the first child was a boy and not just either one of the children.
 
Pretty dumb question and answer still imo.

Also, the problem in the video, if you pick the second time utterly randomly, without consideration for which one you picked first, its 50/50
 
ktimekiller said:
Pretty dumb question and answer still imo.

Also, the problem in the video, if you pick the second time utterly randomly, without consideration for which one you picked first, its 50/50
Click to expand...

No it isn't. The probability doesn't change. Unless you're talking about changing the problem. In which case... no shit. You change the problem you change the answer.
 
No, I don't mean that.

Consider if the setup was like this. The placement of the car in the three cup is random instead of fixed. Then you repeat the following infinite amount of times.

Remove a cup without car throughout all trials. Two cups remain, one with car, one without. Pick one of two.

Evenly pick between left and right throughout all the repetitions.

Before you say I changed the problem, consider that in the original, no matter which cup you picked, essentially, one cup without a car is removed.

Now, if the placement of the car between the three cups XXX is random for infinite number of trials, and the choice is spread 50/50 between the left and right remaining cups, you end up with a 50/50 chance of getting a car.

What I changed was not the problem, but the method in which I pick one after a cup that cannot be a car is removed.
 
No, you are changing the problem. You're taking out the choice. What you're doing is no different than having the person pick one, turn around, remove a random loser and move the cups so that when the person turns around again he doesn't know which one he picked before.

Thats a completely different problem.
 
Why should he bother consider which one he picked previously, if he knows that the one removed didn't was not the one with the car?
 
Because that knowledge makes all the difference. When you pick one of the three, theres a 2/3 chance it was in the other cups. So when you find out which of those other two wasn't the winner, the other cup has that 2/3 chance to itself now. Essentially, when you pick a cup with the strategy of switching in mind, you're 'picking' the other two cups rather than the one you actually picked in the first round.
 
Thats not changing the problem still, its how you are deciding which one to pick.

I am merely isolating the fact that a third cup even existed at all after it was removed.
 
Which is changing the problem. You're no longer given the choice to switch. You're forcing someone to chose one way, and then forcing them to chose a different way in the next trial.

Which, by the way, still doesn't change the probability of each game individually. When you switch you still have a 2/3 chance to win, if you stay you have a 1/3 chance. The only reason multiple trials came in to the picture was to prove that one way was better.
 
Where did I say forced? The person participating may decide to look at it in that fashion, whos to stop him?
 
If the person still has the free will then he will always chose to switch because it will always be the better option. Unless he lacks the knowledge of which one he chose previously, he will always be more likely to win if he switched.
 
I don't deny the 2/3rd chance of picking the car by choosing the OTHER, but that is only in consideration that there were three cups present before the second choice.
 
Again... NO SHIT. You're changing the problem by removing the entire first part.
 
I didnt remove it, the person deciding decided to forget that it happened at all
 
-_-

no

How is forgetting removing.

If one forgets or is unaware of some other factor that precedes them, the ISOLATED choice present at the exact moment would not have consideration of anything that transpired before for the PERSON in question.
 
THE KNOWLEDGE IS A PART OF THE PROBLEM. You're talking about the little green alien.

Marilyn Vos Savant said:
The original answer is still correct, and the key to it lies in the question, "Should you switch?" Suppose we pause at that point, and a UFO settles down onto the stage. A little green woman emerges, and the host asks her to point to one of the two unopened doors. The chances that she'll randomly choose the one with the prize are 1/2, all right. But that's because she lacks the advantage the original contestant had—the help of the host.
Click to expand...

http://www.marilynvossavant.com/forum/viewtopic.php?t=64
 
If you do not consider humanly faults, and differences, the answer is not practical
 
Because you refuse to consider someone might actually isolate the problem? Forget whether it might be right or wrong to do so, fact is, people will.

EDIT: EVEN well educated people, as your link has us to believe.
 
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