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You have 2 buckets. One full of white marbles and the other full of black marbles (equal amounts). How do you allocate the marbles into two buckets in a way that maximizes your probability of picking 2 white ones when you pick 1 marble from each bucket?

I would assume its 50/50? But how to justify it?

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Hint: You could guarantee one bucket gave a white marble, while making the other bucket about (slightly less than) $50\%$ likely to give a white marble.

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I'm assuming that equal amounts means that, when we allocate the marbles, there must always be a same amount in both buckets, otherwise, as paw88789 says, we could make the probability approach 50% if we put one white marble in one bucket and all the other marbles in the other bucket.

If we must always have equal amounts, i.e. proportions, in each bucket, however, then if $r$ is the proportion of white marbles in one bucket, $1-r$ must be the proportion in the other bucket. Hence the probability of picking white marbles in both buckets will be $r(1-r)=r-r^2$. Setting the derivative equal to zero to maximize, we see the critical value is at $$1-2\hat{r}=0\\\hat{r}=0.5,$$ as you had assumed.

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You can justify if you write the probability equations. You have the probability p1 for one bucket to take in the first attempt a white marble, the same for a probability p2.

We assume marbles are equally to be taken and the total amount is the same in the two buckets, where the total is 2M, where M is the number of total white marbles and the same quantity for black marbles, and M is too the amount of marbles in a bucket.

In one bucket the probability to take an white marble will be $\frac{W1}{M}$ where W1 is the total number of white marbles in the first bucket and $W1\in\{0,1,...,M\}$.

In the second bucket the probability to take a white marble will be $\frac{M-W1}{M}$ because in total the number of white marbles in the two buckets is M as I said before.

The total probability to take both white marbles, from bucket one and bucket two is $P1\cdot P2=\frac{W1}{M}\cdot\frac{M-W1}{M}$. The maximum value will be when $W1(M-W1)$ will be max i.e. if $f(W1)=M\cdot W1 -W1^2$ the maximum value happens in some solution to $f'(W1)=0\ $ i.e.:

$$M-2\cdot W1=0 \to W1=\frac{M}{2}$$

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The original question does not specify the size of the buckets. Therefore, you put one white marble in one bucket and the rest of the marbles into the other bucket.

You have then increased your chances from 50% to ~ 75% on average

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  • $\begingroup$ The chances to get two white with that distribution are about 50%, but this is an increase from the 25% for an unbiased distribution. $\endgroup$ – quid Sep 24 '16 at 15:14
  • $\begingroup$ Can you prove this is the best answer? $\endgroup$ – user428487 Apr 24 '18 at 2:50

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