Which clue will have you asked on this situation?

[jatubio]

This is a game with 3 players.



Could you solve without additional clues?
Is there's a clue that you can ask to help you solving and it didn't give much information to the others players?

[Wangechi]

All clues give as much info to you as they do to your opponents. Everyone has the same info, there's no reason why you should assume they didn't deduct what you did, if you used logic.

In your game, I wouldn't have asked anything and tried to guess instead.

[Notello]

There are 4 combinations left (2-2, 2-5, 5-2 and 5-5 for the remaining numbers). So your winning chance in your next move is p1 = 1/4 = 25%.

Otherwise (p = 1- p1 = 1 - 1/4 = 3/4) the second player can make his guess with 3 combinations left. His odds are p2 = 3/4 * 1/3 = 1/4 = 25% as well.

If he fails (p = 3/4 * 2/3 = 1/2) there are 2 combinations left for the third player. So his winning probability is p3 = 1/2 * 1 /2 = 1/4 = 25%.

If all of you have failed, only one (correct!) solution is remaining and you will win with the next move. Odds for this are either the remaining 25% or 3/4 * 2/3 * 1/2 (all players make wrong guesses with 4, 3 and 2 possible combinations) which has the same result of 1/4 = 25%.

So your winning chance in this situation is 25% + 25% = 50%, your opponnents have both 25%. If you do nothing (=ask an "empty" question) the second player will be the one with 50%. If you ask something which reduces the number of combinations, your odds will go under 50% as well - with only 2 combinations left you won´t even have any chance.

So it´s the best to make a guess - good luck!

[jatubio]

Thank you very much!

I learnt a lot with your explanation about winning chance. So, next time, I'll apply it, hahahaha.

This time, I already asked for a clue, and we all tied

[Elvaslion]

I would go for a blank, because anything relevant you ask would give him the game.
If I'm wrong, I want to learn why

[nuffbought]

There are 4 combinations left (2-2, 2-5, 5-2 and 5-5 for the remaining numbers). So your winning chance in your next move is p1 = 1/4 = 25%.

Otherwise (p = 1- p1 = 1 - 1/4 = 3/4) the second player can make his guess with 3 combinations left. His odds are p2 = 3/4 * 1/3 = 1/4 = 25% as well.

If he fails (p = 3/4 * 2/3 = 1/2) there are 2 combinations left for the third player. So his winning probability is p3 = 1/2 * 1 /2 = 1/4 = 25%.

If all of you have failed, only one (correct!) solution is remaining and you will win with the next move. Odds for this are either the remaining 25% or 3/4 * 2/3 * 1/2 (all players make wrong guesses with 4, 3 and 2 possible combinations) which has the same result of 1/4 = 25%.

So your winning chance in this situation is 25% + 25% = 50%, your opponnents have both 25%. If you do nothing (=ask an "empty" question) the second player will be the one with 50%. If you ask something which reduces the number of combinations, your odds will go under 50% as well - with only 2 combinations left you won´t even have any chance.

So it´s the best to make a guess - good luck!

I have to admit I'm lost with your explanation, my fault, not yours. I'm not a statistician or a mathematician.

How would the second player have 3 combinations left? If the OP takes a clue and gets one number, there's 2 combinations left. If the OP doensn't take a clue, the second player is in the same boat as the OP with 4 combinations. What am I missing?

So now my question is, if the OP takes a clue, they will get the answer to one of the numbers, then have a 50/50 chance of guessing the last number. My instinct would have me do that action, but the thing that makes me doubt doing it is that it would give the game to the next player as they would take a clue and get the last number. Is this wrong?