Ruby - Picking an element in an array with 50% chance for a[0], 25% chance for a[1]

Question

Nothing too complicated, basically I just want to pick an element from the array as if I were making coin tosses for each index and and choosing the index when I first get a head. Also no heads means I choose the last bin.

I came up with the following and was wondering if there was a better/more efficient way of doing this.

def coin_toss(size)
  random_number = rand(2**size)
  if random_number == 0
    return size-1
  else
    return (0..size-1).detect { |n| random_number[n] == 1 }
  end
end

Answer 1

First guess...pick a random number between 1 and 2**size, find the log base 2 of that, and pick the number that many elements from the end.

Forgive my horrible ruby skillz.

return a[-((Math.log(rand(2**size-1)+1) / Math.log(2)).floor) - 1]

if rand returns 0, the last element should be chosen. 1 or 2, the next to last. 3, 4, 5, or 6, third from the end. Etc. Assuming an even distribution of random numbers, each element has twice as much chance of being picked as the one after it.

Edit: Actually, it seems there's a log2 function, so we don't have to do the log/log(2) thing.

return a[-(Math.log2(rand(2**size - 1)+1).floor) - 1]

You may be able to get rid of those log calls altogether like

return a[-((rand(2**size-1)+1).to_s(2).length)]

But you're creating an extra String. Not sure whether that's better than complicated math. :)

Edit: Actually, if you're going to go the string route, you can get rid of the +1 and -1 altogether. It'd make the probabilities more accurate, as the last two elements should have an equal chance of being chosen. (If the next-to-last value isn't chosen, the last value would always be.)

Edit: We could also turn the ** into a bit shift, which should be a little faster (unless Ruby was smart enough to do that already).

return a[-(rand(1<<size).to_s(2).length)]

Answer 2

A non-smart, simple way is:

def coin_toss( arr )
  arr.detect{ rand(2) == 0 } || arr.last
end