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Death by drawing?

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12 Identical Balls

Initially we know the set is 11 identical and 1 unknown which is either light or heavy (11i + 1u).

Weight 4v4.

This yields 2 possible states. When the scale balances (CASE1) we know 8 are identical and 4 are unknown either light or heavy (8i + 4u). When the scale is unbalanced (CASE2) we know 4 balls are potentially light and 4 are potentially heavy and 4 are identical, (4i + 4l + 4h).

CASE 1

Weigh iiu v iuu and exclude 5i+u.

If balanced then we mark the balls iii v iii and our state is 11i + u.

If the left was lighter then mark the balls iil v ihh and our state is 9i + l + 2h

If the right was lighter then mark the balls iih v ill and our state is 9i +2l +h.

CASE 2

Weigh lhh v lhh and exclude 4i + 2l

If balanced mark the balls iii v iii and our state is 10i + 2l

If left was light then mark the balls lii v ihh and our state is 9i +l + 2h

If right was light then mark the balls ihh v lii and our state is 9i +l +2h.

After 2 weighings we see there are only four cases: [9i+l+2h, 9i+2l+h, 10i+2l, 11i+u]

CASE 1: weight the 2 potentially heavy balls. If they balance the potentially light ball is the odd ball and it is light. If they do not balance the heavier ball is the odd ball and it is heavy.

CASE 2: Same as CASE1 by symmetry. Simply exchange light for heavy.

CASE 3: Weight the 2 potentially light balls, whichever is lighter is the odd ball and it is light.

CASE 4: Weight the unknown ball against any ball to determine if it is lighter or heavier.

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12 Identical Balls

this doesn't work if the first weighing was balanced and the second weighing was also balanced.

you need two weighing to distinguish 3 balls with 1 unknown

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12 Identical Balls

This assume the unknown is heavier, it can be lighter.

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12 Identical Balls

Solution is definitely correct, but not explained the best.

1,2,5 v 3,7,12 is correct, the 6 is a typo and if you assume he meant 3 the logic is fine.

I think my explanation is clearer, but OPs is fine.

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Death by drawing?

1 white marble in a bag all the rest in the other.

1/2 * (1 + 9/19)

we want to maximize 1/2 * [w1/(w1+b1) + (10-w1)/(20-w1-b1)] over the integers w=0..5 and b=0..10 and also w+b can not equal 0, which prevents empty bags.

so we want to maximize the average probability of being saved.

we can imagine starting with a distribution like all white in 1 bag and all black in the other and then imagine how to improve the average.

We start with 1/2 * (10/10 + 0/10) = 1/2.

We only have 2 choices either move a white to the black or vice versa. Moving white to black improves the odds because the white marbles are wasted being together with no blacks. By symmetry moving a black to the white bag can only worsen odds by diluting white success and leaving the black failure chance still at zero.

This gives us an intuitive argument for why you want 1 white marble alone which is the minimal number for certain salvation and then a maximally diluted failure bag.

We can also just brute force the maximization over the 65 lattice points which is trivial to implement.

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