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Dividing a pile of coins into two such each pile has same number of tails facing up

19 viewed last edited 1 year ago
Vivekanand Vellanki
0
There are 10 coins all with heads facing upwards. A magician picks a volunteer from the crowd and says the following to the volunteer: - Pick any number of coins and flip them so that tails are facing up - Now, tell me how many coins have tails facing up The magician claims that he can divide the 10 coins into 2 piles such that both piles have same number of coins with tails facing up. And, he will do this blindfolded. The magician is only allowed to separate the coins into 2 piles and he can choose to flip some coins. How does the magician do this?
Vivekanand Vellanki
1
Here's the proof to this technique. There are N coins on a table; with T tails. You need to divide the N coins into 2 piles such that both the piles have the same number of coins with tails facing up. Also, this can be done without even seeing the coins. Here's the approach: Step 1) Pick any random T coins and place them to one side. These are your 2 piles - one set of T random coins; and the other set (N-T) coins Step 2) Now, flip each and every of the T random coins picked. Flip them exactly once - if the coin had heads facing up, it becomes tail; and vice versa Both piles will now have the same number of coins with tails facing up. After Step 1, there are 2 piles - P1 with T random coins, and P2 with (N-T) coins. Since we picked coins at random and placed them in P1, we dont know the number of coins with tails facing up in P1. Since we dont know the number, lets use the symbol Q to represent the number of coins with tails facing up in P1. Pile P1 has the following: Q tails; the remaining (T-Q) heads Pile P2 has (T-Q) with tails facing up. Why? The entire collection has T tails - Q went into P1, the remaining (T-Q) have to be in P2. So, Pile P2 has the following: (T-Q) tails; and the remaining heads. The exact number of heads in P2 does not matter to us. Step 2 flips all coins in pile P1. After step 2, Pile P1 has the following: Q heads (all tails become heads); and (T-Q) tails (all tails become heads) Pile P2 has the following: (T-Q) tails and the remaining heads Both the piles P1 and P2 have the same number of tails (T-Q). This is exactly what the magician claimed he would do.
Vivekanand Vellanki
0
Let's generalise the problem. There are N coins on a table; with T coins facing tails up. You need to divide the N coins into 2 piles such that both the piles have the same number of coins with tails facing up. Also, this can be done without even seeing the coins. Here's the approach: - Pick any random T coins and place them to one side. These are your 2 piles - one set of T random coins; and the other set (N-T) coins - Now, flip each and every of the T random coins picked. Flip them exactly once - if the coin had heads facing up, it becomes tail; and vice versa Both piles will now have the same number of coins with tails facing up. Try it out on your loved ones and pretend to be a magician. I'll post the proof tomorrow.