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Get final status from the drawing sequence

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There are 4 balls marked as "A", "B", "C" and "D" in the order. So, randomly draw one out and then put it back to the

next draw position. For example,

ABCD -> 3 ©

CABD -> 3 (B)

BACD -> 4 (D)

DACB -> 2 (A)

So, the question is, how to calculate the final stage with a given draw sequence with minimum draw simulation?

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Posted · Report post

There are 4 balls marked as "A", "B", "C" and "D" in the order. So, randomly draw one out and then put it back to the

next draw position. For example,

ABCD -> 3 ©

CABD -> 3 (B)

BACD -> 4 (D)

DACB -> 2 (A)

So, the question is, how to calculate the final stage with a given draw sequence with minimum draw simulation?

OK. It looks like what you do is

[1] remove a ball at random

[2] re-arrange the other balls into their original order [e.g. ACD in line 3] but ... not ACB in the last line. I don't get that part.

[3] replace the removed ball at the beginning [left end] of the line.

OK now what are you asking? Can you rephrase it or give an example? Thanks.

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Posted · Report post

[2] re-arrange the other balls into their original order [e.g. ACD in line 3] but ... not ACB in the last line. I don't get that part.

-> the "C" put "back" to "B"'s position in line 3 and "B" put "back" to "D"'s position in the last line.

OK now what are you asking? Can you rephrase it or give an example? Thanks.

For example,

2,3,1,4,1,2,4,3,1,3,2,2,4

How to quickly get the final sequence is:

ADBC

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