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18 replies to this topic

### #1 superprismatic

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Posted 30 March 2010 - 09:04 PM

You have two bags containing balls.
Bag A contains 2 white balls, bag B
contains 2 black balls. You randomly
pick a ball from bag A and place it
into bag B, then you randomly pick a
ball from bag B and place it into bag
A - this will be called an iteration.
Note that after any number of
iterations the two bags will each
contain 2 balls. What is the
expected number of iterations needed
to go from the initial configuration
of 2 white balls in bag A and 2
black balls in bag B to a final
configuration of 2 black balls in
bag A and 2 white balls in bag B?
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### #2 NeoPark

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Posted 31 March 2010 - 12:55 AM

The minimum number of iteration will be 2 if we pick white balls from bag A and black balls from bag b.
The maximum number of iteration will be infinity if we pick a black ball from bag A and white ball from bag B except the first iteration.
But, in random choice, we might have to calculate the probability of choosing a white ball from bag A after the first iteration.
That's my first shot for this problem. if anyone can have a good idea to solve this problem. I will be happy to hear that.
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### #3 mmiguel1

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Posted 31 March 2010 - 03:07 AM

You have two bags containing balls.
Bag A contains 2 white balls, bag B
contains 2 black balls. You randomly
pick a ball from bag A and place it
into bag B, then you randomly pick a
ball from bag B and place it into bag
A - this will be called an iteration.
Note that after any number of
iterations the two bags will each
contain 2 balls. What is the
expected number of iterations needed
to go from the initial configuration
of 2 white balls in bag A and 2
black balls in bag B to a final
configuration of 2 black balls in
bag A and 2 white balls in bag B?

Good problem.
Hopefully not as controversial as White Balls.
I just started school yesterday and won't have time to debate at passionately as I did in White Balls.
That is not to say your problem is less interesting.

One clarification: In an iteration, after you place a ball from bag A into bag B, and you are selecting from B to place back into A, do you include the ball you put into B in the same iteration, or do you randomly choose from one of the 2 balls initially in B at the start of the iteration?

Spoiler for Here is a try

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### #4 mmiguel1

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Posted 31 March 2010 - 03:26 AM

Spoiler for Next Step Attempt

Edited by mmiguel1, 31 March 2010 - 03:31 AM.

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### #5 bushindo

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Posted 31 March 2010 - 05:34 AM

Haven't been here for so long. I miss superprismatic's problems.
Spoiler for spoiler

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### #6 mmiguel1

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Posted 31 March 2010 - 05:55 AM

Haven't been here for so long. I miss superprismatic's problems.

Spoiler for spoiler

Hello Bushindo, it has been a while. Superprismatic's Walter Penney puzzles were the best.

In this problem, you interpret that once you have placed a ball from A into B, when you select a ball in B to place in A, you may select any of the 3 in B.
Looking at the wording, that may be the correct interpretation.
I assumed each iteration meant swapping a ball between each bag.

"You randomly
pick a ball from bag A and place it
into bag B, then you randomly pick a
ball from bag B and place it into bag
A - this will be called an iteration."

because he uses the word "then", I think I originally had an incorrect interpretation.

May I ask how you calculated the expected value from your graph?

Edited by mmiguel1, 31 March 2010 - 05:56 AM.

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### #7 bushindo

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Posted 31 March 2010 - 07:11 AM

Hello Bushindo, it has been a while. Superprismatic's Walter Penney puzzles were the best.

In this problem, you interpret that once you have placed a ball from A into B, when you select a ball in B to place in A, you may select any of the 3 in B.
Looking at the wording, that may be the correct interpretation.
I assumed each iteration meant swapping a ball between each bag.

"You randomly
pick a ball from bag A and place it
into bag B, then you randomly pick a
ball from bag B and place it into bag
A - this will be called an iteration."

because he uses the word "then", I think I originally had an incorrect interpretation.

May I ask how you calculated the expected value from your graph?

Hi mmiguel1, it really has been a while. Looks like you have been a very productive member on this board. I remember a couple of month ago you had your 1st post, and now you have almost the same post count as me.

Spoiler for spoiler

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### #8 mmiguel1

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Posted 31 March 2010 - 07:44 AM

Hi mmiguel1, it really has been a while. Looks like you have been a very productive member on this board. I remember a couple of month ago you had your 1st post, and now you have almost the same post count as me.

Spoiler for spoiler

Yes, there are some fun problems on here. It's a nice site.

You have a very good, simple solution.
Spoiler for

Edited by mmiguel1, 31 March 2010 - 07:47 AM.

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### #9 Tuckleton

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Posted 31 March 2010 - 09:21 AM

Maybe you guys can help me with a stumbling block that always keeps me from trying these types of questions.

What exactly does it mean when it asks for the expected number of iterations to go from one state to another?

Spoiler for Take this problem as an example

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### #10 superprismatic

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Posted 31 March 2010 - 10:33 AM

One clarification: In an iteration, after you place a ball from bag A into bag B, and you are selecting from B to place back into A, do you include the ball you put into B in the same iteration, or do you randomly choose from one of the 2 balls initially in B at the start of the iteration?

Include the new ball which was placed into B in the same iteration.
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