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September 21, 2025, 05:33:15 am

Author Topic: Just for fun!  (Read 1530 times)  Share 

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QuantumJG

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Just for fun!
« on: July 06, 2010, 10:39:26 pm »
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Before I give you the problem, I will give + karma to the most ingenious solution to getting the answer.

Ok I have $1.00 and you have $0.00, I then give you half of my money and then you give me half of what you just recieved back, I then give you half of what I have and you give me back half of what you have and so on.

I.e. I give you $0.5, you give me $0.25, I give you $0.375, etc.

If we were to settle the transaction in one go, how much money should I give you?

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tram

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Re: Just for fun!
« Reply #1 on: July 06, 2010, 10:48:06 pm »
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wtfffffffffffff i'm trying to work otu a way to do ths without just hacking thu the numbers.........

QuantumJG

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Re: Just for fun!
« Reply #2 on: July 06, 2010, 11:33:30 pm »
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I didn't do that.

I have manually solved the problem too.
« Last Edit: July 07, 2010, 11:05:06 am by QuantumJG »
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QuantumJG

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Re: Just for fun!
« Reply #3 on: July 06, 2010, 11:40:43 pm »
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Is this a more complex version of the question: 1/2 + 1/3 + 1/4 + 1/5 + 1/6 ... = 2?

Have you solved the question manually? =D

That series doesn't equal 2. Actually ithat series is a p-harmonic series and it is divergent.

The geometric series:

1 + 0.5 + (0.5)^2 + 0.5^3 + (0.5)^4 + ... = 2

The easiest way is to just write out several transactions and see the pattern emerge.

But one of my friends brought up the idea of an iterative sequence.
« Last Edit: July 07, 2010, 11:03:40 am by QuantumJG »
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Ahmad

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Re: Just for fun!
« Reply #4 on: July 08, 2010, 09:31:13 pm »
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In an attempt to prevent this topic from dying here's a possible start:
Let be first person's money after move n, be second person's money.



What are some recursive relations?
,
,

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QuantumJG

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Re: Just for fun!
« Reply #5 on: July 08, 2010, 09:57:15 pm »
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Here is a formula to find out what fraction of the first person's money is given to the other.





So to end this puzzle, the first person should give 1/3 of his money to the other person at the start.
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brightsky

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Re: Just for fun!
« Reply #6 on: July 08, 2010, 10:17:22 pm »
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Let



Hence ,

So
« Last Edit: July 08, 2010, 10:27:21 pm by brightsky »
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Mao

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Re: Just for fun!
« Reply #7 on: July 11, 2010, 03:41:40 am »
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Without dealing with series, we look at the steady-state solution which the series must approach1.

At steady state, you have 1-x, and I have x. For one iteration (You give me some, I give some back)

or

The prior: , which solves to x=me=1/3. [the latter gives the same solution]

You give me $ 1/3. Have I earned my karma?

[It's a matter of interpretation though, if the transaction ends on me returning money, you give me $1/3, if the transaction ends on you giving, you give me $2/3, but the system will settle down to alternating between 1/3 and 2/3 for each person.]

Note:
1. It would be fairly easy to show this, as a pair of transactions here can be represented by a 2x2 matrix, and the process here is a dynamical system. Computing the eigenvalues/eigenvectors would show that the dynamical system converges to some eigenvector. I cbf looking it up in the lecture notes right now, but it's definitely doable.
« Last Edit: July 11, 2010, 04:01:39 am by Mao »
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