Today here is another **brain game**. I haven’t posted them for a few weeks and I hope, you have missed my **riddles**. Try to solve this one. There are 50 1-**cent **coins on the table. Initially all **coins **are tails up. Close your eyes, and I will turn over 10 random coins. The task is to divide all the coins into two groups blindly, so that the groups have an equal number of heads up. Mind, that in this brain game you are not a psychic. Maybe, you are going to become one, while you are thinking about the solution?

Please, if you want to **solve **the riddle yourself, be careful and do not look at the comments of this post, because someone could have already written the **solution **there. Good luck! You will find the answer in the next brain game. You are always welcome to ask questions about the task, in case something is not clear. But sorry, I will not answer leading questions.

The answer to the previous brain game is the mail carriers at the **United Nations** Building in New York . EvilArchie was the only one who gave the right solution. Very well done!

At first glances, is this down to probability (not something silly like we can feel the difference on the face of the coins).

Initially, I can say that if it were 2 coins turned over from a pool of 10, then moving 3 coins at random should in theory be the highest likelihood of having 1 head in both piles (3 coins, 7 coins).

Extending that to 50 coins with 10 heads isn’t obvious, so I’ll just guess at move 15 coins into one group, leaving 35 in the original group. That should allow 5 coins in each group on average.

Hmm that doesnt sound right looking at that…

Adjusting my other comment. Moving 3/10 coins is equally likely to have even distribution of heads as moving 4/10 coins. Thus move 3.4 coins is my guestimate to a pool of 10 coins as 3 coins has 21/45 probabilty and 4 coins has 24/45 probability. Not a uniform distriubtion, skewed in favour of fewer coins.

Thus, 3.4 * 5 is a terrible estimate, but move 17 coins.

I’m guessing there’s a way to reach a limit of 50% probability by moving coins backwards and forwards in a super method… Or not. I don’t know this is hard.

You can do this with 100% percent accuracy. Skip the damn probabilities, you’re using your hands. You get to feel the coins. The human hand is more than capable of telling the difference between a heads up and a tails up coin.

Move ten coins to a seperate pile turning these coins over as you do.

If any any of the coins are the heads these will then become tails.

But since you moved ten there will be an equal number of heads in both piles.

Example three of the ten where heads. Leaving seven heads in the 40 pile. Turning over the ten pile gives 3 tails and 7 heads.

stand the pennies on edge then divide into two groups both groups will have zero heads

make a group of 10 coins and a group of 40 coins. Then, turn each of the 10 coins over and you’ve done it.

You may be blind but you can still feel the coins. It shouldn’t not be that hard to tell if it is heads or tails with your fingers.

I feel the coins with my fingers and put the heads together and then turn over 15 more of the tails and put them with the first 10 heads

Turn them all on edge and divide into 2 groups. With no heads up or down, they are in equal amounts.

I think Jerry’s solution is the perfect one. No need to guess.

Rubber and Jerry, this is a good idea, but if this were the solution, the brain game were too easy. Please, try again, I believe, you can do it! The riddle has a particular, “nicer” solution. When you read (or find) it, you will agree with it.

Two people have already answered this riddle correctly, but I will publish their comments later in order not to spoil the game for everybody.

Solving the riddle is simple: You turned over ten random coins, but you did not specify that these coins were part of the 50 one cent coins in the table. Therefore all of your 50 coins are still tails up.

All you have to do is count 25 coins and separate them from the other 25 coins. You will now have two groups of 25 coins that are tails up. Turn over one coin from each group . Now you have one coin in each group that is heads up.

Now both groups have the same number of coins that are heads up.

Alberto Beale

Alberto, you may use only those 50 coins and no extra coins.

Seperate the coins into two groups one of 40 the other of 10.

Turn all of the coins in the group of 10 over.

This will give you two groups with the same number of heads up.

Great puzzel. I drew a total blank then last night, in bed just before I went to sleep the answer hit me. Draw out 10 pennies to form one group, leaving 40 in the other group. Turn each of those 10 pennies over. The result will be an equal number of heads up pennies in each group.

in probability’s sake,

the initial face of each coin is irrelevant.

In flipping a coin, the chance it’s a heads or tails is 50/50.

If you pick up a coin, flip it, then put it to your right, then pick up another coin, flip it, and put it to your left. by the end, you’ll should have an equal number of faced / tails coins in each pile.

Or do we need a more concrete solution?

hi all,

I think the solution is simple.U r sure that there r 10 heads in the group na..

So separate 10 random coins to form a group.. and remaining 40 will be in another group.. So if there are x heads in 10grp there will b 10-x heads in 40grp..

ie, the no of tails in 10grp is equal to the no of heads in 40grp..

So if u flip all the coins in 10grp all the tails will become heads and finally now both groups have same number of heads..

This is a nice puzzle.. and im gonna post this in my blog

( muralizblog.wordpress.com ).. I think u can see some gud puzzles in my blog..

thanx for the author who posted this..