3
$\begingroup$

You are blind and sitting at a table with 100 coins on it. You know exactly 50 coins are showing heads and the rest tails, and you can tell by feel where the coins are, but you are unable to see or feel whether a particular coin is showing heads or tails.

Your task is to divide the coins on the table into two sets each of which has the same number of heads and the same number of tails as the other. How can you do it?

$\endgroup$
2
  • $\begingroup$ divide it into two sets-->flip any set coins--->done tadannn :) $\endgroup$ May 21, 2015 at 8:27
  • $\begingroup$ Hmm ... seems this one was too easy. $\endgroup$ May 21, 2015 at 9:25

2 Answers 2

10
$\begingroup$

Since you can feel coins....

Divide coins into two sets 50-50

and

Flip all the coins of one set.

So you have exactly same number of H and T coins in both sets

Thanks @Albert Masclans for explanation....

I actually solved it for less number lets say 14 coins

so lets say we have H T H | H T H T

and                          T H H H | T T T

above are random arrangements

Now lets divide them along a line | shown above (in anyways you can divide them)

now flip first (left set)

will look like

T H T|H T H T

H T T T|T T T

Now we have 2 H and 5 T in both sets

$\endgroup$
7
$\begingroup$

The solution user2408578 is indeed correct. Allow me to illustrate it somewhat.

As said...

Divide coins into two sets 50-50. You'll have now $t$ Tails in the first set and $50-t$ Heads. And on the second set you'll have $50-t$ Tails (since, there are $50$ Tails in total) and $50-(50-t)=t$ Heads.

Given that, it's clear that you have to...

Flip one of the set's all coins. Since doing that will mean turning all Tails into Heads. So flipping all coins in the second set would mean having $t$ Tails and $50-t$ Heads, as in the first set.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.