riddles

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clintmemo
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Re: riddles

Postby clintmemo » Fri Mar 24, 2017 8:06 am

The obvious way to answer #1 is to start with a single letter and work backwards.

<tries 4 times and fails>
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Lord Foul
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Re: riddles

Postby Lord Foul » Fri Mar 24, 2017 8:09 am

clintmemo wrote:The obvious way to answer #1 is to start with a single letter and work backwards.

<tries 4 times and fails>

I find that result quite startling. :o
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Re: riddles

Postby Burning » Fri Mar 24, 2017 10:44 am

Lord Foul wrote:
tombombodil wrote:4. You sit blind-folded at a large table covered in hundreds and hundreds of coins, but you don't know precisely how many. Since you are blinded-folded you cannot see the coins, but you are told that there are 20 coins sitting tails-side up. The rest are heads-up.

You can move and flip and turn the coins as you please but you cannot see what you do, and, though you can touch the coins, you cannot tell heads from tails by feel.

How can you separate the coins into two piles such that each pile has the same number of heads-up and tails-up coins?

Are you sure that last line is correct? Same number of heads-up AND tails-up coins? :eh:


I am fairly certain that the problem as stated is impossible.

If you start with an equal number of heads as tails, then the problem is easy. Divide the coins into two piles. It is fairly easy to show that the number of tails in pile 1 equals the number of heads in pile 2 (and vice versa). Therefore if you flip all the coins in one pile and none in the other, you've made two piles that have the same number of heads and the same number of tails. You have no idea how many of each is in a pile, you just know the other pile is the same.

But if the starting number of heads doesn't equal the starting number of tails, you don't have this nice symmetry to take advantage of. In fact, it's fairly simple algebra to show that if H = T +2n, then after you have divided into two piles, you have H1 = T2 + n and H2 = T1 + n. So clearly flipping all the coins in one pile will not have the desired effect. It is in fact guaranteed to create an imbalance.

Flipping no coins means that you are depending on just being lucky and getting even amounts on the first division. Flipping some coins in one (or both piles) just adds the randomness of which coins you happen to pick. Swapping coins between piles also introduces more randomness. Flipping some coins before you divide them into the two piles just destroys your foreknowledge of how many tails there are.

I haven't explored rigorously details of the possible combinations of swapping and flipping, so I haven't proved that there isn't some clever trick where the random chances all cancel out, but I'm highly skeptical.
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Re: riddles

Postby Lord Foul » Fri Mar 24, 2017 10:51 am

Burning wrote:
Lord Foul wrote:
tombombodil wrote:4. You sit blind-folded at a large table covered in hundreds and hundreds of coins, but you don't know precisely how many. Since you are blinded-folded you cannot see the coins, but you are told that there are 20 coins sitting tails-side up. The rest are heads-up.

You can move and flip and turn the coins as you please but you cannot see what you do, and, though you can touch the coins, you cannot tell heads from tails by feel.

How can you separate the coins into two piles such that each pile has the same number of heads-up and tails-up coins?

Are you sure that last line is correct? Same number of heads-up AND tails-up coins? :eh:


I am fairly certain that the problem as stated is impossible.

If you start with an equal number of heads as tails, then the problem is easy. Divide the coins into two piles. It is fairly easy to show that the number of tails in pile 1 equals the number of heads in pile 2 (and vice versa). Therefore if you flip all the coins in one pile and none in the other, you've made two piles that have the same number of heads and the same number of tails. You have no idea how many of each is in a pile, you just know the other pile is the same.

But if the starting number of heads doesn't equal the starting number of tails, you don't have this nice symmetry to take advantage of. In fact, it's fairly simple algebra to show that if H = T +2n, then after you have divided into two piles, you have H1 = T2 + n and H2 = T1 + n. So clearly flipping all the coins in one pile will not have the desired effect. It is in fact guaranteed to create an imbalance.

Flipping no coins means that you are depending on just being lucky and getting even amounts on the first division. Flipping some coins in one (or both piles) just adds the randomness of which coins you happen to pick. Swapping coins between piles also introduces more randomness. Flipping some coins before you divide them into the two piles just destroys your foreknowledge of how many tails there are.

I haven't explored rigorously details of the possible combinations of swapping and flipping, so I haven't proved that there isn't some clever trick where the random chances all cancel out, but I'm highly skeptical.

If you take any 20 coins as one pile and flip them then you'll have two piles in which the number of tails-up is equal, but the number of heads-up will be very different. That's why I believe the last line may be in error.
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Re: RE: Re: riddles

Postby tombombodil » Fri Mar 24, 2017 12:53 pm

Lord Foul wrote:
Burning wrote:
Lord Foul wrote:
tombombodil wrote:4. You sit blind-folded at a large table covered in hundreds and hundreds of coins, but you don't know precisely how many. Since you are blinded-folded you cannot see the coins, but you are told that there are 20 coins sitting tails-side up. The rest are heads-up.

You can move and flip and turn the coins as you please but you cannot see what you do, and, though you can touch the coins, you cannot tell heads from tails by feel.

How can you separate the coins into two piles such that each pile has the same number of heads-up and tails-up coins?

Are you sure that last line is correct? Same number of heads-up AND tails-up coins? :eh:


I am fairly certain that the problem as stated is impossible.

If you start with an equal number of heads as tails, then the problem is easy. Divide the coins into two piles. It is fairly easy to show that the number of tails in pile 1 equals the number of heads in pile 2 (and vice versa). Therefore if you flip all the coins in one pile and none in the other, you've made two piles that have the same number of heads and the same number of tails. You have no idea how many of each is in a pile, you just know the other pile is the same.

But if the starting number of heads doesn't equal the starting number of tails, you don't have this nice symmetry to take advantage of. In fact, it's fairly simple algebra to show that if H = T +2n, then after you have divided into two piles, you have H1 = T2 + n and H2 = T1 + n. So clearly flipping all the coins in one pile will not have the desired effect. It is in fact guaranteed to create an imbalance.

Flipping no coins means that you are depending on just being lucky and getting even amounts on the first division. Flipping some coins in one (or both piles) just adds the randomness of which coins you happen to pick. Swapping coins between piles also introduces more randomness. Flipping some coins before you divide them into the two piles just destroys your foreknowledge of how many tails there are.

I haven't explored rigorously details of the possible combinations of swapping and flipping, so I haven't proved that there isn't some clever trick where the random chances all cancel out, but I'm highly skeptical.

If you take any 20 coins as one pile and flip them then you'll have two piles in which the number of tails-up is equal, but the number of heads-up will be very different. That's why I believe the last line may be in error.

This is correct. Last line was miswritten :oops:

The good ol "my brain knew what I meant and so ignored what I wrote"
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Leoff
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Re: riddles

Postby Leoff » Fri Mar 24, 2017 1:53 pm

Can we have a correct re-statement of that problem?

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Re: riddles

Postby Lord Foul » Fri Mar 24, 2017 2:53 pm

Leoff wrote:Can we have a correct re-statement of that problem?

4. You sit blind-folded at a large table covered in hundreds and hundreds of coins, but you don't know precisely how many. Since you are blinded-folded you cannot see the coins, but you are told that there are 20 coins sitting tails-side up. The rest are heads-up.

You can move and flip and turn the coins as you please but you cannot see what you do, and, though you can touch the coins, you cannot tell heads from tails by feel.

How can you separate the coins into two piles such that each pile has the same number of tails-up coins?
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Re: riddles

Postby tombombodil » Fri Mar 24, 2017 3:25 pm

Lord Foul wrote:
Leoff wrote:Can we have a correct re-statement of that problem?

4. You sit blind-folded at a large table covered in hundreds and hundreds of coins, but you don't know precisely how many. Since you are blinded-folded you cannot see the coins, but you are told that there are 20 coins sitting tails-side up. The rest are heads-up.

You can move and flip and turn the coins as you please but you cannot see what you do, and, though you can touch the coins, you cannot tell heads from tails by feel.

How can you separate the coins into two piles such that each pile has the same number of tails-up coins?


Yeah, I edited the original post.
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Re: riddles

Postby tombombodil » Fri Mar 24, 2017 3:26 pm

Lord Foul wrote:
clintmemo wrote:The obvious way to answer #1 is to start with a single letter and work backwards.

<tries 4 times and fails>

I find that result quite startling. :o


;) :mrgreen:
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clintmemo
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Re: riddles

Postby clintmemo » Sat Mar 25, 2017 10:12 am

tombombodil wrote:
Lord Foul wrote:
clintmemo wrote:The obvious way to answer #1 is to start with a single letter and work backwards.

<tries 4 times and fails>

I find that result quite startling. :o


;) :mrgreen:



<looks at the word "startling" again and counts the letters. Once again realizes that isn't the joke that he is not getting.>
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