AxiomOfChoice
AxiomOfChoice
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May 27th, 2014 at 4:56:15 PM permalink
Quote: kubikulann

What does it have to do with the problem here?



The outcome of each birth is a random variable. The expectation of the sum is the sum of the expectations. In other words, 0.5 boys per birth.
AxiomOfChoice
AxiomOfChoice
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May 27th, 2014 at 4:59:47 PM permalink
Quote: kubikulann

This kind of sentence costs points to my students. Shows a misunderstanding of the concept of average. "Each" and "average" are contradictory. The correct form is "Half the mothers, on average, kill one girl."



I disagree with you. I'm guessing that English is not your first language?

Actually, I'm not sure that that excuses it. The expected number of girls killed per mother is 1/2. Expectation and average are the same.

Or, we can represent the number of girls killed by each mother by a different random variable. The expectation of each random variable is 1/2. Once again, "expectation" and "average" mean the same thing.
Tomspur
Tomspur
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May 27th, 2014 at 5:38:51 PM permalink
This sounds like a Game of Thrones problem.....killing kids until you get a boy??? What are we, barbarians?? LOL
“There is something about the outside of a horse that is good for the inside of a man.” - Winston Churchill
kubikulann
kubikulann
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May 28th, 2014 at 8:12:17 AM permalink
Quote: AxiomOfChoice

I disagree with you. I'm guessing that English is not your first language?

Actually, I'm not sure that that excuses it. The expected number of girls killed per mother is 1/2. Expectation and average are the same.

Or, we can represent the number of girls killed by each mother by a different random variable. The expectation of each random variable is 1/2. Once again, "expectation" and "average" mean the same thing.

You clearly did not read or did not understand what I wrote. I'm guessing statistics is not your first language?
I said Each is the wrong term to use. You wrote "each mother", now you try to hide it by writing "each random variable". Naughty trick. Does not work with me.

I never contested that expectation and average are similar concepts.

(Note: they are different, though, because expectation is in probability and average is on a statistical sample. The expectation is a constant, while the average is a random variable.)
Reperiet qui quaesiverit
kubikulann
kubikulann
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May 28th, 2014 at 8:15:27 AM permalink
Quote: AxiomOfChoice

The outcome of each birth is a random variable. The expectation of the sum is the sum of the expectations. In other words, 0.5 boys per birth.

Again, you seem incapable of integrating the concept of conditional probability. It is not 0.5 boy when you have to allow for the case "no birth".
a × 0.5 + (1-a)× 0 is not 0.5 if a<1.
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kubikulann
kubikulann
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May 28th, 2014 at 8:19:32 AM permalink
Quote: thecesspit

The expected chance of boy or girl doesn't change

May you be more precise?
One talks of the expected number of something (a random variable), or of the chance of something happening (an event), but the "expected chance" is not meaningful (in your context).
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kubikulann
kubikulann
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May 28th, 2014 at 8:32:50 AM permalink
Quote: 24Bingo

The answer to the first is pretty obvious, really...

It is not, since most people provide a wrong answer. Maybe I'm not getting the meaning of "obvious"? I thought it meant "evident", "trivial".

Quote: 24Bingo

The second obviously changes things.

;-) "obviously" again?

Your answer is correct for the special case of 50% chance of boys. Yet that assumption was introduced by a later poster, it is not in the original problem. Can you provide the answers for the general case of p : 1-p ?
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thecesspit
thecesspit
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May 28th, 2014 at 10:16:20 AM permalink
Quote: kubikulann

May you be more precise?
One talks of the expected number of something (a random variable), or of the chance of something happening (an event), but the "expected chance" is not meaningful (in your context).



Okay drop the word 'expected'. I think you understood the point made well enough. Not being marked here for a final examination, I may make some slightly loose statements.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
AxiomOfChoice
AxiomOfChoice
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May 28th, 2014 at 11:43:11 AM permalink
Quote: kubikulann

Quote: AxiomOfChoice

I disagree with you. I'm guessing that English is not your first language?

Actually, I'm not sure that that excuses it. The expected number of girls killed per mother is 1/2. Expectation and average are the same.

Or, we can represent the number of girls killed by each mother by a different random variable. The expectation of each random variable is 1/2. Once again, "expectation" and "average" mean the same thing.

You clearly did not read or did not understand what I wrote. I'm guessing statistics is not your first language?
I said Each is the wrong term to use. You wrote "each mother", now you try to hide it by writing "each random variable". Naughty trick. Does not work with me.



In English, "each" is the right word to use in this case.

This is a word problem, and each mother is a word-representation of a random variable. I'm perfectly comfortable using "mother" and "random variable" interchangeably here.

Quote:

I never contested that expectation and average are similar concepts.

(Note: they are different, though, because expectation is in probability and average is on a statistical sample. The expectation is a constant, while the average is a random variable.)



The observed average of a sample is a random variable, only because the observed results are themselves random variables. Average itself is not a random variable, and it is the same as expectation.
AceTwo
AceTwo
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May 28th, 2014 at 1:33:32 PM permalink
I agree with AxiomOf Choice in the Blackjack/Effect of position thread.
I thought that this problem was diffrent, but I did the calculation and he is he right 100%, on the first problem the ratio stays 1:1
He is the math calculation for the problem as stated (ie upto 10 children untill a boy is born)
Boys Girls
1 0 0,5
1 1 0,25
1 2 0,125
1 3 0,0625
1 4 0,03125
1 5 0,015625
1 6 0,0078125
1 7 0,00390625
1 8 0,001953125
1 9 0,000976563
0 10 0,000976563
Weighted Average Number of Boys = 0,999023438
Weighted Average Number of Girls = 0,999023438
Ratio 1:1

And for the second problem (first girl is killed)

Boys Girls
1 0 0,5
1 0 0,25
1 1 0,125
1 2 0,0625
1 3 0,03125
1 4 0,015625
1 5 0,0078125
1 6 0,00390625
1 7 0,001953125
1 8 0,000976563
0 9 0,000976563
Weighted Average Number of Boys = 0,999023438
Weighted Average Number of Girls = 0,499023438
Ratio 2,001956947:1
Not exactly 2 (I thing because we stop at 10)

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