Hello, I wanted to start those maths but I'm in over my head, so I'm turning to the hive mind :

The game consists of a 5*5 grid, and you are rewarded for every row, column, diagonal of 5 cells that you can form. There is no mistery symbols behind the cell, anytime you select a row of 5 cells you get the reward for 1 row.

The twist is you get 8 cell selection, and anytime you make a selection, a random unselected cell is given for free, making the final board 16 cell discovered, 8 you chose, 8 awarded randomly after each choice you made.

The goal is to have 4 row/column/diagonal for maximum reward. 3 is pretty easy and I'd estimate that 20% of the time I manage to get a 4th.

My questions : 1. Is this solvable (4 row/column/diagonal is obtainable) with perfect play everytime ? 2. If not what is the expected rate of 4th reward ?

I don't feel that the random cell is skewed any particular way, so we can assume it is purely random among the unsellected cells.

Thanks !

  • General Discussion Thread

    This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you must post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed.

    I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

    parent

  • The problem is the game gives you:

    25 * 23 * 21 * 19 * 17 * 15 * 13 * 11 choices

    and

    24 * 22 * 20 * 18 * 16 * 14 * 12 * 10 random draws (if they are actually random).

    Especially with the randomness I expect this to be way to difficult to calculate by hand. You best bet would probably to write a simulation for it.

    Some thought:
    Similar to tic-tac-toe starting in the middle should be best. You contribute to 4 completions.

    Otherwise I think having a strategy where the randomness selects were you go is pretty strong. Were in the beginning you try and get "control" of the lanes by placing in the diagonals where the random stones are placed. And then try and play for complete, while the randomness selects which rows/columns to go for. But this strategy makes it really difficult to predict outcome.

    Generally I think luck plays an important role.

    For a 4 row game you need at least 4 cells overlap (and even then you can't have to many unused cells, which means you have to get lucky with the last few randoms).

    I think for a 5 row game you would need 17 stones instead of 16 for it to be possible.

    My educated guess is: No it's not possible to get a 4 game every game.

    For winrate you would need the best strategy as well as a simulation.

    20% Winrate seems a bit high for me (people are bad for estimating probabilities), why don't to go and make an actual list of wins and losses.

    parent

  • With perfect play, and assuming it is random, it's about 26% solvable. 0.2623603517190757

    from functools import cache
wins=[31<<(5*i) for i in range(5)]+[1082401<<i for i in range(5)]+[17043521,1118480]
def p(x):
    for i in range(5):
        print("".join("." if (x>>j)&1==0 else "x" for j in range(5)))
        x>>=5
@cache
def get_win_prob(k):
    b=k.bit_count()
    if b==16:
        return float(sum(w&k==w for w in wins)>=4)
    if b&1==0:
        return max(get_win_prob(k | (1<<i)) for i in range(25) if k & (1<<i) ==0)
    else:
        l=[get_win_prob(k | (1<<i)) for i in range(25)if k & (1<<i) ==0]
        return sum(l)/len(l)
print(max(get_win_prob(1<<i) for i in [0,1,2,6,7,12]))
    parent