Alice and Bob are playing connect four. Bob is tired of losing all the time, so he suggests changing the rules. He wants to drop 2 pieces every turn. Alice thinks this would make it too easy for him, so she wants to change the win conditions. They agree that she only has to get 3 in a row to win, while Bob needs 5 in a row. Alice goes first.
Assuming perfect play, who will win?
Standard connect four rules apply, unless something else is specified
The game is played on a 7 wide and 6 high grid
Bob will make 2 moves every time it's his turn
Alice will only make 1 move when it's her turn
Bob wins if he gets 5 in a row
Alice wins if she gets 3 in a row
If there is no moves left (the grid is full) without a winner, the game is a draw
Alice would win because she starts.
Bob might get 2 turns(pair), but needing 5 in row(odd) would still take him 3 turns to set a full chain same as Alice. Also, needing a chain of 5 will often make the size of the board a big problem blocking his chains. Of course having 2 turns can allow you to do more clever things so I might be wrong too.
After reading the question again it would seem highly possible that a PERFECT play would result in a draw. Bob's long chains would quickly fill up the board and easily be blocked by Alice. While Bob having 2 turns could also easily block all of Alice's chains.
I'm not entirely sure what determines "perfect play" here but I'll do my best.
First off, a 7x6 board has 42 spaces, with 3 pieces being played per turn this means the game will end in 14 turns with 14 pieces played by Alice and 28 by Bob.
Alice can't win. So can she force a draw?
Essentially, by only being able to place 1 piece per move and needing 3 in a row to win, for Alice to win she needs to already have 2 pieces in a row (duh Cpt. Obvious). With 2 pieces in a row she has 2 possible outs, one on either side. Because Bob can play 2 pieces per turn he can always cover those 2 outs. The only way Alice could possibly have more than 2 outs is to have something like...
- - - - -
- - - - -
- - - A A
B - a B B
A - B A B
where Alice has just dropped the "a" and has 3 outs. But as you can see this requires multiple Alice pieces in a scenario that would be easily spotted early and cut off by Bob.
How can Alice force a draw? Essentially, she needs to cut Bob off from the center. Because she moves first her first move has to be the middle. This means Bob can't possibly win horizontally on the bottom row. Alice then needs to essentially build up pieces in the middle 3 columns (to shut off both rows and diagonals) only straying to shut off a column when Bob has 3 consecutive vertically (he can play 2 on top in a row and win). I'm fairly certain in this way Alice can force a draw.