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# cool riddle

13 replies to this topic

### #1 bociniki

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Posted 13 December 2007 - 01:51 AM

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?
Spoiler for Solution

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### #2 Writersblock

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Posted 13 December 2007 - 04:15 AM

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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### #3 bociniki

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Posted 13 December 2007 - 05:31 PM

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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### #4 Writersblock

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Posted 13 December 2007 - 06:13 PM

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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### #5 bociniki

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Posted 14 December 2007 - 02:14 AM

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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### #6 bonanova

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Posted 14 December 2007 - 06:59 AM

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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Vidi vici veni.

### #7 bociniki

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Posted 14 December 2007 - 03:32 PM

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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### #8 Writersblock

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Posted 14 December 2007 - 06:23 PM

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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### #9 Writersblock

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Posted 14 December 2007 - 06:33 PM

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?

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### #10 bociniki

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Posted 14 December 2007 - 06:50 PM

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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