This famous brainteaser has a counter-intuitive answer that might surprise you! Can you solve it?
Imagine you have THREE playing cards. One has a blue back on both sides. One has a red back on both sides. The third has a blue back on one side and a red back on the other.
Place all three cards into a hat and shake the hat to mix them up. Without looking, reach into the hat, pull out a single card, and look at only ONE side of it. You see a red back. Obviously it isn’t the blue/blue card, therefore it must be either the blue/red card or the red/red card. Here’s the question:
What are the chances that it’s the red/red card? Think carefully!
Most people guess that the answer is 1 in 2 (fifty-fifty). That’s incorrect! While it’s true that you have eliminated one card completely (the blue/blue card) and there are only two cards left to choose from, one of those cards is more likely to be the one selected than the other. The red/red card has two possible sides you could be looking at right now, and the blue/red card only has one side you could be looking at right now (remember, you’re looking at a red back).
The chances of you holding the red/red card are 2 in 3. Don’t believe me? Try the experiment yourself many times in a row. Since you probably don’t have double-sided playing cards, make up some cards out of regular slips of paper with “blue” and “red” sides marked on them. After going through the procedure, you’ll find that you’re holding a card with both sides the same roughly 2/3rds of the time.
Sometimes you’ll have the blue/blue card and sometimes you’ll have the red/red card. But only 1/3 of the time will you have the red/blue card (which is exactly what chance would have predicted in the first place). Learn more here. Like this new article? Hit the LIKE button above to share it with your friends!