kubikulann Joined: Jun 28, 2011
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May 28th, 2014 at 4:34:00 PM permalink
I edited it to prevent bad feelings. His answer WAS wrong. Go reread it. He says the proportion changes in the first case.
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AxiomOfChoice Joined: Sep 12, 2012
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May 28th, 2014 at 5:01:59 PM permalink
Quote: kubikulann

I edited it to prevent bad feelings. His answer WAS wrong. Go reread it. He says the proportion changes in the first case.

Oh, I see. Yes I misunderstood what he was saying (he answered 3 problems, including one which you didn't ask, which threw me off) Admittedly I also answered a question which you didn't ask.

You do at least see how the question can be answered by repeatedly applying the identity that E(X+Y) = E(X) + E(Y), right? In this case, each birth is a random variable (representing number of boys born) which maps to 0 half the time and 1 half the time. Thus each birth has expectation of 1/2.

Total number of boys is E(X1 + X2 + X3 + ...) where each Xi is a birth.
By applying that identity, this is the same E(X1) + E(X2) + ... which is 1/2 + 1/2 + ... which is 1/2 of the number of births. Therefore the expected number of boys = 1/2 the number of births, regardless of how many births there are, or how you decide whether or not to have another birth. This is very similar to a betting system question where you decide to quit or not quit based on how you are doing in a session; it does not change the fact that your total expectation is the sum of the expectations of all your bets.

As for the 2nd question, we can apply the same identity. Number of surviving girls = Girls born - girls killed.
Expected number of surviving girls = E(girls born - girls killed) = E(girls born) - E(girls killed). There's that identity again.

E(girls killed) is easy -- again, we let Xi be the number of girls that the ith mother kills and use the same identity. So E(girls killed) = E(X1 + X2 + ... + Xn) = E(X1) + E(X2) + ... + E(Xn) = 1/2 + 1/2 + ... + 1/2 = 1/2 the number of mothers.

E(girls born) would be easy if it were not for the max of 10 children total -- it would be 1 per mother (using the identity again). The max of 10 lowers it slightly, since mothers no longer have an average of 2 children each. Instead, 1 in 1024 mothers stops after 10 girls. That group would normally average 12 children, so the average number of children per mother is 2 - 2/1024 = 1023/512.

So each mother has, on average, 1023/1024 boys and 1023/1024 girls, and kills 1/2 of a girl. That leaves us with 2047/2048 boys and 511/1024 girls for a ratio of 511:1023 girls:boys. IMO the max of 10 just makes the problem unnecessarily ugly. 1:2 is such a nicer answer.

Note that the only math that I did was repeatedly applying E(X+Y) = E(X) + E(Y). Over and over and over, just applying the same identity. It turns this into a simple arithmetic (addition and subtraction) problem.

The blackjack problem is the same.
24Bingo Joined: Jul 4, 2012
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May 28th, 2014 at 5:02:46 PM permalink
This post has been deleted because I thought kubikulann was talking about me when he was talking about MangoJ.

I made a sign error in my second answer, putting down the number of girls killed where I meant to put down the survivors. The actual number of girls per boy is (1 - my answer).
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
kubikulann Joined: Jun 28, 2011
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May 28th, 2014 at 5:23:45 PM permalink
Quote: AxiomOfChoice

You do at least see how the question can be answered by repeatedly applying the identity that E(X+Y) = E(X) + E(Y), right?

Certainly. My solution is using that sort of argument. I take all the first-borns, all the second-born etc. In each layer, there must be a proportion p/q of boys and girls. So the total maintains the same proportion.

Quote:

In this case, each birth is a random variable (representing number of boys born)

Yes, because the sex of the next child is independent from the past. That is not exactly the same in the blackjack situation, because of the dependency on players' decisions.

Quote:

That leaves us with 2047/2048 boys and 511/1024 girls for a ratio of 511:1023 girls:boys. IMO the max of 10 just makes the problem unnecessarily ugly. 1:2 is such a nicer answer.

The general answer is p (1-q10) boys for q²(1-q9) girls. (p being the prob of a boy)

With a max of two children (think China's policy), then the ratio is 1:3 (for a p=0.50). Also a nice answer, no?
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AxiomOfChoice Joined: Sep 12, 2012
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May 28th, 2014 at 5:38:41 PM permalink
Quote:

In this case, each birth is a random variable (representing number of boys born)

Quote:

Yes, because the sex of the next child is independent from the past. That is not exactly the same in the blackjack situation, because of the dependency on players' decisions.

No, no, no. NO!!!! This is the point; this is what I have been saying over and over again.

E(X+Y) = E(X) + E(Y). Always. Even if X and Y are dependent. Even if they are correlated. You can ignore the dependencies. That is what makes the question easy! It's what makes the hat question easy. If you pick your hat before I pick my hat, my probability of getting my hat is dependent on whether you got your hat, but it doesn't matter! You still have n people, and each E(Xn) = 1/n, so E(X1 + X2 + ... + Xn) = E(X1) + ... E(Xn) = 1, so the expected number of people to get their own hat back is 1. It doesn't matter than the Xn's are all dependent. That is the whole point of the identity. It's not particularly interesting for iid variables; the fact that it works for dependent, correlated, and different variables is what makes it so powerful! It's why I don't need to do a lot of work to figure out the answer to the hat problem, it's just n * 1/n = 1.

In other words, it's not just that E(2X) = 2E(X). That is boring and not that useful.

It's that E(X+Y) = E(X) + E(Y) for all X and Y. That is so much more powerful!
kubikulann Joined: Jun 28, 2011
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May 29th, 2014 at 4:26:18 AM permalink
Quote: AxiomOfChoice

No, no, no. NO!!!! This is the point; this is what I have been saying over and over again.

E(X+Y) = E(X) + E(Y). Always. Even if X and Y are dependent. Even if they are correlated. You can ignore the dependencies. That is what makes the question easy! It's what makes the hat question easy. If you pick your hat before I pick my hat, my probability of getting my hat is dependent on whether you got your hat, but it doesn't matter! You still have n people, and each E(Xn) = 1/n, so E(X1 + X2 + ... + Xn) = E(X1) + ... E(Xn) = 1, so the expected number of people to get their own hat back is 1. It doesn't matter than the Xn's are all dependent. That is the whole point of the identity. It's not particularly interesting for iid variables; the fact that it works for dependent, correlated, and different variables is what makes it so powerful! It's why I don't need to do a lot of work to figure out the answer to the hat problem, it's just n * 1/n = 1.

In other words, it's not just that E(2X) = 2E(X). That is boring and not that useful.

It's that E(X+Y) = E(X) + E(Y) for all X and Y. That is so much more powerful!

Everything you write here is perfectly correct.

What I'm saying is that you cannot use this if your premises are E(x|a) and E(y|b). These cannot be added just like that, they have to be weighted. Also, E(x|b) and E(y|a) should be accounted for.

In another thread, you acknowledge this, but mix the above argument with another: that E(x|a) and E(y|b) are both zero, so the sum is zero, weighting or not. Well, that does not prevent the necessity of weighting, does it?
And, finally, the question was to ascertain that they were effectively zero in the problem under analysis. Which proved right under one assumption and false under another.
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MangoJ Joined: Mar 12, 2011
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May 29th, 2014 at 6:44:10 AM permalink
Quote: kubikulann

I edited it to prevent bad feelings. His answer WAS wrong. Go reread it. He says the proportion changes in the first case.

Yes, the answer to the second problem was wrong. My fault, it was late in the evening, and I'm not a mathematician.

(a) A good mathematician should have pointed a changing of birth strategy would never result in any other distribution of boys vs. girls. His professioni is to provide answers of absolute truth, however abstract the proof may be.
(b) A good math teacher would have spotted the false argument, formula, or assumption. His profession is to teach and not to judge right or wrong without giving hint to the right direction.

While (a) has been achieved in this thread numerous time (and I feel ashamed of not seeing it even when fighting with sleep), there is no trace of (b) at all. Which, in my modest opinion, doesn't speak high for the "teacher".
kubikulann Joined: Jun 28, 2011
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May 29th, 2014 at 8:27:20 AM permalink
I'm not here to teach. On the contrary, it is a break from work, to be able to talk to people who like math, on an equal foot.
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AxiomOfChoice Joined: Sep 12, 2012
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May 29th, 2014 at 11:07:26 AM permalink
Quote: kubikulann

What I'm saying is that you cannot use this if your premises are E(x|a) and E(y|b). These cannot be added just like that, they have to be weighted.

Yes, but when the expectations are 0, the weightings become irrelevant.

You are basically saying, no, TC' is not TC + 0 + 0, it is TC + 0 + f(TC)*0, and telling me that I need to do a bunch of number crunching with some complicated function f before I multiply the result by 0.

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