It depends on your definition of optimal. If you optimize for time, then pressing each button one at a time until a Coke comes out. Plus it doesn't take any brain cells so I can get back to coding quicker. : )
- The highest likelihood of getting your choice with a single pick
- The least amount of money spent to guarantee getting your choice
- pepsi is coke or random
- coke is pepsi or random
- random is coke or pepsi
If you want Coke and you start with Pepsi, you'll get back Coke (1/2 + 1/2x1/2) or Pepsi (1/2x1/2). However, you could pick it and get back Pepsi 10 times in a row.
If you start by picking Random, you get back Coke (1/2) or Pepsi (1/2). If you get back Pepsi, you know
1. Random = Pepsi 2. Coke = Random (since it can no longer be Pepsi, since Random is) 3. Pepsi = Coke
So, you get back Coke on your first attempt (1/2), or you get back Coke on your second attempt (1/1).
So picking random first, then the correct one, optimizes for lowest upper bound on the number of choices to get what you want. Which is _actually_ what I meant by "The least amount of money spent to guarantee getting your choice", but didn't actually express correctly.
Why would you push it ten times in a row? If you get a Pepsi from it on your first press, you now know:
- The button labeled Pepsi is actually Random (it can't be the actual Pepsi button, since none of the labels are correct, and it's also not the Coke button)
- Therefore, the button labeled Coke is actually Pepsi (it can't be Coke by rule, and it's also not Random since we know where Random is now)
- Therefore, the button labeled Pepsi is actually Coke (elimination)
So the max is still 2 steps, but you have 75% chance of getting a Coke on the first try instead of just 50% chance. Same lower bounds, but better expected value.
I agree, and didn't mean to suggest otherwise. I was just referring to the specific probability computations being made.