Riddle Brothers and Sisters is wrong
posted by heyhoo
- December 15 2011 12:58:12 PM
I know this forum is suposed to post new riddles, but I just wanted to inform that the solution to the Riddle Brothers and Sisters is wrong.
In fact I think there is 1/2 chance in both houses. The error is in:
"There are 3 possibilities for the children in the first house:
Younger Older
Girl Boy
Boy Girl
Boy Boy"
There in fact 4 possibilities:
Younger Older
Girl(loves math) Boy (loves history)
Boy(loves history) Girl (loves math)
Boy(loves history) Boy (loves math)
Boy(loves math) Boy(loves history)
The fact that the story differentiates the childs of the house on the left, using the loving math and history thing should only help to understand that there are 4 possibilities.
However, even if there was no way of diferentiating them, you should always do it by calling Child 1 and Child 2.
Since there are 4 possibilities and only 2 of them has a girl the probability is 1/2 just like the house on the right.
P.S. you can consider only 3 possibilities, but the last one (Boy Boy) has a higher probability (twice as high). And doing the math it will always come to the same output: 1/2 chance that the house on the left has a girl.
In fact I think there is 1/2 chance in both houses. The error is in:
"There are 3 possibilities for the children in the first house:
Younger Older
Girl Boy
Boy Girl
Boy Boy"
There in fact 4 possibilities:
Younger Older
Girl(loves math) Boy (loves history)
Boy(loves history) Girl (loves math)
Boy(loves history) Boy (loves math)
Boy(loves math) Boy(loves history)
The fact that the story differentiates the childs of the house on the left, using the loving math and history thing should only help to understand that there are 4 possibilities.
However, even if there was no way of diferentiating them, you should always do it by calling Child 1 and Child 2.
Since there are 4 possibilities and only 2 of them has a girl the probability is 1/2 just like the house on the right.
P.S. you can consider only 3 possibilities, but the last one (Boy Boy) has a higher probability (twice as high). And doing the math it will always come to the same output: 1/2 chance that the house on the left has a girl.
Reply by heyhoo
- December 15 2011 01:04:37 PM
Sorry for the bad english and the tables...
And I wanted to add that if you think there is any flaw in my reasoning, please reply.
And if no one rejects it, is there any way of changing the solution of the riddle?
And I wanted to add that if you think there is any flaw in my reasoning, please reply.
And if no one rejects it, is there any way of changing the solution of the riddle?
Reply by Caprico
- December 16 2011 06:53:39 AM
Your reasoning is incorrect. You have considered a repeated choice of boy-boy:
Young Old
Boy(loves history) Boy (loves math)
Boy(loves math) Boy(loves history)
These two options are probability-wise the same. You can't count them twice. If you've studied probability as a subject, you'll understand this particular counter-intuitive riddle better. This is similar to the Monty Hall Riddle... Read it up... :) The given solution is correct.
Young Old
Boy(loves history) Boy (loves math)
Boy(loves math) Boy(loves history)
These two options are probability-wise the same. You can't count them twice. If you've studied probability as a subject, you'll understand this particular counter-intuitive riddle better. This is similar to the Monty Hall Riddle... Read it up... :) The given solution is correct.
Reply by heyhoo
- December 16 2011 02:44:48 PM
I don't want to sound too arrogant, but I'm taking a masters in applied mathematics and I've had plenty of probability courses... (too many btw)
I think the riddle tries to be similar to the Monty Hall Riddle, but they are different.
The reason I counted twice the possibility "Boy Boy" is because that possibility has twice the probability of the other two.
Let's consider the 3 possibilities of the solution:
Younger Older
Girl Boy
Boy Girl
Boy Boy
The probability of the first possibility is: the probability of the 2nd Child being a girl (1/2) and being younger (1/2). Which means that the probability of "Girl Boy" is 1/2 * 1/2 = 1/4.
The probability of the second possibility is: the probability of the 2nd Child being a girl (1/2) and being older (1/2). Which means that the possibility "Boy Girl" has a probability of 1/2 * 1/2 = 1/4.
The probability of the third event is the probability of the 2nd Child being a boy (1/2).
So, you see that there is a probability of 1/2 of being a Girl in the house on the left.
I think the riddle tries to be similar to the Monty Hall Riddle, but they are different.
The reason I counted twice the possibility "Boy Boy" is because that possibility has twice the probability of the other two.
Let's consider the 3 possibilities of the solution:
Younger Older
Girl Boy
Boy Girl
Boy Boy
The probability of the first possibility is: the probability of the 2nd Child being a girl (1/2) and being younger (1/2). Which means that the probability of "Girl Boy" is 1/2 * 1/2 = 1/4.
The probability of the second possibility is: the probability of the 2nd Child being a girl (1/2) and being older (1/2). Which means that the possibility "Boy Girl" has a probability of 1/2 * 1/2 = 1/4.
The probability of the third event is the probability of the 2nd Child being a boy (1/2).
So, you see that there is a probability of 1/2 of being a Girl in the house on the left.
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