# A game of n players [duplicate]

There are n players. Each player has to choose an integer. The winner is the one with the lowest integer that has not already been chosen by another player.

What is the optimal strategy?

e.g. if there are 3 players and 2 choose the same number, the winner is the third player. However if all three choose a different number, the winner is the one who has chosen the lowest integer.

• I believe this is a pretty well known problem. For example: this and this would solve it... So I am voting to close this since it is a math question and not a puzzling question. – eye_am_groot Nov 13 '18 at 19:53
• @kaine uhh, I think it's only similar... this one is any $n$ while that one is $0<n<6$ – Hugh Nov 13 '18 at 20:10
• @Hugh this is asking for a more general case of that question without adding anything more to it. That question made for a good puzzle. Expanding it to any value of n does not make for a good puzzle though a very good math topic. – kaine Nov 13 '18 at 20:22

I recommend playing this on a computer.

because

first of all, press - and hold down the 9 key until you’ve got at least a page.

then

copy that page (don’t be dumb and copy the minus sign), and begin spamming command-v (the paste button)

When you feel like it,

copy the entire thing, and continue pasting until your computer dies to lag. Congratulations!

What I’m trying to say is

it doesn’t have to be a positive integer, so go wild with negatives.

So, the winner will always be

the person with the best computer.

• What if I say rot13("artngvir vasvavgl")? – Dorrulf Nov 13 '18 at 20:31
• @Dorrulf (rot13) V qba’g guvax Vasvavgl vf na vagrtre... – Excited Raichu Nov 13 '18 at 20:33
• Yeah, I was considering that too. But it's a short hand representation of what you're doing ;) – Dorrulf Nov 13 '18 at 20:35
• This answer would have been quite suitable for a comment – Bass Nov 13 '18 at 20:42
• @Bass that's quite a lot to put in a comment, and I'd have to rot13 a majority of it, but... okay... – Excited Raichu Nov 13 '18 at 20:45