Coin Flip Conundrum

Imagine this scenario: You are given a bag full of discs that have on one side an integer value and on the reverse side an integer double the first side. You can pull discs out and choose to keep the value you see or turn the disc over and take that value instead, but you must take the new value if you flip the coin over. Whatever value you end up with, you will receive that many gold coins. What strategy can you employ to maximize the number of gold coins you get?

First, note that never switching, always switching, or randomly switching all have the same expected value. (How can you show this?)

One obvious strategy is to always switch when the first number you see is odd. Since one side is always two times the other, if the first side you see is odd, that means the reverse must be the double-value side.

A less obvious strategy is to pick an arbitrary integer (somewhere below where you feel the discs' values lie) and switch whenever the value on the first face is less than that arbitrary integer. If you don't believe this works, write a computer program to convince yourself that it does. (Why does this work? What information would you want to know, and how would you choose the threshold value?)

If you know the full range of numbers on the discs you can create very specific criteria of when to switch and not switch based on the probability that the first side you select is the doubled value or not. (How can you generalize this?)

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