its like betting a 2 game parlay, but you already won the first game...the odds NOW of you hitting the parlay and getting paid is 50-50...whereas before the first game was decided, (boy) there were 4 possible outcomes...w/l, w/w, l/l, and l/w...well you got the w which helps your odds to w/l or w/w ...50-50
%^_
Geoff, I promise this is not a theory. It is a counterintuitive logic problem, but I promise I'm not giving you false information. I knew the question would throw people off when I posted it. Please try to see what I'm saying, I'm not lying and the math is not incorrect.
YOU CAN NOT DESIGNATE WHICH COIN GETS TURNED TO HEADS!! That is the point of saying "you are not told whether the son is the oldest or youngest child." For every set of 2 children, sometimes you will get GG. The only thing the statement of the problem does is eliminate the GG possibility. BG, GB, and BB (all with probability 1/3) are still possible. In the coin example I gave (which I encourage you to do), we assure ourselves that at least 1 H is present by discounting all TT results. This is very different from designating one of the coins as H, as we would only have one coin left to flip. If you designate the coin, that's the same as saying that the youngest child is male, which of course leaves the gender of the other child at 50/50.
Again, I am not giving you misinformation. It is quite difficult to understand if you don't "see it." But the answer is certain. There are no college professors that would dispute the simple combinatrics and elementary proofs that lead to the answer.
I can only imagine how vehemently people would deny the answer to the Monty Hall Problem.
Geoff, I promise this is not a theory. It is a counterintuitive logic problem, but I promise I'm not giving you false information. I knew the question would throw people off when I posted it. Please try to see what I'm saying, I'm not lying and the math is not incorrect.
YOU CAN NOT DESIGNATE WHICH COIN GETS TURNED TO HEADS!! That is the point of saying "you are not told whether the son is the oldest or youngest child." For every set of 2 children, sometimes you will get GG. The only thing the statement of the problem does is eliminate the GG possibility. BG, GB, and BB (all with probability 1/3) are still possible. In the coin example I gave (which I encourage you to do), we assure ourselves that at least 1 H is present by discounting all TT results. This is very different from designating one of the coins as H, as we would only have one coin left to flip. If you designate the coin, that's the same as saying that the youngest child is male, which of course leaves the gender of the other child at 50/50.
Again, I am not giving you misinformation. It is quite difficult to understand if you don't "see it." But the answer is certain. There are no college professors that would dispute the simple combinatrics and elementary proofs that lead to the answer.
I can only imagine how vehemently people would deny the answer to the Monty Hall Problem.
This is the crux of the dispute between the 50 percenters and the 67 percenters. If you get to set aside the coin that gets turned to heads everyone would agree the probability of the other coin turning to heads would be 50%.
However, we don't know which coin turned to heads. Neither do we know which child is the boy in our logic problem. Therefore BG does not equal GB. They have to be taken separately because they are separate probabilities and probabilities is what the question is asking.
I will vehemently deny the answer to the Monty Hall Problem if it isn't presented properly, and in the 3-4 times I've heard it presented, usually at poker tables, it's been presented wrong. Certain rules have to be included in the presentation.
In fact, there is a way in which it can reasonably be presented where the player should definitely NOT change his pick.
Try this: Monty Hall owns the show. If you get the prize, it comes out of his pocket. He almost never shows what's behind another door and then gives a contestant a chance to change their mind.
Now, on the day of a huge superprize, he suddenly, after you make your choice, opens another door and asks if you want to stick by your first choice.
I know you'll be smart enough not to change your mind then, right?
FWIW, I think Zit is wrong in answering your question(s). It doesn't matter that you see the son or know his name, he could still be the B in a BG, GB or BB, which is 2/3. Neither his name or seeing him identifies him as either the first or last in the combinations.
If I am wrong I will be interested to hear why. The additional information you provided, name and visual, is irrelevant and changes nothing about the original question.
this example, no doubt 50%
the part that is confusing is that if you peek at both and tell us ONE of them is heads (not revealing which hand), you are left with 3 possible scenarios:
right hand: heads ; left hand: heads
right hand: heads ; left hand: tails
right hand: tails ; left hand: heads
they are saying that in 2/3 of the scenarios, the "other" coin is tails.....
....the logic is easy to comprehend....much harder to believe.
I think the key point is to realize you are looking at a SET of data (in this case gender of 2 children) not 2 individual data points.
Saying it is 50/50 of B/G of all individual data points is CORRECT.
Saying it is 50/50 on a SET of 2 data points given 1 is a B of unknown birth order is INCORRECT.
The CORRECT answer is 2/3 as previously explained.
Saying it is 50/50 of B/G of all individual data points is CORRECT.
Saying it is 50/50 on a SET of 2 data points given 1 is a B of unknown birth order is INCORRECT.
The CORRECT answer is 2/3 as previously explained.
there are NOT 3 possible out comes...there are 4 if you include the known in each set....
the first set puts the known in the first of the series....the second set puts the known in the second place in the series...it is still 50/50
BB
BG
is a possible outcome OR
GB
BB
is a possible outcome....all 4 have an even chance of happening
since BB is the same as BB in each, and GB is the same as BG in each...you have a 50/50 chance that the unknown child is a boy or a girl....
in closing, I leave my challenge open to anyone who wants to put this to practical use....we go out to an event, we survey people who have 2 children, one being a boy.....you pay me 2:1 every time the unknown child is a boy, and I will pay you 1 to 2 everytime it is a girl....
the first set puts the known in the first of the series....the second set puts the known in the second place in the series...it is still 50/50
BB
BG
is a possible outcome OR
GB
BB
is a possible outcome....all 4 have an even chance of happening
since BB is the same as BB in each, and GB is the same as BG in each...you have a 50/50 chance that the unknown child is a boy or a girl....
in closing, I leave my challenge open to anyone who wants to put this to practical use....we go out to an event, we survey people who have 2 children, one being a boy.....you pay me 2:1 every time the unknown child is a boy, and I will pay you 1 to 2 everytime it is a girl....
BG does not equal GB
BB does equal BB
therefore there are 3 posssible outcomes not 4
to the point of your bet (bet it $1 per outcome):
for simplicity...
99 dads interviewed with 2 kids (1 of which is a boy)
33 have 2 boys - you get $66
66 have 1 boy 1 girl - you pay $66
thus 2:1 is the odds of 2/3
Out of those 99 dads, with 2 kids, ALL 99 dads having at least 1 boy....
You are honestly sitting here tellimg me that you thoroughly believe that the out of 99 second children, 66 will be girls and 33 will be boys?????
Given your mathematics....this world would be overrun with women...why isnt that the case?
I honestly have no other arguement....my offer stands to ANYONE wanting to take on the challenge....send me an email and well will work out the details....
In other words...out of 99 random children....everytime, 66% will be women and 33% will be men?
because this series has a known outcome....
in other words:
if the boy was the oldest:
BG
BB
if the boy was the youngest:
BB
GB
there are no other possibilities....so, based on the information, the result would fall under the first OR second series...you cannot combine the 2
if it falls under the first, you have a 50/50 chance the other child is a girl
if it falls under the second, you have a 50/50 chance the child is a girl
everyone in this thread is wanting to COMBINE the two into one equation, you cant because you have a KNOWN RESULT ....
it's 50/50....
