[bushindo]

This is a variation on superprismatic's excellent puzzle White balls/ Black balls.

Suppose that we have two bags: bag A and bag B. Bag A contains 3 white balls, while bag B contains 3 black balls. We define an iteration as the act of randomly selecting a ball from bag A and transferring it to bag B, and then randomly selecting a ball from the four available balls in bag B, and transferring it back to bag A. Thus, at the end of each iteration, there are 3 balls in each bag.

Suppose we start with 3 white balls in bag A, and 3 black balls in bag B, and apply the iteration above 10 times. Now, if we open up bag A, what is the probability that we will see 3 black balls?

[Guest]

This is a variation on superprismatic's excellent puzzle White balls/ Black balls.

Suppose that we have two bags: bag A and bag B. Bag A contains 3 white balls, while bag B contains 3 black balls. We define an iteration as the act of randomly selecting a ball from bag A and transferring it to bag B, and then randomly selecting a ball from the four available balls in bag B, and transferring it back to bag A. Thus, at the end of each iteration, there are 3 balls in each bag.

Suppose we start with 3 white balls in bag A, and 3 black balls in bag B, and apply the iteration above 10 times. Now, if we open up bag A, what is the probability that we will see 3 black balls?

4 States:

S1: A has WWW, B has BBB

S2: A has WWB, B has BBW

S3: A has WBB, B has BWW

S4: A has BBB, B has WWW

P[s1 -> S1] = 1/4

P[s1 -> S2] = 3/4

P[s2 -> S1] = 1/12

P[s2 -> S2] = 7/12

P[s2 -> S3] = 1/3

P[s3 -> S2] = 1/3

P[s3 -> S3] = 7/12

P[s3 -> S4] = 1/12

P[s4 -> S3] = 3/4

P[s4 -> S4] = 1/4

I will build a table of the probability of being in a given state after a given iteration.

Rules for row generation:

P[s1|Iter] = P[s1->S1]*P[s1|Iter-1] + P[s2->S1]*P[s2|Iter-1]

P[s2|Iter] = P[s1->S2]*P[s1|Iter-1] + P[s2->S2]*P[s2|Iter-1] + P[s3->S2]*P[s3|Iter-1]

P[s3|Iter] = P[s1->S2]*P[s1|Iter-1] + P[s2->S2]*P[s2|Iter-1] + P[s3->S2]*P[s3|Iter-1]

P[s4|Iter] = P[s1->S2]*P[s1|Iter-1] + P[s2->S2]*P[s2|Iter-1] + P[s3->S2]*P[s3|Iter-1]

Iteration : {P[s1|Iter] , P[s2|Iter] , P[s3|Iter] , P[s4|Iter]}

0: {1, 0, 0, 0}

1: {1/4, 3/4, 0, 0}

2: {1/8, 5/8, 1/4, 0}

3: {1/12, 13/24, 17/48, 1/48}

4: {19/288, 143/288, 29/72, 5/144}

5: {25/432, 409/864, 737/1728, 73/1728}

6: {559/10368, 4787/10368, 71/162, 239/5184}

7: {101/1944, 14179/31104, 27629/62208, 2989/62208}

8: {19027/373248, 169055/373248, 41717/93312, 9149/186624}

9: {28267/559872, 505525/1119744, 1004489/2239488, 110881/2239488}

10: {675127/13436928, 6056459/13436928, 754597/1679616, 334283/6718464}

Answer is 334283/6718464 or about 4.97559%

Edited April 1, 2010 by mmiguel1

[bonanova]

As the number of iterations increases, the configuration

loses its memory of how it started and tends to randomness.

The number of ways six things can be taken three at a time is 20.

The configuration in question is unique. There is only one of them.

It's probability after 10 iterations [a sizeable number] is therefore

[approximately] 1/20.

[bushindo]

As the number of iterations increases, the configuration

loses its memory of how it started and tends to randomness.

The number of ways six things can be taken three at a time is 20.

The configuration in question is unique. There is only one of them.

It's probability after 10 iterations [a sizeable number] is therefore

[approximately] 1/20.

Congrats to mmiguel1 and bonanova, who got the answer. I have a question about bonanova's method. While the answer is correct in this case, would the approach change at all if I change my state movement mechanism

From this: "an iteration as the act of randomly selecting a ball from bag A and transferring it to bag B, and then randomly selecting a ball from the four available balls in bag B, and transferring it back to bag A"

To this: "an iteration as the act of randomly selecting a ball from bag A and another ball from bag B, and then switching the two balls"

Intuition says that the final probability of 3 black balls in bag A will change with the change in state movement mechanism, but it is not apparent from the quick and dirty method.

[Guest]

Congrats to mmiguel1 and bonanova, who got the answer. I have a question about bonanova's method. While the answer is correct in this case, would the approach change at all if I change my state movement mechanism

From this: "an iteration as the act of randomly selecting a ball from bag A and transferring it to bag B, and then randomly selecting a ball from the four available balls in bag B, and transferring it back to bag A"

To this: "an iteration as the act of randomly selecting a ball from bag A and another ball from bag B, and then switching the two balls"

Intuition says that the final probability of 3 black balls in bag A will change with the change in state movement mechanism, but it is not apparent from the quick and dirty method.

It would certainly change for the results of the first iteration. There would be no possibility of returning to the initial state of 3 white balls and 3 black balls. Eliminating 3 white and 3 black as an initial state for the second iteration changes probabilities of each of the possible outcomes of the second iteration as well. It stands to reason (at least my feeble reason)that the probabilities would change for all successive iterations. However as more iterations accumulate the difference between the two methods would become less apparent.

[Guest]

Congrats to mmiguel1 and bonanova, who got the answer. I have a question about bonanova's method. While the answer is correct in this case, would the approach change at all if I change my state movement mechanism

From this: "an iteration as the act of randomly selecting a ball from bag A and transferring it to bag B, and then randomly selecting a ball from the four available balls in bag B, and transferring it back to bag A"

To this: "an iteration as the act of randomly selecting a ball from bag A and another ball from bag B, and then switching the two balls"

Intuition says that the final probability of 3 black balls in bag A will change with the change in state movement mechanism, but it is not apparent from the quick and dirty method.

I can find out by altering my previous work.

P[s1 -> S2] = 1

P[s2 -> S1] = 1/9

P[s2 -> S2] = 4/9

P[s2 -> S3] = 4/9

P[s3 -> S2] = 4/9

P[s3 -> S3] = 4/9

P[s3 -> S4] = 1/9

P[s4 -> S3] = 1

I messed up writing this down last time, forgot to change values after copy, paste.

P[s1|Iter] = P[s2->S1]*P[s2|Iter-1]

P[s2|Iter] = P[s1->S2]*P[s1|Iter-1] + P[s2->S2]*P[s2|Iter-1] + P[s3->S2]*P[s3|Iter-1]

P[s3|Iter] = P[s2->S3]*P[s2|Iter-1] + P[s3->S3]*P[s3|Iter-1] + P[s4->S3]*P[s4|Iter-1]

P[s4|Iter] = P[s3->S4]*P[s3|Iter-1]

Iter: {P[s1|Iter], P[s2|Iter], P[s3|Iter], P[s4|Iter]}

0: {1, 0, 0, 0}

1: {0, 1, 0, 0}

2: {1/9, 4/9, 4/9, 0}

3: {4/81, 41/81, 32/81, 4/81}

4: {41/729, 328/729, 328/729, 32/729}

5: {328/6561, 2993/6561, 2912/6561, 328/6561}

6: {2993/59049, 26572/59049, 26572/59049, 2912/59049}

7: {26572/531441, 239513/531441, 238784/531441, 26572/531441}

8: {239513/4782969, 2152336/4782969, 2152336/4782969, 238784/4782969}

9: {2152336/43046721, 19374305/43046721, 19367744/43046721, 2152336/43046721}

10: {19374305/387420489, 174339220/387420489, 174339220/387420489, 19367744/387420489}

The answer is then:

19367744/387420489 or about 4.99915%

It seems like bonanova's estimation would be valid in either case.

Interesting