| It sounds like enough information to me. The fact that we are
told that we CAN figure out which ball remains means that the
rules (which presumably are deterministic only on the 3 choices:
white/white, white/black, black/black) uniquely determine the
final ball (with the given starting conditions). There is
no probability involved here.
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| You can not determine the color of the last ball. Proof follows.
Let XY/Z denote the rule that says if you pull out two balls
whose colors are {X,Y}, then you must put back in one ball of color Z.
I will show that various rule sets allow you to determine the color
of the final ball, but that this resulting color is not always the
same.
There are 8 possible rule sets (2^3), but I will only discuss 6 of them.
We presume that we always start with at least 2 balls.
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Case 1: BB/B BW/B WW/B
In this case, no matter what you start with, you must end with
a black ball in the urn.
Proof: Keep decreasing the number of balls in the urn by 1 until
precisely 2 balls remain. Whatever these 2 balls are, remove them
and put back in a Black ball according to the rules. Thus a
single black ball remains.
Case 2: BB/B BW/B WW/W
In this case, no matter what you start with (except all white balls),
you must end with a black ball in the urn.
Proof: Just observe that none of the rules can exhaust the number
of black balls. Then, if there is at least one balck ball to start
with, you must wind up with one at the end.
In the given case of 123 black balls and 128 white balls, this rule
set would guarantee that a black ball remains.
Case 3: BB/B BW/W WW/B
In this case, if you start with 123 black balls and 128 white balls,
you must end up with a single black ball.
Proof: Note that the parity of the white balls in the urn is always
preserved. Thus, since we start with an even number of white balls,
we cannot wind up with a single white ball in the urn.
Case 4: BB/B BW/W WW/W
In this case, (as in case 2), assuming there is at least one white
ball to start with, you must end up with a single white ball in the
urn.
Proof: Observe that none of the rules can exhaust the number of
white balls.
Case 5: BB/W BW/B WW/W
As in case 3, the parity of the black balls is preserved.
Thus, starting with 123 black balls and 128 white balls,
it is clear that you must end up with a single black ball in the urn.
Proof: Since you started with an odd number of black balls,
there can never be 0 black balls in the urn. Thus there must be
a single black ball at the end.
Case 6: BB/W BW/W WW/W
In this case, no matter what you start with, you must end with
a white ball in the urn.
Proof: Keep decreasing the number of balls in the urn by 1 until
precisely 2 balls remain. Whatever these 2 balls are, remove them
and put back in a white ball according to the rules. Thus a
single white ball remains.
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Thus, using some rule sets, one can determine that a black ball must remain.
With other rule sets, one can determine that a white ball must remain.
Since the rule set is not specified, the final color cannot be determined.
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| The bin contains 123 black 128 white balls to start.
The answer is that the last ball will be white.
The rule is that you select two balls and always throw a white ball back in
the bin if at least one of the two is white.
Hence you gradually have only white balls in the bin. (If you select two
black ones, through one back).
Seems pretty simple, once I thought about it. Unless I missed something . . .
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| The rule
BB/W BW/B WW/B
does not guarantee either a white or a black ball as the last one.
Consider the following sequence of moves:
OP
Apply rule BW/B 128 times. This gives 123 black balls remaining
Apply the pair (BB/W, BW/B) 61 times. This gives a single black ball.
versus:
Apply rule WW/B 1 time. This gives 124 black balls, 126 white
Apply rule BW/B 126 times. this gives 124 black balls
Apply the pair (BB/W, BW/B) 61 times. This gives 2 black balls
Apply the rule BB/W once. This gives a single white ball.
Thus, either outcome is possible.
(Similarly for the remaining rule)
I
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