A Very Odd Game

Question

Barry, John, and Caleb, all perfect logicians, were very bored one day and they decided to come up with a new game (they haven't got a name for that game yet). The rules were quite simple :

1. The first player gets to take either 1, 2, 3, 4 or 5 from the 40 available pens.

2. Then, the second player is also able to pick either 1, 2, 3, 4, or 5 from the remaining pens.

3. The third player is able to do the same

4. Repeat steps 1, 2 and 3 until the player who takes the last pen loses

Now comes the question :

Imagine you are Barry and whether Barry or the other players go first is entirely up to you.

Is it always possible to win at this game?

Players aren't allowed to work together in this game

Answer 1

For the question as it is written (10AM, 8/14/2018), the solution is that

there is no way to force a win for this game without collusion.

Based on my solution to @Bass's question here,

This game is usually won by "forcing" another player to take a specific candy (in this case, it's the 8th last candy, the 15th, 22nd, 29th, 36th). However, the only way to force another player to take that candy is to force the next player in line to take it. Based on my reasoning in the linked question, this strategy guarantees a win for the 3rd player (the player after the player you forced), but does not guarantee a win for you since the 3rd player can still screw you over. There is a maximum likelihood winning strategy (still to be found in @Bass' question), but it is not guaranteed - nobody's chances of winning will be 100% without collusion.

Answer 2

You go first and take 3. From here, you can move in sixes toward 1 – whatever number $n$ the other player picks, you pick $6-n$. After 12 moves, there's one pen left and it's the opponent's turn.

Answer 3

If I understand the game correctly

Then yes, it is possible for the first player to win always
Case 1: First player keeps picking 1, making the other player also pick 1 each time. Player 1 will pick the 39th pen, and player 2 the 40th.
Case 2: First player keeps picking 2, forcing the other player also to keep picking 2. Finally player 1 picks the 37th and 38th pens, while player 2 picks the 39th and 40th.
Case 3: First player keeps picking 3, forcing the other player also to keep picking 3. Finally player 1 picks the 35th, 36th and 37th pens, while player 2 picks the 38th, 39th and 40th.
Case 4: First player keeps picking 2, forcing the other player also to keep picking 2. Finally player 1 picks the 33, 34, 35, 36th pens, while player 2 picks the 37, 38, 39, 40th pens.
Case 5: First player keeps picking 5, forcing the other player also to keep picking 5. Finally player 1 picks the 31, 32, 33, 34, 35th pens, while player 2 picks the 36, 37, 38, 39, 40th pens.

Answer 4

For Bonus-

Barry should choose to go last, and choose 13-n pens, where n is the sum of pens picked by the first two players, after 3 rounds, only 1 pen remains and player one loses, then he can beat the other player by @jafe 's method.

Answer 5

Bonus

If there are 3 players, they can always conspire against Barry and make sure he loses. If Barry goes first, they just need to make sure that in every "round", 10 pens are removed. If Barry goes second, same thing (first player takes 4 then whatever Barry takes, third player takes 6 minus that number). If Barry goes third, first 2 players take 5 each, and it's as if Barry is now "first" with 30 pens left.