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Difficult Logic Problems
Flipping Coins solution - explaination needed

Fri Feb 05, 2010 9:02 pm  by julioprimo

There are a hundred coins sitting on the table, ten are currently heads and nintey are currently tails. You are sitting at the table with a blindfold and gloves on. You are able to feel where the coins are, but are unable to see or feel if they heads or tails. You must create two sets of coins. Each set must have the same number of heads and tails as the other group. You can only move or flip the coins, you are unable to determine their current state. How do you create two even groups of coins with the same number of heads and tails in each group?

Create two sets of ten coins. Flip The coins in one of the sets over, and leave the coins in the other set alone. The first set of ten coins will have the same number of heads and tails as the other set of ten coins.

Can somebody hep me better understand the solution? It would seem that if by chance I (blindfolded) create two sets of coins 0h:10t - 0h:10t and flip one set thus making 10h:0t - 0h:10t that I do not have the same number of heads and tails in each group. There are situations where this would work, but not 100% of the time. Am I missing a key point in the puzzle maybe?

Many thanks!
Mon Nov 29, 2010 9:26 pm  by sudheerkuttiyat

Yes, The question should be ..."there are twenty coins, with 10 showing heads and 10 showing tails..... "

The solution does not work for 100 coins with only 10 showing heads initially

Splitting the twenty to two groups and flipping all coins in one group will make both groups look the same.
Wed Sep 11, 2013 7:02 am  by nik

For 10 and 90 divisions, you create sets of 10 and 90 only. Then flip all coins of any one of the set. You will get same number of heads facing up in both sets.
say the set 90 has [b]x [/b]coins with heads up. It means the set of 10 has 10-x coins with heads up. If you flip the set of 10 then it will now have 10-(10-x)=[b]x[/b] coins with heads up, which is equal to number of coins with heads up in the set of 90.
Mon Sep 01, 2014 12:16 pm  by ankitnamdeo34

one solution to this problem which i think is that make the set of 2 ten coins selected randomly and then do this in each set :flip 0 times the coin 1 ,flip 1 time the coin2 and so on .By doing so we will have the two groups each having equal number of heads and tails. please tell me if there is any fault in this
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