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Write a method for doing an in-place shuffle of an array.
**

The shuffle must be "uniform," meaning each item in the original array must have the same probability of ending up in each spot in the final array.

Assume that you have a method get_random(floor, ceiling) for getting a random integer that is >= floor and <= ceiling.

A common first idea is to walk through the array and swap each element with a random other element. Like so:

def get_random(floor, ceiling)
rand(floor..ceiling)
end
def naive_shuffle(the_array)
# for each index in the array
(0...the_array.length).each do |first_index|
# grab a random other index
second_index = get_random(0, the_array.length - 1)
# and swap the values
if second_index != first_index
the_array[first_index], the_array[second_index] = the_array[second_index], the_array[first_index]
end
end
end

**However, this does not give a uniform random distribution.**

Why? We could calculate the exact probabilities of two outcomes to show they aren't the same. But the math gets a little messy. Instead, think of it this way:

Suppose our array had 3 elements: [a, b, c]. This means it'll make 3 calls to get_random(0, 2). That's 3 random choices, each with 3 possibilities. So our total number of possible *sets of choices* is 3*3*3=27. Each of these 27 sets of choices is equally probable.

But how many possible *outcomes* do we have? If you paid attention in stats class you might know the answer is 3!, which is 6. Or you can just list them by hand and count:

a, b, c
a, c, b
b, a, c
b, c, a
c, b, a
c, a, b

But our method has 27 equally-probable sets of choices. 27 is not evenly divisible by 6. So some of our 6 possible *outcomes* will be achievable with more *sets of choices* than others.

We can do this in a single pass. time and space.

A common mistake is to have a mostly-uniform shuffle where an item is less likely to stay where it started than it is to end up in any given slot. Each item should have the same probability of ending up in each spot, including the spot where it starts.

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