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cool riddle

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You are a prisoner. Your caprtor leads you into a room and shows you fifty pennies on a table. Thirty of them are heads up and the rest are tails up. You are then blind folded, and your fingers are taped so that you can't feel if the pennies are heads up or tails up. Then your captor tells you that you must get two piles with the same number of pennies heads up in each. How do you do it?

You turn over thrity pennies.

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Posted · Report post

I don't think your solution works, or your riddle is not set up properly - one of the two.

I assume you meant getting equal piles of 25. If you randomly turn over 30 pennies there is a great chance that you will not have two equal piles of heads and tails. You'd have better chances just flipping 5 and hoping you hit 5 out of the 30 heads.

If you meant two equal piles of any size, you start out with the correct solution already there. You don't need to flip ANY to win because you have two piles of 20 heads and tails. Or, if you must flip some over, just flip the minimum and you have two equal piles by disregarding the others.

Either way, your solution doesn't fit the proposition.

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This one is pretty confusing but it does work. As it says in the riddle, “You must have two piles the same number of pennies with HEADS UP in each.” Just to make it simpler, I will give you a hint. It works with any number of pennies and any number of them can be heads up. So, you could have 5 pennies with 2 pennies heads up. Still, the solution would be the same.

You have 2 pennies heads up and 3 pennies tales up.

You must start with all the pennies in one pile. So let’s say that you pick up a heads up penny and flip it over. You now have 1 penny heads up in the first pile and none in the second. Now you flip your second penny. If you get a tales, you flip it and it becomes a heads so you now have 1 penny heads up in each pile. Again, it does not matter how many pennies you have in each pile, only how many pennies heads up. Sorry if my original puzzle was unclear. Hopefully this will clarify it.

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Ok, yes - I misread the proposition.

Still, how do you split the 50 pennies into two piles so that you ensure an equal number of heads in each? If you flip 30 at random and can ensure that you never flip a penny twice, you could end up with anywhere between zero and forty heads showing without knowing how many you have. How are you going to then make 2 equal piles? I'd dare you to try that for big stakes.... Again, if you don't have to physically make the piles, then the solution is met from the start. You have to assume some volitive act on the part of the captive to make the piles or it's a meaningless challenge.

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How stupid of me. I meant to say that the pennies you flip over you then put in a new pile.

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Another approach is to make two piles with 0 pennies in each pile.

Each pile will have 0 pennies with heads showing.

*scurrying back inside the box*

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I guess I really did a bad job on this riddle. I sort of implied but never actaully stated that you must have all the pennies in those two piles. Thanks for all the input!

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If you did mean for there to be two piles with all the pennies in them and an equal number of heads in each pile, we get back to my original objection - which is that you cannot be sure by flipping 30 random pennies that you will have an equal number of heads in each pile. By flipping 30 random pennies with no repeats you can have anywhere between 40 and zero heads showing when you get done. How on earth do you guarantee that you will have an equal number in each pile?

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So let’s say that you pick up a heads up penny and flip it over. You now have 1 penny heads up in the first pile and none in the second. Now you flip your second penny. If you get a tales, you flip it and it becomes a heads so you now have 1 penny heads up in each pile.
This doesn't make sense to me. Will you clarify?

If I have 5 pennies and 3 are heads and 2 are tails, and I grab one at random and flip it for pile 1, then I have a 2 in 5 chance of having a heads. For pile 2, I have either a 2 in 4 or a 1 in 4 chance of having heads depending on the first flip. To make this complete to your answer (of 30 out of 50 flipped) I'd need to flip a third penny, which would give me a either a 2 in 3, 1 in 3, or zero chance of hitting a heads on the flip. It's entirely possible that I flip tails the first time then heads, then tails. How do I have the same number of heads in each pile now? Unless you are talking about making 3 piles, you don't even have 2 equal piles in this situation. You can with 50, as you'd have 25 and 25, but you still have the very large random chance in your flips that makes it impossible to guarantee an equal number of heads in the two piles.

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Once again we will go with me example of 5 pennies with 2 of them heads up, all in one pile.

So you have one pile that has 5 pennies in it. You pick up one penny, flip it over, and put it in a NEW pile. Pick up the second penny and put it in pile two. So if you pick up a penny that is heads up the first time, you have 4 pennies in your first pile with 1 heads up and you have 1 penny in your second pile with 0 heads up. If your second penny is tails, you pick it up and flip it over and put it in pile number two. Now you have 3 pennies in pile one and 1 heads up, in pile number two, you have 2 pennies with 1 heads up. This does work no matter how many pennies you have and no matter how many heads up you have and no matter what pennies you pick up. As I have previoiulsy mentioned, you need the same number of pennies HEADS UP in each pile not the same number of pennies total.

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Very good riddle, but you're right -- it could have been told better. The question is: Without looking at the coins, can you divide them into two (not necessarily equal) piles that each contain the same number of heads?

Answer: One easy solution is to consider all 50 pennies to be in one pile, then randomly select 30 pennies for a second pile, turning all those pennies over. Since you started with exactly 30 heads, you are guaranteed to have the same number of heads showing in each pile.

Why does this work?

Consider that you randomly select N heads out of the original 30, leaving 30-N heads in the first pile. Since you selected 30 coins, N of which are heads, this means that you also have selected 30-N tails into the second pile.

Huh. Look at that. You have 30-N heads in the original pile and 30-N tails in the new pile. If only there were some way to reverse the heads and tails in the new pile...

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the same number of pennies HEADS UP in each pile not the same number of pennies total.

Ah! I think this is where it was unclear. So it's perfectly valid to have one pile of one penny and one pile of 49 pennies, correct?

The way you presented it it seemed you wanted equal numbers in the piles. I see where you are going with this now. Nice one.

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Posted · Report post

That is exactly right!

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This problem was actually posed to one of my friends in his interview today. The problem got leaked and he asked me to try to solve it before he went in for his interview. I found it very interesting after I could get to the bottom of it and was thinking to post it on this forum. But looking at the mess in the explanation I thought I would rather drop in the general solution!

So here it goes:

Problem

Say you have N coins with H of them being heads up. Now the task is to make 2 piles of these N coins, so that they have equal number of heads. And, by the way, you are blindfolded.

Solution

Now the solution is make a pile of H coins at random. Flip them (so that all H becomes T and all T becomes H). And you have solved the problem.

Explanation:

Lets say that the pile of H random coins that you chose has X no. of Heads and (H-X) number of Tails. Thus the pile 2 has (H-X) number of Heads. Thus if you flip all the coins in pile 1, then you have X no. of Tails and (H-X) no. of Heads. Which in turn is exactly equal to the number of Heads in pile 2. Problem solved!!

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