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Here is a variant of the standard connect 4.

There are seven connect 4 grids (size 6 by 7) numbered 1 to 7. Each grid has the columns numbered 1 to 7. The person who starts selects any one of the seven grids and plays a single move in any column of that grid. (Gravity obviously exists in each grid)

Now onwards, a player has to play in the grid corresponding to the column number of the previous move by the opponent.

For example, suppose Alice starts by playing in the 3rd column of the 6th grid. Now Bob has to play in the 3rd grid. Let's say he plays in the 2nd column. Now Alice has to play in the 2nd grid. And so on....

Alice can choose to play in the 2nd column of this grid also, in which case Bob will also have to play back in the same (2nd) grid.

Once a person completes 4 in a row in one of the seven grids, he/she is the winner of that particular grid. However play will continue as per above mentioned rules. Both players are allowed to make more moves in the same grid, but any more 4 in a rows created in the same grid are of no use.

If Bob say plays in the 7th column of the 4th grid and suppose the 7th grid is completely filled (all 42 spaces), then Alice can play a move in any grid (just like the first move of the game). Same goes with Bob also.

The person to win the most number of grids (of the 7) wins the competition.

If Alice and Bob play perfectly, who wins?

P.S. This isn't just a random game, I know the solution to it.

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    $\begingroup$ could you illustrate with a picture of "seven connect 4 grids (size 6 by 7) numbered 1 to 7" please? having a hard time to picture it :( what is size 6 by 7? $\endgroup$
    – Oray
    Commented Dec 30, 2016 at 13:35
  • $\begingroup$ @Oray, size 6 by 7 is a standard-size Connect Four grid. Just imagine 7 of these grids. $\endgroup$
    – user33097
    Commented Dec 30, 2016 at 14:09
  • $\begingroup$ +1 because you know the solution and you aren't asking us to solve chess or something. $\endgroup$
    – greenturtle3141
    Commented Dec 30, 2016 at 16:26
  • $\begingroup$ sorry but the text is not that understandable. You need to show everything with illustration: You say "each grid has columns", and "selects any one of the seven grids and plays a single move in any column of that grid". Grid is rows? or the empty circle? If grid is the empty circles, then the column numbers represents each grid on that column... you are supposed to say "each column has grids" if so, not "each grid has columns"... you need to fix the text, otherwise noone would able to answer it! $\endgroup$
    – Oray
    Commented Dec 30, 2016 at 18:45
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    $\begingroup$ @Oray This is entirely understandable to me. A Connect Four grid is the 6 row, 7 column vertically oriented grid into which each player drops their pieces. In OP's game, you are playing on 7 of these grids at once, numbered 1 to 7. The first player to play chooses what grid they will play on, and (of course) chooses which column to play in. Whichever column that player played in, the next player must play on the same numbered grid in whatever column they choose, and so on. Each play's column# will be the grid# for the following play, unless a grid fills fully. $\endgroup$
    – Rubio ♦
    Commented Dec 30, 2016 at 20:40

1 Answer 1

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Bob wins.

Alice plays anywhere. Assume it's grid 3 - if it's not, replace every instance of "3" with whatever number it is.

Bob now plays in column 3 on every single move. This gives Alice grid 3, but while she has to keep playing there, Bob can build up his columns on all the other boards. Eventually, Bob will get four of his other boards' columns up to four in a row.

So can Alice delay this by playing in board 3, column 3, where she's already won? Well, no.

Alice playing in board 3, column 3 isn't a problem because of parity. If she plays there, Bob plays in column 3 again. Eventually, that column fills up, with final state being alternation between the two chips. Since Alice was the first to play in B3C3, Bob was the last. So that column is filled up and it's Alice's turn, giving Bob an easy win.

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  • $\begingroup$ What if Alice goes for horizontal 4 in a row? $\endgroup$
    – greenturtle3141
    Commented Dec 31, 2016 at 0:30
  • $\begingroup$ @greenturtle3141: That doesn't matter. Alice plays freely on board 3, claiming it. But since Bob keeps playing in other boards' column 3, Alice has to keep filling up board 3 even after she's won it. This gives Bob 4 in a row on column 3 on every other board. $\endgroup$
    – Deusovi ♦
    Commented Dec 31, 2016 at 0:32

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