## Welcome to BrainDen.com - Brain Teasers Forum

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-) |

# cool riddle

### #1

Posted 13 December 2007 - 01:51 AM

### #2

Posted 13 December 2007 - 04:15 AM

I assume you meant getting equal piles of 25. If you randomly turn over 30 pennies there is a great chance that you will not have two equal piles of heads and tails. You'd have better chances just flipping 5 and hoping you hit 5 out of the 30 heads.

If you meant two equal piles of any size, you start out with the correct solution already there. You don't need to flip ANY to win because you have two piles of 20 heads and tails. Or, if you must flip some over, just flip the minimum and you have two equal piles by disregarding the others.

Either way, your solution doesn't fit the proposition.

### #3

Posted 13 December 2007 - 05:31 PM

You have 2 pennies heads up and 3 pennies tales up.

You must start with all the pennies in one pile. So let’s say that you pick up a heads up penny and flip it over. You now have 1 penny heads up in the first pile and none in the second. Now you flip your second penny. If you get a tales, you flip it and it becomes a heads so you now have 1 penny heads up in each pile. Again, it does not matter how many pennies you have in each pile, only how many pennies heads up. Sorry if my original puzzle was unclear. Hopefully this will clarify it.

### #4

Posted 13 December 2007 - 06:13 PM

Still, how do you split the 50 pennies into two piles so that you ensure an equal number of heads in each? If you flip 30 at random and can ensure that you never flip a penny twice, you could end up with anywhere between zero and forty heads showing without knowing how many you have. How are you going to then make 2 equal piles? I'd dare you to try that for big stakes.... Again, if you don't have to physically make the piles, then the solution is met from the start. You have to assume some volitive act on the part of the captive to make the piles or it's a meaningless challenge.

### #5

Posted 14 December 2007 - 02:14 AM

### #6

Posted 14 December 2007 - 06:59 AM

Each pile will have 0 pennies with heads showing.

*scurrying back inside the box*

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #7

Posted 14 December 2007 - 03:32 PM

### #8

Posted 14 December 2007 - 06:23 PM

*did*mean for there to be two piles with all the pennies in them and an equal number of heads in each pile, we get back to my original objection - which is that you cannot be sure by flipping 30 random pennies that you will have an equal number of heads in each pile. By flipping 30 random pennies with no repeats you can have anywhere between 40 and zero heads showing when you get done. How on earth do you guarantee that you will have an equal number in each pile?

### #9

Posted 14 December 2007 - 06:33 PM

This doesn't make sense to me. Will you clarify?So let’s say that you pick up a heads up penny and flip it over. You now have 1 penny heads up in the first pile and none in the second. Now you flip your second penny. If you get a tales, you flip it and it becomes a heads so you now have 1 penny heads up in each pile.

If I have 5 pennies and 3 are heads and 2 are tails, and I grab one at random and flip it for pile 1, then I have a 2 in 5 chance of having a heads. For pile 2, I have either a 2 in 4 or a 1 in 4 chance of having heads depending on the first flip. To make this complete to your answer (of 30 out of 50 flipped) I'd need to flip a third penny, which would give me a either a 2 in 3, 1 in 3, or zero chance of hitting a heads on the flip. It's entirely possible that I flip tails the first time then heads, then tails. How do I have the same number of heads in each pile now? Unless you are talking about making 3 piles, you don't even have 2 equal piles in this situation. You can with 50, as you'd have 25 and 25, but you still have the very large random chance in your flips that makes it impossible to guarantee an equal number of heads in the two piles.

### #10

Posted 14 December 2007 - 06:50 PM

So you have one pile that has 5 pennies in it. You pick up one penny, flip it over, and put it in a NEW pile. Pick up the second penny and put it in pile two. So if you pick up a penny that is heads up the first time, you have 4 pennies in your first pile with 1 heads up and you have 1 penny in your second pile with 0 heads up. If your second penny is tails, you pick it up and flip it over and put it in pile number two. Now you have 3 pennies in pile one and 1 heads up, in pile number two, you have 2 pennies with 1 heads up. This does work no matter how many pennies you have and no matter how many heads up you have and no matter what pennies you pick up. As I have previoiulsy mentioned, you need the same number of pennies HEADS UP in each pile not the same number of pennies total.

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users