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You're playing tic-tac-toe and you're blindfolded.

The objective is to never lose, winning when given the chance under the classic ruleset of the game. Furthermore, an observer who is only aware of the classic rules should watch you play a perfect game.

The player who succeeds in placing three of their marks in a horizontal, vertical, or diagonal row is the winner. It is a solved game, with a forced draw assuming best play from both players.

Since you're blindfolded, you're unable to see the moves your opponent is making and to make things worse, your opponent could be any employing any strategy at any skill level.

Every turn, prior to making a move, you may ask the arbiter a question which can only be answered with a yes or no. These questions must also be focused on either the present or future state of the board.

⚠️ IMPORTANT

  • When playing X, you may only ask one question for the game's duration.

  • When asking about future states of the board, the first player that makes a move such that the answer can no longer be proven correct, loses.
    (in this sense, the question is treated as a move)

After you attempt to make a move, you're allowed to try again if your opponent had marked that spot.

What's your strategy?

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    $\begingroup$ Welcome to PSE! Can you please give examples of questions that are possible (or at least clarify your constraints)? As it stands, I can potentially see some ways to cheat the system. $\endgroup$
    – Benjamin Wang
    Commented Jul 1 at 19:45
  • $\begingroup$ Sure, you can ask before the first turn if the opponent has marked in a corner or not by the end of the second turn and you need to take into account that the arbiter can answer both ways. I would like to know what questions have you thought of. I've been looking for cheats for some time and i haven't found any. $\endgroup$
    – Thheo_sc2
    Commented Jul 1 at 19:55
  • $\begingroup$ If you are looking for easy wins with 'X', you should take into account 2 things: 1) if you are in a winning positions as per the regular game, you need to make a winning move everytime and 2) the first 'O' might break your plans because it could be anywhere. if you were to ask a question like is that exact square occupied by the end of opponent's second move and if the answer is yes, you win elsewhere. @BenjaminWang sorry forgot to tag. got one more comment $\endgroup$
    – Thheo_sc2
    Commented Jul 1 at 20:32
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    $\begingroup$ @JonathanAllan 1. you can assume yes, although it doens't really matter since you'd look forward for any move at any point from the opponent. 2. same as for 1, the opponent can make any move, you could assume the opponent only plays random moves and that would translate to any kind of knowledge in some line of play. 3. you need to create a strategy for both X and O. $\endgroup$
    – Thheo_sc2
    Commented Jul 2 at 18:08
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    $\begingroup$ @JonathanAllan i might have not understood the second question, but i'm pretty sure we can say he uses their knowledge of the answer. not sure if relevant, but the problem statement has been changed and the opponent can't ask questions. (it was edited wrongly at some point) $\endgroup$
    – Thheo_sc2
    Commented Jul 2 at 18:20

2 Answers 2

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As x, winning in the 3rd round can be guaranteed as follows:

Start with asking: Will O play a side and a corner in their first 2 turns?

yes: Play corner, opposing corner, center
- if blocked in round 2: play a side adjacent to your corner and win in round 3
no: Play center, side, opposing side
- if blocked in round 2 win by playing 2 opposite corners
- if blocked in round 3 play a corner adjacent to your side and win in round 3
(2 options, only 1 might be blocked since O played sides in round 1 and 2 )

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  • $\begingroup$ Hope you don't mind, but i now changed the problem's statement. The way i intended was that if you place corner then adjacent side and the O is in the opposite corner, you lose the chance to win under the classic ruleset(it was a winning position, but made a losing move), but that wasn't outlined at all. Now an observer of the game should see you play perfectly. Nonetheless the question still works i think, just not in 3 moves, but i'd like an answer for O too. $\endgroup$
    – Thheo_sc2
    Commented Jul 3 at 16:33
  • $\begingroup$ On second thoughts, with the new rules, if you are blocked in round 3 by the first O. You will have lost the opportunity to win, had the opponent not been restricted. $\endgroup$
    – Thheo_sc2
    Commented Jul 3 at 16:52
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Edit: Incorrect answer. I achieve 'never lose' but not 'win when possible'.

I think I have a working strategy.

Step 1:

Play as X.

Step 2:

Before your first turn, ask this: "If my opponent wins, will it be along a horizontal line?"

Step 3:

Regardless of the answer, play in the centre.

Step 4:

After the opponent's first move, play in a corner. This leaves 4 possible board states (removing reflections and rotations). Lines where the O player can win are marked in red. the four possible board states after one turn from both players

Now note that regardless of which possibility is true, the O player has only two lines they can win on. Additionally, one of these has become unplayable due to the question. Regardless of what O plays here, X always has a move which blocks O's final possible win line.

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  • $\begingroup$ I think you overlooked the fact that you have to win when given the chance. That is if O plays on the side first, that's already a winning position for X and will need to set up a double attack. As it stands, i don't think you can tell which way to do the double attack. You'd also want to make a strategy both for X and O for a complete answer. there might be ambiguity in the statement for this matter. $\endgroup$
    – Thheo_sc2
    Commented Jul 3 at 11:44
  • $\begingroup$ Not that it matters for the conclusions of this answer, but, you can't ignore (some) positions that are 90 degree rotations or reflections across either diagonal. Since your question differentiates the axes based on "winning for O" and "not winning for O" any transformation that swaps O's position relative to these results in a different board state. $\endgroup$
    – GoblinGuide
    Commented Jul 3 at 14:14

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