@protyushchowdhury9025

Are repetitions allowed? If not, then choosing  6 is not a checkmate. If yes, your strategy is the best strategy. Following explanation assumes that repetitions are not allowed.

If I go first, my turn will come in every odd turn. So for me to win, possible turns are 9,7..and so on but it is not possible to win in 7 turns, because no 7 numbers will add up to 50 (even if you take 7 largest numbers i.e. 10,9,8,7,6,5,4 they sum up to 49). Hence only possible chance for the first player to win is in the 9th turn of the game. In 9 turns, all the numbers from 1 to 10 has been chosen except one number. And that number must be 5 if 9 numbers should sum up to 50. Hence the first player can never win this game, if the second player choses 5 in any of his turns. This is also evident from your answer. Since the first player has already chosen 6 in his first turn, when the second player choses 5 in any of his turns, the first player can't complement it to make 11, since 6 is already chosen.

@estring123

pick 6

losing numbers if its ur turn: 39, 28, 17, 6

@PatrickPichardo

Can you mathematically prove this visually with a decision tree?

@NjdjeJjiijj

I could only thought If last move is 50 at my turn then in the privious total..it must be 50-k ie around 40+..so my goal is to bring friend at a number between 40 and 49 so that in next move I win..but I couldn't derive how I should pick a specific number at first am I half witt or low IQ for the job?