Aces and Jack
Let's play a game! On a table, there are three cards: two aces and one jack. You point to a card, and ask ONE yes-no question. If you are pointing to an ace, I will answer your question truthfully. If you are pointing to a jack, my answer to your question will be a lie. After my answer, you must flip over a card on the table. If the card is an ace, you win! If it is a jack, you lose.
Just by guessing randomly, you have a probability of of winning. Can you develop a strategy that makes your chance of winning better?
Answer as the highest probability that one can win this game.
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Point at the middle card and ask, "Is the leftmost card an ace?" If the answer is yes, flip over the leftmost card. If the answer is no, flip over the rightmost card.
Here is the reason this question works:
Scenario 1: The three cards are arranged as follows: Ace Ace Jack. If you point to the middle card, I will answer truthfully (because the middle card is an ace), and say "Yes". You will choose the leftmost card, and win.
Scenario 2: The three cards are arranged as follows: Jack Ace Ace. If you point to the middle card, I will answer truthfully (because the middle card is an ace), and say "No". You will choose the rightmost card, and win.
Scenario 3: The three cards are arranged as follows: Ace Jack Ace. If you point to the middle card, I will answer with a lie (because the middle card is a jack), and say "No". You will choose the rightmost card, and win.
In other words, if the middle card is an ace, I will truthfully reveal which of the outside cards is the second ace. However, if the middle card is a jack, it won't matter which of the outside cards you choose, because both are aces. So it is wise to assume that the middle card is an ace, and try to discern the location of the second ace. Even if the middle card is a jack, no problem, you still win.
So, you win no matter what by asking this question (and following good logic afterwards). So, with this question, you have a probability of 1 of winning.