# Help Solving a Probability Problem

Four siblings are sitting down to eat some mashed potatoes for lunch: Ethan has 1 ounce of mashed potatoes, Macey has 2 ounces, Liana has 4 ounces, and Samuel has 8 ounces. This is not fair. A blend consists of choosing any two children at random, combining their plates of mashed potatoes, and then giving each of those two children half of the combination. After the children's father performs four blends consecutively, what is the probability that the four children will all have the same amount of mashed potatoes?

Thanks I was confused on how to solve this problem

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A fair share is $3\frac{3}{4}$ ounces. How many selections of pairs can get everybody there? There are $\binom{4}{2}=6$ ways to choose a pair of kids, so $6^4$ sets of pairs (because order matters).

It looks to me like you have to choose one pair of kids for the first mix, choose the opposite pair for the second mix, pick one of the first pair and one of the second pair for the third mix, and pick the other of the first pair and the other of the second pair for the fourth mix. You should check that there is no other way to achieve the desired result. If my route is the only one, there are $6$ choices for the first mix, $1$ for the second, $4$ for the third, $1$ for the fourth for a total of $6*1*4*1=24$ ways to get there and a chance of $\frac{24}{1296}=\frac{1}{54}$

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Instead of the original {1,2,4,8} ounces, let's consider the equivalent values (multiplying everthing by 4) {4,8,16,32} values for each kid. Each mix consists of taking a pair of values and averaging them. The desired uniform distribution has 15 for each one. If we express each number in binary representation:

A (32)     1 0 0 0 0 0
B (16)       1 0 0 0 0
C  (8)         1 0 0 0
D  (4)           1 0 0

Target         1 1 1 1


We see that each mix consists of summing the bits and shifting one position to the right. This shows that a sequence of 4 mixes will lead to the desired result if and only if each mix combine a pair that doesn't have a 1 in the same positions. And this implies that the valid sequences are:

Step 1: choose any pair

Step 2: choose the complementary pair (probability 1/6)

Step 3: choose any pair not previoulsy chosen (probability 4/6)

Step 4: choose the complementary pair of the previous (probability 1/6)

Hence, the probability is 4/216 = 1/54 ~ 0.0185

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