/12/

>12 coins placed in a row
>2 players take turns removing 1, 2 or 3 at a time
>whoever takes the last coin wins
>You are player 2
What is the winning strategy?

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bounce two coins off dem titties and take her home while some nerds figure out their stupid game.

nice bobs

not even math, but it's obvious. Always take however many coins will add up to 4.
So they take 2, you take 2. They take 3, you take 1. They take 1, you take 3.

NIM

i slap the other guy in the face and declare myself the winner

only the strongest survive

nice boobs

i take the coins and buy some beer

I sucked on some titties a few days ago. First time in a while.

When was the last time Veeky Forums sucked on titties?

You mirror his moves. It will result in you picking up the last coins.

Unless he picks 3 you have to pick 1. so basically take his number + your number=4

mathforum.org/library/drmath/view/54203.html

In a scenario where he picks one and you pick one, until the last pair of coins where you lose. ALWAYS 4 total for both picks, bumping bumps.

When I was a few months old

;_;

there was a Cyberchase episode about this over 100 years ago when we were all babby middle schoolers.
is correct

Is there an ideal strategy for player 1? Or will 2 always win playing optimally?

Can be solved easily using dynamic programming.

It could be set up as game theoretical matrix and solved using linear algebra techniques. So yes, it's math-related.

You take 4-(the numver they took). Each 2 turns, 4 coins are removed, and each time, you remove the last one. On the sixth turn, you reach 12.

giv

Those are really nice boobs. I love boobs so much.

Playing through The Azran Legacy, huh?

Let the other player always start by removing n coins and you remove 3-n. You'll always win.