[Guest]

So, you are blindfolded and wearing thick gloves, and someone put twenty coins on the table. Ten of them are on heads and ten of them are on tails, but you wouldn't know which because of your blindfold and gloves. You could flip or move the coins if you want. Your goal is to group them into two with ten coins each. The two groups should have the same number of heads and the same number of tails. How do you do it?

[Guest]

Randomly divide them into 2 groups of 10 each.

Now if the first group has X tails, then it has (10-X) heads. This implies also that the second group has (10-X) tails and X heads. Now simply flip all the coins in one of these groups and you have the desired result: same number of heads and tails in each.

[Guest]

I would just take the blindfold off!

Randomly divide them into 2 groups of 10 each.

Now if the first group has X tails, then it has (10-X) heads. This implies also that the second group has (10-X) tails and X heads. Now simply flip all the coins in one of these groups and you have the desired result: same number of heads and tails in each.

[Guest]

put them in your mouth, assuming you're not germ-a-phobic

[Guest]

Arrange the coins into 2 random groups of 10. In each group of 10, each coin has one head, one tail. No matter how you arrange the coins, there will be 10 heads and 10 tails in each group. The instructions don't specify that you should count only the side of the coins facing up.

[Guest]

Since you start off with half the coins on tails and half on heads; you can randomly divide them into two piles of 10 and each set of ten will be the compliment of the other set of 10. i.e. if you randomly take 10 coins that happen to contain 3 tails and 7 heads, the other pile will have to have 3 heads and 7 tails. Therefore the solution is to just take any 10 coins and flip them over. They will then be just like the 10 coins you didn't flip over.

[Guest]

I'm not sure I could flip a coin while wearing thick gloves; I doubt that I'd be capable of successfully picking up a coin, especially if I were blindfolded too.

[Smith]

Arrange the coins into 2 random groups of 10. In each group of 10, each coin has one head, one tail. No matter how you arrange the coins, there will be 10 heads and 10 tails in each group. The instructions don't specify that you should count only the side of the coins facing up.

early,

The instructions don't explicitly prevent one from removing the blindfold either, but there are certain assumptions one can make.

I was thrown by one assumption I made that perhaps I should NOT have made. I thought the goal was to end up with an EVEN number of heads and tails IN EACH GROUP, whereas the idea of having the same number of heads in each group, and the same number of tails in each group, is a more solvable puzzle.

[Guest]

So, you are blindfolded and wearing thick gloves, and someone put twenty coins on the table. Ten of them are on heads and ten of them are on tails, but you wouldn't know which because of your blindfold and gloves. You could flip or move the coins if you want. Your goal is to group them into two with ten coins each. The two groups should have the same number of heads and the same number of tails. How do you do it?

[Guest]

Hi

I am new

I do not have an answer

but how do I receive the solution.

TXS

Robsbrain

[Guest]

Put them all on their edges, then both groups will have 0 heads and 0 tails.

[fabpig]

Hi

I am new

I do not have an answer

but how do I receive the solution.

TXS

Robsbrain

Hi Robsbrain.....welcome to the den!

Sometimes, puzzles go unanswered for quite a while (Peace*out has had 1 on the go since last June!), but quite often they're solved fairly quickly. If you really want to know the answers, click on the spoiler "show" button to reveal what people have guessed. (Their guesses aren't necessarily correct, btw!)

Edited March 11, 2011 by fabpig

[Guest]

Hi

I am new

I do not have an answer

but how do I receive the solution.

TXS

Robsbrain

Hi Rob,

Welcome to BrainDen. The idea of this forum is that you have to work it out

If your brain isn't working today, you might get the answer by clicking the box labelled "Show" after the words "Spoiler for ..."

I hope this helps you.

Cheers,

Tony

[Guest]

Sometimes, puzzles go unanswered for quite a while (Peace*out has had 1 on the go since last June!)...

How do I find that puzzle, please?

Edited March 11, 2011 by TonyB

[fabpig]

How do I find that puzzle, please?

It's in the "New Word Riddle" section, entitled "100 days to live"

[Guest]

It's in the "New Word Riddle" section, entitled "100 days to live"

Thank you kindly, sir.

[Guest]

similar post of sent a month before,whatever

first take 10 coins out as it is and now reverse every coin in the another pile you will have same no of coins with heads in each pile

I know the answer but I can't understand it's working

Can anyone HELP!!

Edited March 12, 2011 by Shivam

[Guest]

..it says u must have exact number of heads and tails in each group...which means both groups must have 5 heads and 5 tails or 7 heads and 3 tails in both gropus...hm..my thinking time is over..but Im still trying to understand how to divide them..what if there are all tails or head?....

hi Im new here...trying to activate my brain a little bit...

Edited March 12, 2011 by Nano7

[Guest]

soryy in my 1. gusssing spoiler I have some misstakes...only thing true is that both gropus need 5 heads and 5 tailis.. the other stuff is nonsence...Im still trying to divde them...lol..this will take some time....

[Guest]

The answer was you divide them randomly into two groups. Then you flip every coin on the other side because if you had 6 heads 4 tails on one side, you have 4 heads 6 tails on the other. Flipping everything on the other side, you would have 6 heads on both and 4 tails on both. That is true for all combination like 5 and 5, 7 and 3, 8 and 2, 9 and 1, 10 and 0. Hope this is clear to you!

enrightmcc, Shivam, and Adhish Majumdar all got it right !

Thanks, Smith!

thanks too for the other funny answers.

Welcome, Robsbrain!