### Posting a Puzzle

Humans beings have a curious relationship with reality. The following simple puzzle points at something deeper than optical illusions.

I have three cards in a box. One card is red on both sides. One card is blue on both sides. One card is red on one side and blue on the other side. At random I pull a card from the box and show you one side. It is red. What are the chances the other side is also red?

Typically, after some thought, figuring it must be either the red/red card or the red/blue card, one answers 50/50, since it is equally likely that I picked either.

Wrong.

Solutions welcome. I'll provide in a little while.

I have three cards in a box. One card is red on both sides. One card is blue on both sides. One card is red on one side and blue on the other side. At random I pull a card from the box and show you one side. It is red. What are the chances the other side is also red?

Typically, after some thought, figuring it must be either the red/red card or the red/blue card, one answers 50/50, since it is equally likely that I picked either.

Wrong.

Solutions welcome. I'll provide in a little while.

## 11 Comments:

66%

Outstanding!

A true mathematician would have said 2/3, but that's totally picking nits. It's clear you solved the problem.

I'm impressed.

You do realize some of our readers may like a little more than the number alone.

I seem to recall a computer, after thousands of years, saying:

42.People were less than satisfied, much like the math teachers who write at the top of the exam:

SHOW YOUR WORK!!

NO WORK = NO CREDIT!!

Willing to share how you did it? If not, that's okay.

To quote John Lennon, "It came to me on a flaming pie"

seriously, though, if one side is red (the side you can see) it clearly can't be the Blue-Blue card. So that takes that completely out of the equation. Look at the permutations of the remaining cards, RED-BLUE and RED-RED

as a table:

visible RED RED RED

Back side RED RED BLUE

I always think of things in percentages.

Yes, the trick is to realize the RED-RED card gives two permutations where the other side of the card is also red, as you could be looking at either side of it.

Bullseye. You two have spared me the task of explaining. Between AZ showing the three possibilities and Sirocco noting the "trick" about the red card, this one is done.

I promise that the majority of people do not see what Sirocco pointed out and swear it's 50/50.

"a flaming pie" !

I might have use that some day.

Heh,

I LIVE this equation every morning. Except in my case it is two drawers with socks with holes vs. socks without. In actual testing, however the odds of drawing the sock with a hole far outstrip mathematical probability.

For an extra degree of difficulty figure for dress socks vs. casual socks.

I would think that wealthy Republicans could afford socks without holes.

Liza, as a republican he probably has a 52 in Hi-def plasma TV, an H2, plenty of guns and ammo, but won't spend money on those basic needs. Just like this administration will throw money at the war, but won't spend a dime on healthcare for all, etc.

I know the answer is 42...what was the question?

Wasn't the question something like "What do you get if you multiply six by nine?"

Yes,permutations...yes, I still remember...9th grade... Only that I didn't graduate high school in US!

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