Ace and Jacks

If we think of the three cards as X, Y and Z, picking any one at random has a 2/3 chance of being a jack, therefore you have a 2/3 chance of being lied to when you ask a question. If you point to X and ask "is Y the ace?" and the answer you are given is "yes", then Z has a 2/3 chance of being the ace, and if the answer is "no", then Y has a 2/3 chance of being the ace.

You can point to the card on the left and ask a question like, "Does $2+2=4$ ?"

If the answer is yes, which occurs with probability $\frac{1}{3}$ (since that's the probability the card on the left is the ace), you can flip the left card and it has to be the ace, so then the probability of winning is 1.

If the answer is no, which occurs with probability $\frac{2}{3}$ , the left card can't be the ace, so you can guess between the other 2, and have a $\frac{1}{2}$ chance of winning.

The probability of winning is therefore $\frac{1}{3} \times 1+\frac{2}{3} \times \frac {1}{2}=\boxed{\frac{2}{3}}$ .

The three scenarios are: 1) Jack Jack Ace, 2) Jack Ace Jack, and 3) Ace Jack Jack. You do not know which scenario it is, so we will be looking at all the possible answers you could get from a question.

Question 1 : [No matter what the scenario, you will get a No as the answer]. This is a question like, (pointing to the middle card) "Are you a jack?"

Subcase 1.1: The cards are arranged Jack Jack Ace. The answer is No.

Subcase 1.2: The cards are arranged Jack Ace Jack. The Answer is No.

Subcase 1.3: The cards are arranged Ace Jack Jack. The Answer is No.

Result 1 : All three answers are No. The cards are randomized before you, so you do not know which scenario is in front of you right now. Therefore, you must guess, which gives you a $\frac{1}{3}$ probability.

Question 2 : [No matter what the scenario, you will get a Yes as the answer]. This is a question like, (pointing to the middle card) "Are you an ace?"

Subcase 2.1: The cards are arranged Jack Jack Ace. The answer is Yes.

Subcase 2.2: The cards are arranged Jack Ace Jack. The Answer is Yes.

Subcase 2.3: The cards are arranged Ace Jack Jack. The Answer is Yes.

Result 2 : All three answers are Yes. The cards are randomized before you, so you do not know which scenario is in front of you right now. Therefore, you must guess, which gives you a $\frac{1}{3}$ probability.

Question 3 : [For two scenarios, the answer is No, while the answer is Yes for the third]. This is a question like, (pointing to the middle card) "Is the card to the left an ace?"

Subcase 3.1: The cards are arranged Jack Jack Ace. The answer is Yes.

Subcase 3.2: The cards are arranged Jack Ace Jack. The Answer is No.

Subcase 3.3: The cards are arranged Ace Jack Jack. The Answer is No.

Result 3 : Two answers are Yes, while one is No. So if we ask the question and get a Yes, we know exactly where the ace is. So for Subcase 3.1, we have a probability of $1$ . However, if the answer is No, we have narrowed down the ace to 2 spots. Therefore, the probabilities of both Subcase 3.2 and Subcase 3.3 are $\frac{1}{2}$ . To find the total probability for this question, we find the average of the three probabilities, which is $\frac{1+\frac{1}{2}+\frac{1}{2}}{3}=\frac{2}{3}$ .

Question 4 : [For two scenarios, the answer is Yes, while the answer is No for the third]. This is literally just the flip of Question 3, and yields the same result ( $\frac{2}{3}$ ), so I'm not going to go over it.

You may have had different questions than I had, but your question will land in one of these four possibilities, with either a probability of $\frac{1}{3}$ or $\frac{2}{3}$ . Therefore, the highest probability you can have is $\frac{2}{3}$ .

(Please tell me if there is a better way of doing this :) )