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### Number and algebra

### Geometry and measure

### Probability and statistics

### Working mathematically

### Advanced mathematics

### For younger learners

# Sociable Cards

Joshua from Chesham Preparatory School had a similar idea and found a precise solution through trial and error:

Solution for counters on different cards: 1. Random start but this was not useful. 2. We found out that aces, twos and threes (low cards) weren't helpful as they put the counters onto the same or close cards. 3. We tried 2 different methods to resolve the problem of having 1 counter per card. 4. Our successful method was: - the same high number one after each other - then low numbers (which would be skipped) We carried on this technique until none of the counters could move on. List of cards: 4 tens, 4 aces, 2 twos, 4 nines, 2 twos, 3 threes, 4 eights, 3 fours, 1 three, 4 sevens, 1 four, 2 fives, 4 sixes and 2 fives. This method worked successfully!

It is worth noting that if two counters end up on the same spot after a few steps, they are going to end up in the same place when the game is over. This might make the trial and error a bit faster.

The Maths Club at King Solomon Academy noted:

Our results have proven that it is possible to land all four counters on the same final card. This we believe is due to the starting cards, which are all odd or are all even. However we found out that if you have a mixture of both odd and even in the first four cards our results will be varied. Firstly, if the sequence of the first four cards are odd, odd, even, even, then the three of the counters will land on the same card and one will land on a separate card. Secondly, we realised that if the first four cards were odd, odd, odd, even then all of them will land on the same card apart from one.

Strangely, we found out that our predictions were proven wrong as we thought that if the four starting cards were odd, even, even, even, three counters would land on the same card. The experiment completely proved this theory wrong as the results show that all cards landed on the same final card.

The way the Maths Club came up with theories, tested them and then tried to further and improve is a useful technique. Well done to you all!

Can anyone develop a more rigourous approach?

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Age 11 to 14

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Ed from St Peter's College noticed a pattern:

Most of the counters landed on the same card but there was one counter which landed on a different card. We repeated this game three times and in total four counters landed on differnt cards. In one instance the counters all finished on the same card. This happens because every card has a different number, so they all land on the same numbered cards, so that again would change their path to the same card. From this we have gathered that counters do land on the same card most times. :D

Tom from Wilson School developed a strategy:

You could put all the high numbers near the start and the low numbers nearer the end so that some will land on the end one some may not, it would be best to do a mixture of aces at the end. However if you did only aces near the end you would get all your counters to the end. The last card would have to be 1 more than you have to move. If it is one until the end you would have to put a number two there; if you were two from the end, a number 3 and so on.Joshua from Chesham Preparatory School had a similar idea and found a precise solution through trial and error:

Solution for counters on different cards: 1. Random start but this was not useful. 2. We found out that aces, twos and threes (low cards) weren't helpful as they put the counters onto the same or close cards. 3. We tried 2 different methods to resolve the problem of having 1 counter per card. 4. Our successful method was: - the same high number one after each other - then low numbers (which would be skipped) We carried on this technique until none of the counters could move on. List of cards: 4 tens, 4 aces, 2 twos, 4 nines, 2 twos, 3 threes, 4 eights, 3 fours, 1 three, 4 sevens, 1 four, 2 fives, 4 sixes and 2 fives. This method worked successfully!

It is worth noting that if two counters end up on the same spot after a few steps, they are going to end up in the same place when the game is over. This might make the trial and error a bit faster.

The Maths Club at King Solomon Academy noted:

Our results have proven that it is possible to land all four counters on the same final card. This we believe is due to the starting cards, which are all odd or are all even. However we found out that if you have a mixture of both odd and even in the first four cards our results will be varied. Firstly, if the sequence of the first four cards are odd, odd, even, even, then the three of the counters will land on the same card and one will land on a separate card. Secondly, we realised that if the first four cards were odd, odd, odd, even then all of them will land on the same card apart from one.

Strangely, we found out that our predictions were proven wrong as we thought that if the four starting cards were odd, even, even, even, three counters would land on the same card. The experiment completely proved this theory wrong as the results show that all cards landed on the same final card.

The way the Maths Club came up with theories, tested them and then tried to further and improve is a useful technique. Well done to you all!

Can anyone develop a more rigourous approach?

All you need for this game is a pack of cards. While you play the game, think about strategies that will increase your chances of winning.

This is a game for two players. Does it matter where the target is put? Is there a good strategy for winning?

Can you beat Piggy in this simple dice game? Can you figure out Piggy's strategy, and is there a better one?