1call2many
Joined: Mar 7, 2014
• Posts: 31
March 7th, 2014 at 10:19:50 AM permalink
Sorry that this is disjointed, i thought i was posting after an earlier point in the thread. i'm new to this and will learn how to quote-the post i am responding to. Anyhow you all should get the gist of what i am thinking and or asking.

I f you get 50% when you know it is the eldest son, the same can be calculated if you knew it was the youngest son ( you would rule out GG and BG in this case).

So why do you even need that info, its already there. Without being told, I know that the chances of a boy being either the eldest or the youngest out of two children is 100%. Both the eldest and the youngest son solutions end in the 50% not the 33.33% chance.

In other words how can you not have the information you need to get to 50%.
BleedingChipsSlowly
Joined: Jul 9, 2010
• Posts: 972
March 9th, 2014 at 8:52:26 AM permalink
Perhaps this reasoning will help: consider two tossed coins with the outcomes hidden under cups. You are guaranteed at least one of the coins is "heads." I hope you will find it easily reasoned that the chance of both being "heads" is 1/3. (HH, HT, TH) A cup is removed and you are shown that one of the coins is "heads." Obviously, the probababiliy that both are "heads" is then 1/2. The difference is not knowing the result of either coin and knowing the result of one coin.

Edit: spelling
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
Doc
Joined: Feb 27, 2010
• Posts: 7169
March 9th, 2014 at 8:58:14 AM permalink
Quote: 1call2many

In other words how can you not have the information you need to get to 50%.

I posted the answer to that last night in the other thread you started on this same topic.
BleedingChipsSlowly
Joined: Jul 9, 2010