There are four balls in a hat: a blue one, a white one, and two red ones. Now I draw simultaneously two balls, look at them, and announce that at least one of them is red.

What is the chance that the other is red as well?

Apparently there are two interpretations, two approaches and two answers.

The concise summaries   of both are presented by Jer and Steve.

Clearly, only one is correct – and the puzzle as presented does not need additional assumptions or psychological insight.

Prior to disclosing my support to J or SH   -  I  aim to convince the erring party that there are no  "buts"  and "on the other hand" –  just the text as presented.

Maybe, after considering another example, someone will change his mind,  and if not I will definitely address his remarks.

The 3 colors in the text are redundant – there are 2 RED & 2 NON-RED.

Please solve the following example, using both approaches and see for yourself .

<begin>

There is a deck of 52 cards, perfectly shuffled. I ask Jer to pick up randomly (all done under the watchful eyes of SH) 2 cards and to place them face down on the table  and then I pose the following question: 

 If  I,  after looking at both cards announce that at least one of them is red- what are the chances that both cards are red?

Is the said procedure equivalent to simply discarding one red card from the deck and stating that the chances of drawing a RED  card are 25/51?
<end>

Please do not try to justify any answer on simulation based on procedures  not strictly adhering  to my text per se, no modifications, no additional  interpretation

 

 

 

Posted by Ady TZIDON on 2017-05-20 00:41:00