Seven Counters


All you need for this game is a friend and seven counters (or coins, biscuits, matches, or whatever you like).

It's a game for two players. Place the seven counters in a pile and decide who will go first. (In the next game, the other player will have the first turn.) Each player takes turns to take away either one or two counters. The player who takes the last counter wins.

Have fun playing the game a few times. 

Can you find a winning strategy, that is a way to win right from the start, no matter what move your opponent makes?

If you'd like to play some similar games, have a go at Got It and Approaching Midnight. Can you see how these games are linked? (You may like to read our Playing the Same Game article.)

Share your thoughts and discoveries


It is so fun

My workings

P1: 1 1 1

P2: 2 1 1 wins

P1: 1 2 wins

P2: 2 2

P1: 1 2

P2: 2 2 wins

P1: 2 1 2 wins

P2: 1 1

If you do the first move you have to keep the number of counters left in stack not on a prime number.

That's interesting Ryan! Can you explain it further?

First player can win always.
What he has to do is start playing one, so it will rest 6 counters. Second player will leave 5 or 4 counters. First player should leave 3 counters in both cases. Second player could leave 2 or 1 counters. First player wins.

Thanks Luna, that's great! Can you generalise this strategy for any number of counters?

Luna doesn't appear to be coming back, so I'll chime in. If the number of counters ...

is 1 or 2 more than a multiple of 3, take away enough counters to get it down to a multiple of 3. Eventually, the pile will have 7 or 8 counters left. Luna has solved this for 7. For 8, take away 2 counters, leaving 6, which as Luna has shown guarantees you a win.

If the staring number is a multiple of 3, graciously allow your opponent to go first and proceed as above.

Thanks for following on from Luna's comments, Steve. I have hidden most of your post under a 'reveal' button so that users can choose to read your thoughts, rather than us giving the game away immediately.

Okay, so you take one counter to get the total to 6 ...

When your opponent takes either 1 or 2 counters, you take enough to leave exactly 3 counters remaining, then whatever your opponent does you can take all remaining counters and winning. I know Luna put this solution before, but I have a generalization. If there are n counters, and each player can take 1,2,3,4.. up to a counters per turn, then this is the strategy. If n is a multiple of a+1, then be player 2. Whatever player 1 does, make your turn such that the number of counters remaining is either 0 or a multiple of a+1. It is always possible to do so. If n is not a multiple of a+1, be player 1, and when it's your turn, then always make your turn such that the number of counters remaining is either 0 or a multiple of a+1. Such turns are again always possible. Always employing the strategy will guarantee you a win as long as you get to be your desired player. If you do not get your desired player you can still win as long as your opponent makes one deviation from this strategy. Always try to make the remaining number of counters a multiple of a+1 when it's your opponents turn. For this to be impossible, you must have not got your desired player(1 or 2) AND your opponent has to follow this strategy 100%. This will be pretty rare, and I estimate the probability of you winning will be about 90% against people who don't know this strategy and 50% against a person who knows this strategy.


Thank you for your detailed post, Ishwar, which outlines a generalisation for this game very clearly. I have hidden most of your comment under a 'reveal' button so that other people need to choose to read it, rather than us giving away the strategy too easily.

We found that whoever goes first wins but they should take 1

In a two player game if you have three counters at the end of your go you will always win.

If you start, always take 1 counter. After your opponent has had a turn, take the opposite amount of counters that they did. Then they are left with three counters and you have won

If you are player one and you want to win then 1st you have to take one and if they take 2 then you can take 1 and you are a guaranteed winner
However if player 2 takes 1 on their first go then you have to take two and you are a guaranteed winner

first if it starts with a multiple of 3 you want to go second but every other time you want to go first. the key is to make sure u end with a multiple of 3 in the middle.

You take one, they take two you take one and you have won or make them on 3 in anyway

To win the seven counters problem, you must start. On your first go, take the amount (1 or 2) that will leave you with a multiple of three. Then on everyone of your go's, take away the amount that will leave you with a multiple of 3 ( again!). Keep doing this, and eventually, there will be two or one left on your go; take them and you win!

The strategy is for if you're going first then take 1 then take 1 and finally you take 2 as a strategy to win.