Reply to thread

Message

<blockquote data-quote="The Sigil" data-source="post: 352947" data-attributes="member: 2013">One last time, REAL SLOWLY...1.) Woman with kids:4 possibilities.MMMFFMFFAll have equal odds.&nbsp; 50% chance of having one child of each gender.2. Man with picture of son:4 possibilitiesMMMFFMFFAll have equal odds.&nbsp; When he shows a picture of his son, we do NOT then have &quot;MX&quot; and we do not have &quot;XM.&quot;&nbsp; We have &quot;MX or XM&quot; which is much different.&nbsp; All we learn is that FF is impossible.&nbsp; In other words, all we learn is that we are chatting with someone from one of the first three groups.&nbsp; That leaves us with a 67% chance of having one child of each gender.3. Pick, reveal, ask to switch:9 possibilities (&quot;choice picked&quot; and &quot;right choice&quot;):AAABACBABBBCCACBCCWe have a one in three chance of being right (AA, BB, CC).&nbsp; When a choice is revealed to be wrong, provided it is NOT the choice we have chosen (bloody loophole-searchers), we get the following:AA (B or C)AB (C)AC (B)BA (C)BB (A or C)BC (A)CA (B)CB (A)CC (A or B)The above all have equal probability - and we are still right only 1/3 of the time.&nbsp; It does NOT look like this for equal probability:AA (B)AA (C)AB (C)AC (B)BA (C)BB (A)BB (C)BC (A)CA (B)CB (A)CC (A)CC (B)It DOES look like this with the following unequal probabilities:AA (B) - 1/18AA (C) - 1/18AB (C) - 1/9AC (B) - 1/9BA (C) - 1/9BB (A) - 1/18BB (C) - 1/18BC (A) - 1/9CA (B) - 1/9CB (A) - 1/9CC (A) - 1/18CC (B) - 1/18Now let us trace those who wish to quibble the specifics of the problem (&quot;it said #3 was taken away&quot; or &quot;he picks #1&quot;).If he picks A, we are left only with....AA (B) - 1/18AA (C) - 1/18AB (C) - 1/9AC (B) - 1/9And if we know C is taken away, we are left only with...AA (C) - 1/18AB (C) - 1/9This tells us that you are right half as often as you are wrong.&nbsp; NOT 50-50.Again, go do 120 experiments with dice and come back if you are still sure you are right.--The Sigil</blockquote>

[QUOTE="The Sigil, post: 352947, member: 2013"] One last time, REAL SLOWLY... 1.) [u]Woman with kids:[/u] 4 possibilities. MM MF FM FF All have equal odds. 50% chance of having one child of each gender. 2. [u]Man with picture of son:[/u] 4 possibilities MM MF FM FF All have equal odds. When he shows a picture of his son, we do NOT then have "MX" and we do not have "XM." We have "MX or XM" which is much different. All we learn is that FF is impossible. In other words, all we learn is that we are chatting with someone from one of the first three groups. That leaves us with a 67% chance of having one child of each gender. 3. [u]Pick, reveal, ask to switch:[/u] 9 possibilities ("choice picked" and "right choice"): AA AB AC BA BB BC CA CB CC We have a one in three chance of being right (AA, BB, CC). When a choice is revealed to be wrong, provided it is NOT the choice we have chosen (bloody loophole-searchers), we get the following: AA (B or C) AB (C) AC (B) BA (C) BB (A or C) BC (A) CA (B) CB (A) CC (A or B) The above all have equal probability - and we are still right only 1/3 of the time. It does NOT look like this for equal probability: AA (B) AA (C) AB (C) AC (B) BA (C) BB (A) BB (C) BC (A) CA (B) CB (A) CC (A) CC (B) It DOES look like this with the following unequal probabilities: AA (B) - 1/18 AA (C) - 1/18 AB (C) - 1/9 AC (B) - 1/9 BA (C) - 1/9 BB (A) - 1/18 BB (C) - 1/18 BC (A) - 1/9 CA (B) - 1/9 CB (A) - 1/9 CC (A) - 1/18 CC (B) - 1/18 Now let us trace those who wish to quibble the specifics of the problem ("it said #3 was taken away" or "he picks #1"). If he picks A, we are left only with.... AA (B) - 1/18 AA (C) - 1/18 AB (C) - 1/9 AC (B) - 1/9 And if we know C is taken away, we are left only with... AA (C) - 1/18 AB (C) - 1/9 This tells us that you are right [b]half[/b] as often as you are wrong. NOT 50-50. Again, go do 120 experiments with dice and come back if you are still sure you are right. --The Sigil [/QUOTE]

Verification