# Puzzles

## More doubling cribbage

Brendan and Adam are playing lots more games of

high stakes cribbage: whoever
loses each game must double the other players money. For example, if Brendan has £3 and Adam has £4 then Brendan wins, they will have £6
and £1 respectively.

In each game, the player who has the least money wins.

Brendan and Adam notice that for some amounts of
starting money, the games end with one player having all the money; but for other amounts, the games continue forever.

For which
amounts of starting money will the games end with one player having all the money?

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If Adam has £\(a\) and Brendan has £\(b\), we will write this as £\(a\):£\(b\).

First, we can take
\(a\) and \(b\) to have no common factors, as dividing both by a common factor gives an equivalent starting point. For example,
£3:£6 and £6:£12 will have exactly the same behaviour (imagine £6 and 3 £2 coins).

£\(a\):£\(b\) is also clearly equivalent to £\(b\):£\(a\) (but with the two players swapping places).

A game starting £\(a\):£\(b\) will end with one player having the money if \(a+b\) is a power of two. This is because:

If \(a+b=2^n\), then if \(n=1\), either the game has already ended or it sits at £1:£1 and is about to end. If \(n>2\), then the starting
position can be written as £\(2^n-k\):£\(k\) with \(k<2^n-k\). After another game this will be £\(2^n-2k\):£\(2k\). This is equivalent to
£\(2^{n-1}-k\):£\(k\). Therefore by

induction, if \(a+b=2^n\) then the
game ends with one player having all the money.

It can also be shown by induction, that if a game ends then it must be £\(a\):£\(b\)
with \(a+b=2^n\).

#### Extension

What would happen if the losing player has to *triple* the winning player's money?

## Doubling cribbage

Brendan and Adam are playing high stakes cribbage: whoever loses each game must double the other players money. For example, if Brendan has £3 and Adam has £4 then Brendan wins, they will have £6 and £1 respectively.

Adam wins the first game then loses the second game. They then notice that they each have £180. How much did each player start with?

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Working backwards: before the second game, Brendan must have had £90 and so Adam had £270.

Before the first game, Adam must have had **£135**, so Brendan had **£225**.

#### Extension

After the next game, one player will have all the money and no more games can be played. Hence £135 and £225 lead to a finite number of games being played.

If the player with the most money always loses, which starting values £\(A\) and £\(B\) will lead to finite and infinite numbers of games?