2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ "...assume all players are perfect logicians" All players or John? $\endgroup$ – u_ndefined Aug 14 '18 at 3:14
  • $\begingroup$ Well John and the additional player. You just need to make a strategy as Barry to win @u_ndefined $\endgroup$ – Kevin L Aug 14 '18 at 3:14
  • $\begingroup$ Oh, and also, you didn't mention who goes first. Who goes first? $\endgroup$ – u_ndefined Aug 14 '18 at 3:17
  • $\begingroup$ It is up to you to decide whether you or the other players go first as long as you can make sure that you win @u_ndefined $\endgroup$ – Kevin L Aug 14 '18 at 3:18
  • 1
    $\begingroup$ I agree. I'll try showing ur solution to my friends. Hopefully they'll give me my 10 dollars :) @El-Guest $\endgroup$ – Kevin L Aug 21 '18 at 4:11
2
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ +1 for the nice answer. Figure out the bonus question and I'll accept your answer $\endgroup$ – Kevin L Aug 14 '18 at 5:14
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Well you can't just make the other player choose the number you want. Assume they are logicians (I guess). I'll edit the question @nikki $\endgroup$ – Kevin L Aug 14 '18 at 3:10
  • 1
    $\begingroup$ Oh okay, I assumed second player was forced to pick the same number as 1st player $\endgroup$ – nikki Aug 14 '18 at 3:13
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ What if 13 - n > 5? $\endgroup$ – sedrick Aug 14 '18 at 6:48
  • $\begingroup$ oh, didn't notice that, back to the drawing board i go.. $\endgroup$ – Shahriar Mahmud Sajid Aug 14 '18 at 6:49
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Lol. I didn't even consider that possibility. I'll clarify that players aren't allowed to team up $\endgroup$ – Kevin L Aug 14 '18 at 6:35
  • $\begingroup$ Wait never mind. I thought whoever took the last wins. $\endgroup$ – sedrick Aug 14 '18 at 6:37
  • $\begingroup$ It's pens not coins $\endgroup$ – u_ndefined Aug 14 '18 at 6:43
  • $\begingroup$ Oh whoops my bad lol $\endgroup$ – sedrick Aug 14 '18 at 6:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.