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