You want to place the first coin.

There are two key insights in this question that must be reached, otherwise the solution does not hold. Here’s the first: you must place the first coin in the middle of the table.

First imagine a round table so small that only one coin can fit on it. In this case, the person who goes first will win. Now imagine the table grows so that it is large enough to fit more than one coin on it. Given that the table is round and the first coin has been placed in the middle, another ring of coins can be placed around the original coin. This ring of coins will contain 6 coins (try for yourself with some coins or poker chips, it can also be proven with some geometry but let’s not go there!). Because there is an odd number of coins on the table, the player who goes first wins again.

But what if the second player does not place their coins neatly beside each other on the table? Here is the second insight: wherever the second player places his coin, the first player places his next coin opposite it, on the other side of the centre coin. Because the table and the distribution of the coins in symmetrical, wherever the second player places his coin, there will always be a space opposite for the first player to place another coin. The first player’s coins are dark blue.

Now imagine that the table is bigger still, and it can hold another round of coins. This second round of coins will contain 12 coins. Again, the first person to place a coin wins. No matter how large the table becomes, the additional rings of coins will always have an even number and symmetrical properties. The first player always puts his coins opposite the second player’s coins. There will always be an odd number of coins on the table at the end so the person who goes first will always put the last coin on the table. The second player will always lose.