OK here goes. Two players play the following game with a perfectly round table and perfectly round coins. They take turns placing these coins on the table. Last person to put down a coin wins. Should you go first and what is the optimal stratergy? All coins must be completely on the table and no putting 1 coins on top of another.

Could you explain how large is the table, why it has to be round, and how many coins?Originally posted by VirakOK here goes. Two players play the following game with a perfectly round table and perfectly round coins. They take turns placing these coins on the table. Last person to put down a coin wins. Should you go first and what is the optimal stratergy? All coins must be completely on the table and no putting 1 coins on top of another.

This is assuming that there are infinite amount of coins to be placed and it's the act of physically placing them in the space which is important. As written it's a little ambiguous.

Given that tables tend to be quite a bit larger than a typical coin, I wouldn't have thought going first would make much of an advantage. Either player can earlier on play close to other coins, or more spaced out quite easily.

assuming coins are placed directly adjacent to others...

for starters, with any set of round, (for the most part) two dimensional objects, after the first is placed, say, in the center, the next six will completely surround the first. after that, twelve. after that, 18, 24, 30, 36.... That being the case, on a completely round table, it makes sense to go first (and then of course last) being that you will place the last coin in each concentric ring up to and including the very largest at or near the edge of the table.

I'm not sure it matters so much about who goes first. The important part is at the end, when there is room for only a few more coins. Each player will be trying to place their coin so that there is an even number of places thier opponent can go. If comes down to a single open space, if its small enough you can guarentee yourself a win by putting a quarter in the middle of that space.

I know it's not really a complete strategy, just a few thoughts. Interesting game. Seems like it could cause a few arguments in real life...

"That's not all the way on the table!"

"Is too!"

"Look its hanging over by a hair!"

If it'll stay on the table it counts.Originally posted by perihelionI'm not sure it matters so much about who goes first. The important part is at the end, when there is room for only a few more coins. Each player will be trying to place their coin so that there is an even number of places thier opponent can go. If comes down to a single open space, if its small enough you can guarentee yourself a win by putting a quarte ...[text shortened]... ..

"That's not all the way on the table!"

"Is too!"

"Look its hanging over by a hair!"

Say it has to completely on the table. It doesn't matter though because if you allow it to hang over by a certian amount then that is equivalent to having a slightly larger table. Going first is importent if optimal stratergy is used (Couldn't get it either for a good 6 months before I heard a really similiar problem).

Nice puzzle!Originally posted by VirakOK here goes. Two players play the following game with a perfectly round table and perfectly round coins. They take turns placing these coins on the table. Last person to put down a coin wins. Should you go first and what is the optimal stratergy? All coins must be completely on the table and no putting 1 coins on top of another.

(Not giving out the solution to let you have a bit of fun)

I think I've got something. You definitely want to go first and place a coin in the centre of the table. On all your subsequent turns, you place your coin opposite your opponent's coin on the same diameter. To be a bit clearer:

Let "O" be the centre of the table (where the first coin was placed)

Let "A" be your opponent's coin

Let "B" be your coin

Then OA = OB, and AOB lie on a straight line (diameter). This guarantees that if your opponent can place a coin, you can too, hence you always place the last coin. The only place where this is not possible is in the centre of the table (if the centre of a coin is within one coin radius of the centre, the "opposite" coin would overlap which is illegal). Placing your coin in the centre avoids this problem, and ensures you always follow your opponent.

that sounds reasonable but I think there may be solutions whereOriginally posted by PBE6I think I've got something. You definitely want to go first and place a coin in the centre of the table. On all your subsequent turns, you place your coin opposite your opponent's coin on the same diameter. To be a bit clearer:

Let "O" be the centre of the table (where the first coin was placed)

Let "A" be your opponent's coin

Let "B" be your c ...[text shortened]... lacing your coin in the centre avoids this problem, and ensures you always follow your opponent.

you place the coin off center enough that the last coin will fall off

thereby losing. Just an overall strategy I am thinking of here, not

anything specific. Maybe it is sufficient if the first coin is placed

anywhere within a certain coin diameter sized circle surrounding the

center thus allowing for a window of error.

yes....has to be that. jus like its in a game of carrom. the round in the center has 19 coins at the start of the game. But obviously i think here we r assuming that the coins are of the same size.Originally posted by PBE6I think I've got something. You definitely want to go first and place a coin in the centre of the table. On all your subsequent turns, you place your coin opposite your opponent's coin on the same diameter. To be a bit clearer:

Let "O" be the centre of the table (where the first coin was placed)

Let "A" be your opponent's coin

Let "B" be your c ...[text shortened]... lacing your coin in the centre avoids this problem, and ensures you always follow your opponent.

Assumption: Coins are of equal diameter.Originally posted by gaurav2711yes....has to be that. jus like its in a game of carrom. the round in the center has 19 coins at the start of the game. But obviously i think here we r assuming that the coins are of the same size.

First coin in the centre wins as you can fit 6 around it, then 12, etc, all even numbers.

If the assumption is not correct then the one who puts their biggest coins down first wins. ðŸ˜›

yes, as already stated, that is the answer.Originally posted by OffWhiteAssumption: Coins are of equal diameter.

First coin in the centre wins as you can fit 6 around it, then 12, etc, all even numbers.

If the assumption is not correct then the one who puts their biggest coins down first wins. ðŸ˜›