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[OT] mathematical query
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<blockquote data-quote="The Sigil" data-source="post: 352524" data-attributes="member: 2013"><p>As far as I can tell, it's an "order of information question" - and perhaps I was wrong. But I don't think so.</p><p></p><p>Telling me that one child is male does NOT have zero bearing on the other child in this case - because you haven't told me WHICH child is male. That means that the other child's birth is NOT randomized to 50-50 <em>because I cannot identify it as a discrete event</em> in the sequence. Flipping two coins and saying, "one of them came up heads" is not the same as telling me the "the first one came up heads."</p><p></p><p>Here is my reasoning...</p><p></p><p>I am told that you have two children.</p><p></p><p>I can then create four possibilities:</p><p></p><p>MM, MF, FM, FF.</p><p></p><p>Each of these has equal probability.</p><p></p><p>I think everyone agrees with me to this point.</p><p></p><p>By showing me a picture, you essentially tell me that the FF possibility is impossible. Now, we must figure out how to "disperse" the 25% probability that was tied up in the FF possibility before. Why should the entirety of the 25% now be assigned to MM instead of split equally among MM, MF, FM? I miss this point.</p><p></p><p>Here is a more rigorous derivation:</p><p></p><p>The answer to "I have two children" may be expressed as:</p><p>FF (25%)</p><p>FM (25%)</p><p>MF (25%)</p><p>MM (25%)</p><p></p><p>Now let us add the condition of "I show you a picture of my son" - what is the chance of you showing me a given child.</p><p>Let us look at some strict probabilities (unnormalized)...</p><p></p><p>FF (25%) * no picture (0%) = 0.00</p><p>FM (25%) * picture of M2 (100%) = 0.25</p><p>MF (25%) * picture of M1 (100%) = 0.25</p><p>MM (25%) * picture of M1 (50%) = 0.125</p><p>MM (25%) * picture of M2 (50%) = 0.125</p><p></p><p>Hence, we now have four solutions that still fit our original question (2 children, showed a picture of a male) with the following (unnormalized) probabilities:</p><p></p><p>FM/M2 - 0.25</p><p>MF/M1 - 0.25</p><p>MM/M1 - 0.125</p><p>MM/M2 - 0.125</p><p></p><p>The sum of probabilities where there is one male and one female is 0.50. The sum of probabilities where there are two males is 0.25.</p><p></p><p>Hence, we have 0.50 vs. 0.25.</p><p></p><p>That looks like 2:1 odds (67%) to me.</p><p></p><p>If I have made a mistake please explain it to me.</p><p></p><p>--The Sigil</p><p></p><p>P.S. - I know some of you are saying the probability tree should look like this:</p><p></p><p>Picture (100%) * Older (50%) * Other is Male (50%)</p><p>Picture (100%) * Older (50%) * Other is Female (50%)</p><p>Picture (100%) * Younger (50%) * Other is Male (50%)</p><p>Picture (100%) * Younger (50%) * Other is Female (50%)</p><p></p><p>but I'm not sure/convinced that this is the correct order to use.</p></blockquote><p></p>
[QUOTE="The Sigil, post: 352524, member: 2013"] As far as I can tell, it's an "order of information question" - and perhaps I was wrong. But I don't think so. Telling me that one child is male does NOT have zero bearing on the other child in this case - because you haven't told me WHICH child is male. That means that the other child's birth is NOT randomized to 50-50 [i]because I cannot identify it as a discrete event[/i] in the sequence. Flipping two coins and saying, "one of them came up heads" is not the same as telling me the "the first one came up heads." Here is my reasoning... I am told that you have two children. I can then create four possibilities: MM, MF, FM, FF. Each of these has equal probability. I think everyone agrees with me to this point. By showing me a picture, you essentially tell me that the FF possibility is impossible. Now, we must figure out how to "disperse" the 25% probability that was tied up in the FF possibility before. Why should the entirety of the 25% now be assigned to MM instead of split equally among MM, MF, FM? I miss this point. Here is a more rigorous derivation: The answer to "I have two children" may be expressed as: FF (25%) FM (25%) MF (25%) MM (25%) Now let us add the condition of "I show you a picture of my son" - what is the chance of you showing me a given child. Let us look at some strict probabilities (unnormalized)... FF (25%) * no picture (0%) = 0.00 FM (25%) * picture of M2 (100%) = 0.25 MF (25%) * picture of M1 (100%) = 0.25 MM (25%) * picture of M1 (50%) = 0.125 MM (25%) * picture of M2 (50%) = 0.125 Hence, we now have four solutions that still fit our original question (2 children, showed a picture of a male) with the following (unnormalized) probabilities: FM/M2 - 0.25 MF/M1 - 0.25 MM/M1 - 0.125 MM/M2 - 0.125 The sum of probabilities where there is one male and one female is 0.50. The sum of probabilities where there are two males is 0.25. Hence, we have 0.50 vs. 0.25. That looks like 2:1 odds (67%) to me. If I have made a mistake please explain it to me. --The Sigil P.S. - I know some of you are saying the probability tree should look like this: Picture (100%) * Older (50%) * Other is Male (50%) Picture (100%) * Older (50%) * Other is Female (50%) Picture (100%) * Younger (50%) * Other is Male (50%) Picture (100%) * Younger (50%) * Other is Female (50%) but I'm not sure/convinced that this is the correct order to use. [/QUOTE]
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