Tricky tricky coins

[Nathelion]

This thread is more fun than business, I already know the answer to the problem in question, so don't feel compelled to post just to be nice/help out. If you are annoyed that I post this in here and think I should move these things to off-topic, please say so and I will.

You are sitting in a room with a stack of coins in front of you. All the coins are identical. You know that exactly ten of the coins in the stack are tails up. Your task is to make two piles with an equal number of coins that are tails up in each.

However, just as you are about to start the lights go out! It is now completely dark in the room. You can't feel the difference between the two sides of the coins. How do you carry out your task?

 

[Born2bwire]

Make one stack of ten coins and another stack of the leftovers. Then flip all the coins in the stack of ten.

 

[BrownTown]

heh, people need to come up with some more complicated questions, it seems like the fools who were posting dozens of wrong answers around here finally gave up on it and left.

 

[Nathelion]

Yeah. Too bad, I thought it was amusing

 

[pcy]

Just give us some harder questions then...


Peter

 

[esun]

How about writing solutions to all of the problems on this page: http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtml

That'd keep you busy for a loooong time.

 

[Nathelion]

Hahaha I imagine it will. Thx for the link!

 

[brikis98]

I actually like this problem... a keeper for when I interview people

 

[stogez]

Originally posted by: Born2bwire
Make one stack of ten coins and another stack of the leftovers. Then flip all the coins in the stack of ten.

Is that actually the answer? How do you know that some of the coins you used to make the stack of ten weren't already tails up?
Am I just not getting it? Thats quite possible too

 

[Born2bwire]

Originally posted by: stogez

Originally posted by: Born2bwire
Make one stack of ten coins and another stack of the leftovers. Then flip all the coins in the stack of ten.

Is that actually the answer? How do you know that some of the coins you used to make the stack of ten weren't already tails up?
Am I just not getting it? Thats quite possible too

There are N coins and we know that there are T coins that are tails. We divide up into two stacks, one that has T coins and the other has N-T coins. Now there are m coins in the stack of T coins that are tails up. We don't know what m is nor do we care, all we know is that m<=T. That means that there must be T-m coins that are tails up in the other stack and T-m coins that are heads up in the T stack. If we flip all the coins in the T stack, we now have T-m coins that are tails up and m coins that are heads up. Thus, we now have two stacks with equal numbers of tails sides up (T-m). This solution works for any arbitrary number of tails up (T).

 

[TuxDave]

Originally posted by: stogez

Originally posted by: Born2bwire
Make one stack of ten coins and another stack of the leftovers. Then flip all the coins in the stack of ten.

Is that actually the answer? How do you know that some of the coins you used to make the stack of ten weren't already tails up?
Am I just not getting it? Thats quite possible too

Give it a whirl and you'd be amazed how well it works.

 

[gorobei]

interesting.

 

[firewolfsm]

I had to read that answer a few times...but I got it.

 

[jjzelinski]

Kind of sad that this still makes zero sense to me. It would be one thing if you were working with probability but the question requires a precise number as opposed to a range. This just seems arbitrary despite B2W's explanation (which I'm sure is fine.) If you have no idea how many tails up you grabbed to begin with what would it matter if you turned those same coins over?

 

[Born2bwire]

Originally posted by: jjzelinski
Kind of sad that this still makes zero sense to me. It would be one thing if you were working with probability but the question requires a precise number as opposed to a range. This just seems arbitrary despite B2W's explanation (which I'm sure is fine.) If you have no idea how many tails up you grabbed to begin with what would it matter if you turned those same coins over?

Let's say we have a stack of say 25 coins and we know that there are 10 tails up. So if I take any 10 coins and place them into one stack, we now have a stack of 15 and another stack of 10 coins each. We do not know how many coins are tails in either stack but we still know that there are a total of 10 tails. In the second stack of ten, there are some number m of tails. Let's say it's 3. So in the second stack there are m=3 tails and 10-m=7 heads. That means that in the other stack, there must be 10-m = 7 coins that are tails. So if we flip the stack of 10 coins, we will now have 10-m=7 coins tails and m=3 coins head. Thus both stacks now have 7 tails. The only trick is that you make the second stack equal to the number of tails that you know you have. Since you can only have one of two sides up there is a symmetry in the situation that allows for an easy solution.