This is one of my favorite logic puzzles. I like the "girl born on Friday" variation. The following puzzles will lead you to the mind blowing solution:
A. I flip 2 coins, if one of them is heads, what is the chance the other is heads?
B. I flip 2 coins, if the first one is heads, what is the chance the other one is heads.
If you say, "one is a girl" then the domain of possibilities are Boy-Girl, Girl-Boy, Girl-Girl, but not Boy-Boy(question A in the coin flip). If you say the oldest is a girl, then we only have Girl(older)-Girl and Girl(older)-Boy as the possibilities (question B in the coin flip)
By identifying one (the older, the bigger, the stronger) you change the domain of possibilities. So the question is: Does "born on Friday" identify enough to change the domain of possibilities?
I would agree with Peter that it does. My reasoning (and hopefully a reasonable answer) will come later.
Continuing... (warning: I may be wrong or confusing, proceed at your own risk)
I said "identify enough to change the domain of possibilities", I think a clearer statement would be "identify by providing another mutually exclusive condition." Thus, by saying the oldest is a girl, you now have the mutually exclusive conditions of older/younger. If you say the girl is blond or a vegetarian or a serial murderer, it doesn't change the problem, since there is no exclusivity.
However, identifying the day of the week of birth is mutually exclusive with the other days.
Let's simplify the problem to "girl born in the morning (a.m.)" Now we can list all the possibilities: ('a' stands for am, 'p' for pm; B, for boy, G for girl) Ba-Ba Ba-Bp Bp-Ba Bp-Bp
Ba-Ga Ba-Gp Bp-Ga Bp-Gp
Ga-Ba Ga-Bp Gb-Ba Gb-Bp
Ga-Ga Ga-Gp Gb-Ga Gb-Gp
We eliminate the first 4 and all other places a Ga doesn't show up, giving us 7 possibilities with a girl born in the morning. Of these, 3 will yield 2 girls or 3/7 chance of having 2 girls (given an a.m. girl).
With the "girl born on Friday" problem, instead of am and pm, substitute Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, and Sunday. Where there were 4 possibilities in each group above, there will be 49 (7x7). Find all the ones with a girl born on Friday and, of those, find which have two girls. By my calculation (always questionable) I come out with the chance of two girls as 13/27, which seems reasonable since it is close to the 1/2 than the 1/3. (If the girl was born on any day except Friday, it would be closer to 1/3).
My question is this. I have two kids. The girl is dead. what is the probability I have two girls?
4 comments:
This is one of my favorite logic puzzles. I like the "girl born on Friday" variation. The following puzzles will lead you to the mind blowing solution:
A. I flip 2 coins, if one of them is heads, what is the chance the other is heads?
B. I flip 2 coins, if the first one is heads, what is the chance the other one is heads.
A 1/3
B 1/2
And of course the two girls problem is a probability between those two answers
Continuing...
If you say, "one is a girl" then the domain of possibilities are Boy-Girl, Girl-Boy, Girl-Girl, but not Boy-Boy(question A in the coin flip). If you say the oldest is a girl, then we only have Girl(older)-Girl and Girl(older)-Boy as the possibilities (question B in the coin flip)
By identifying one (the older, the bigger, the stronger) you change the domain of possibilities. So the question is: Does "born on Friday" identify enough to change the domain of possibilities?
I would agree with Peter that it does. My reasoning (and hopefully a reasonable answer) will come later.
Continuing...
(warning: I may be wrong or confusing, proceed at your own risk)
I said "identify enough to change the domain of possibilities", I think a clearer statement would be "identify by providing another mutually exclusive condition." Thus, by saying the oldest is a girl, you now have the mutually exclusive conditions of older/younger. If you say the girl is blond or a vegetarian or a serial murderer, it doesn't change the problem, since there is no exclusivity.
However, identifying the day of the week of birth is mutually exclusive with the other days.
Let's simplify the problem to "girl born in the morning (a.m.)" Now we can list all the possibilities: ('a' stands for am, 'p' for pm; B, for boy, G for girl)
Ba-Ba
Ba-Bp
Bp-Ba
Bp-Bp
Ba-Ga
Ba-Gp
Bp-Ga
Bp-Gp
Ga-Ba
Ga-Bp
Gb-Ba
Gb-Bp
Ga-Ga
Ga-Gp
Gb-Ga
Gb-Gp
We eliminate the first 4 and all other places a Ga doesn't show up, giving us 7 possibilities with a girl born in the morning. Of these, 3 will yield 2 girls or 3/7 chance of having 2 girls (given an a.m. girl).
With the "girl born on Friday" problem, instead of am and pm, substitute Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, and Sunday. Where there were 4 possibilities in each group above, there will be 49 (7x7). Find all the ones with a girl born on Friday and, of those, find which have two girls. By my calculation (always questionable) I come out with the chance of two girls as 13/27, which seems reasonable since it is close to the 1/2 than the 1/3. (If the girl was born on any day except Friday, it would be closer to 1/3).
My question is this. I have two kids. The girl is dead. what is the probability I have two girls?
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