A man tells you he has two children. He then starts talking about his son. He does not tell you whether the son is the oldest child or the youngest child. What is the probability that his other child is a girl?
This is the OP. All it says to me is a probabilty question. Age does not mean anything.
ETA the other explanations made my head hurt, it should be very simple the way it was worded.<!-- / message -->
This is the OP. All it says to me is a probabilty question. Age does not mean anything.
ETA the other explanations made my head hurt, it should be very simple the way it was worded.<!-- / message -->
It's not about drawing a second card, but about forming a pair. There are 4 types of pairs (BB RB BR RR), 2 of which unite to form 50% of all pairs. Knowing that one card is red only eliminates 1 of 4 possibilities. You're left with 3 possibilities, 2 of which unite to form 66.666...% of all pairs.
I draw two cards for you and hand you one - it's red. You don't know if it was the first or second one I drew. I could've drawn a RB or BR pair... you'll never know.
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I have a way to put this all to bed...festering zit is willing to put his money where his mouth is...so am i, we both are here in texas...lets put it to the test...I will meet him out at the bar...inthere we will seek out people that have 2 kids...and at least one of them being a boy....then i will guess whether or not the other is a boy or girl....when i am wrong, I will give him 200, when i am right he pays me 300....that is the 2/3 odds that he is claiming is the correct answer. If the math is true, and we ask enough people, at the end of the night we should basically be even....otherwise we keep what we won off each other...what do you say Zit, up for it? I'll even buy the beers...not trying to call you out, but it sounds fun to put your equation to the test.
the problem is you are putting the boy into the equation...the boy has no place in the question...the question does not ask anything about the boy,the question is asking about the other child.....what are the odds on the other child's sex....the boy is irrelevant to the question...he might as well be talking about his umbrella as it pertains to the other child....
i really dont know how else to explain it, other than my challenge is out there...2 to 3 as zit posted the odds....im willing to risk 200 a guess to someone's 300 that i can correctly guess the sex of the 'other' child if the people have 2 kids total and one is a boy....pm me if i have any takers.
That was my reasoning at first, too, but it's wrong.
The boy is part of the pair. You're being asked about half of the pair. Not the "second half", just half of it.
You flip a coin 2,000,000 times and note the results in groups of two flips - so 1,000,000 pairs. About 500k pairs of flips be either Heads/Tails (250k) or Tails/Heads (250k). 250k pairs of flips will be Heads/Heads. Another 250k pairs of flips will be Tails/Tails.
I tell you that I flipped a coin twice. One flip was Heads - could've been the first, could've been the second. What are the odds that my other flip was Tails? 500k (Heads/Tails or Tails/Heads) in 750k (Heads/Tails + Tails/Heads + Heads/Heads).
the question says nothing about the pair....it is singling out the 'other' child....it specifically says 'other' child...no where does it ask anything about the boy in the outcome of the question....i mean it, i will take anyone and go to the bar with this challenge....dont try to explain anything else, because you are not reading the question properly...what is the sex of the other child, boy or girl...i am willing to put the 2/3 math to practical use...then I will report back how much money was won on either side...
The fact that he doesn't tell you oldest or youngest makes it about the pair. Contrary to what many have said, that is not only a salient point, but THE salient point.
Matt's explanations are both accurate, simple, basic and clear. More so than Festering Zit's (no offense).
Had it posited the older or younger son, you would have eliminated two of the four possible combinations, not one. It's contrary to logic (ironic since it is a logic question) b/c the chance of a male or female is 50%, but it isn't about the individual, it is about the pair.
Matt's explanations are both accurate, simple, basic and clear. More so than Festering Zit's (no offense).
Had it posited the older or younger son, you would have eliminated two of the four possible combinations, not one. It's contrary to logic (ironic since it is a logic question) b/c the chance of a male or female is 50%, but it isn't about the individual, it is about the pair.
Geoff, this line of reasoning is wrong b/c you are focusing on the second. If you know the son is older or younger, you would be correct. However, keeping it with roulette. You have (excluding greens) a 25% chance of ending up
BR
RB
BB
RR
If I told you that I spun A (not first or second) black, you remove the possibility of RR, leaving
BR
RB
BB
or 2/3.
Taking it further,
I spun 6 times and grouped them into pairs (spin 1 and 2, 3 and 4 and 5 and 6).
I tell you that I received statistically accurate reflection of two independent events (spins).
you watched the first spin of the first pair and see black
you watched the second spin of the second pair and see black
you watched the first spin of the third pair and see black
how many of the spins that you didn't see have to be red to be a statistically accurate reflection of a pair of independent events?
BR
BB
RB
is incorrect, because you know the second spin could NOT be BR, because you know the second spin was B, so you only have 2 possible results, RB and BB, and we both agree to eliminate RR, because There is a known B in the series of 2.......50/50
so as you eliminate RR you MUST ALSO eliminate BR...because the seocnd spin is a known B.....
or if you flip it and it's the first spin
BR
BB
you have to eiminate RB this time because the first spin is a known B it also i agreed that RR is eliminated as well because we know that the first spin in the series was B.........50/50
you are leaving an option in there that shouldnt be a variable as it is a known.....
You don't know whether or not the boy was the oldest or youngest. Thus, if the order represents the order in terms of age, you could have
BG
GB
BB
In the roulette example, all you know is that I spun a black, you don't know whether it was the first or the second spin. Thus, in order of spin, you can still have
BR
RB
BB
You have to stop focusing on first or second, you don't know whether you saw/heard the first or second spin/oldest or youngest child.
BG
GB
BB
In the roulette example, all you know is that I spun a black, you don't know whether it was the first or the second spin. Thus, in order of spin, you can still have
BR
RB
BB
You have to stop focusing on first or second, you don't know whether you saw/heard the first or second spin/oldest or youngest child.
The RX is slowly being DIVIDED....
the 1/2s and the 2/3s!!!!
CIVIL WAR!!!!
(us 50% people like to keep things simple...some of us understand the 2/3 theory, but too much thought going into this....let's settle this with a capping contest!)
fwiw, whichever side "loses", they will still think they are correct.
the 1/2s and the 2/3s!!!!
CIVIL WAR!!!!
(us 50% people like to keep things simple...some of us understand the 2/3 theory, but too much thought going into this....let's settle this with a capping contest!)
fwiw, whichever side "loses", they will still think they are correct.
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i guess we will never agree on this...i dnt know how else to put unless i just show you guys in the real world....now use your common sense, would you partake ina challenge, where i was guessing the sex, male or female of someones child and pay me 3 to 2 everytime the second child is a male, and i pay you 2 to 3 everytime the second child is a female? because according to your math, a male has 2/3's a more populous time of showing up randomly, than a girl....no, you wouldnt do that because your not interpreting the question and employing the correct equation.
