[Guest]

There are 20 coins on a table and half of them are showing heads and half of them are showing tails. The room is completely dark and you are wearing gloves. You need to divide the coins into two groups only by moving and flipping them. The challenge is, after dividing the number of heads & tails in each group should be the same as the number of heads & tails of the other group. For example, if one group has 3 heads, then the other group should also have 3 heads. How this can be done?

[Guest]

Divide the coins into two groups, then flip around all the coins in one of these groups.

[Guest]

Divide the coins into two groups. Then flip ALL the coins in one of the groups. This should leave the desired effect.

[Guest]

DIVIDE THE COINS INTO @ GROUPS AND THEN FLIP ONE GROUP

THIS IS ALREADY POSTED BUT GOOD, FOR NEWBIES GET CHANCE

[Guest]

can this be gerneralized? lets say i have 18 dice, each showing an equal number of 1-6. i want to split them into two groups, such that the numbers 1-6 are equally prevelent in both groups, and can only move or rotate dice (blindfolded of course).

note that these are stanard dice; ie 1 is on the oppsite side of 6, 3 is on the oppisite side of 4, and 5 is on the oppisite side of 2.

[Guest]

Somone care to explain how this works? I don't follow...

[Guest]

so, you have x number of tails and x number of heads. if you split the group in half, you now have m heads and n tails in one group, and m tails and n heads in the other. if you flips over the coins of one group or the other, you'll have an equal number of heads and tails in both groups.