If Richard Feynman applied for a job at Microsoft (2002)

Re:If Richard Feynman applied for a job at Microsoft (2002) by lagadu on Friday July 23, 2021 @01:26PM (#27933307)

You press random and taste, lets say it comes out pepsi. This means that this button has to be the pepsi button because if it was the random button then the label would be correct, like you mentioned. This settles the pepsi button.

Regarding the other two buttons, one is random and the other is coke but their labels are pepsi and coke, giving us only two possible (order-independent) states which I'll illustrate with the format drink(label)/drink(label):

Coke(Pepsi)/Random(coke)

Coke(Coke)/Random(Pepsi)

The second case is invalid because it would require for the coke label to be correct, leaving only one possible case: the button labeled pepsi would have to give out coke.

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Re:If Richard Feynman applied for a job at Microsoft (2002) by Arnavion on Friday July 23, 2021 @02:55PM (#27934337)

Both methods are fine.

Let's say you press the Random button first. If it is actually the Coke button, you get a Coke, and you can stop. Otherwise it's actually the Pepsi button, then you press the Pepsi button since that is actually the Coke button. So you get a Coke in one or two button presses.

Basically there are only two permutations of buttons possible, because all other permutations are disqualified by the "not the same as before" rule. Since the goal is specifically to get a Coke in the fewest button presses, you just have to not start with the Coke button since that is the least likely to produce a Coke.

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