## Difference Payoff

You have 100 stones and ten buckets. After you place the stones in the buckets, two buckets will be chosen at random. You will then be paid a number of dollars equal to the difference in stones between the two buckets.

1. How do you arrange the stones to maximize your expected winnings?
2. If you want to guarantee you will win at least something (no two buckets hold the same number of stones), how can you maximize your winnings?  And how can you maximize your guaranteed amount (the smallest amount you can win)?

Your stones are now colored. You have 50 blue and 50 green stones. After placing them in the ten buckets, two buckets will be chosen at random, and you will be paid a number of dollars equal to the difference between the blue and green stones combined in the two buckets.

1. How do you arrange the stones to maximize your expected winnings?
2. Is it possible to distribute the blue and green stones to guarantee you will win at least something?  How can you maximize this?

Source: Original.

Solutions were received from Joseph DeVincentis, Saw L.B., Jimmy Chng Gim Hong, Claudio Baiocchi, John Hewson, Denis Borris, Matthew Newell, Sudipta Das, and Jozef Hanenberg.
1. Place all 100 stones in one bucket.  Proof from Jospeh DeVincentis:
Let A, B, C, ..., J represent the numbers of stones in the buckets, with A >= B >= C >= ... J. Your expected winnings are an average of A-B, A-C, ..., A-J, B-C, B-D, ..., B-J, ..., I-J. That average is (9A + 7B + 5C + 3D + E - F - 3G - 5H - 7I - 9J)/45. The best way to distribute the stones among A, B, C, ..., J is clearly to put all the stones in A, leading to expected winnings of 900/45 = \$20
2. From Joseph DeVincentis:
a. If we must always win something, we must put a different number of stones in each bucket. The formula above still applies to our expected winnings, so we put as few as possible in 9 buckets (0 through 8 stones) and all the rest in A (64 stones). This gives expected winnings of 14 and 2/3 dollars.
b. To maximize the guaranteed amount, we must maximize the smallest difference between buckets. A difference of 2 between all buckets is possible: 0, 2, 4, 6, 8, 10, 12, 14, 16, 18 with 10 stones left over we can put into the bucket with 18, or, to maximize the chance you will get at least \$3, distribute them 1-2-3-4 among the biggest 4 buckets making them 13, 16, 19, 22.
3. From Claudio Baiocchi:
With two colors, the best expected winning (17.777) is given by:
(50,0), (0,0), ..., (0,0), (0,50);
but the same expected winning is also given e.g. by:
(50,0),(0,5),(0,5),(0,5),(0,5),(0,5),(0,5),(0,5),(0,5),(0,10)
that guarantees 10 as minimum amount.
4. From John Hewson:
It is possible to distribute the blue and green stones to guarantee you will win at least something. To maximize this, arrange the green stones in 8 buckets containing 6, 6, 6, 6, 6, 6, 6, 6, 8 respectively, and 25 blue stones in each of the remaining 2 buckets. The minimum you will win occurs when selecting two buckets containing 6 green stones, that is 12 dollars.

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