Two people are dividing a piece of cheese. First, the first person cuts the cheese into two pieces, then the second person cuts any of the pieces into two, and so on, until there are five pieces. Then the first person takes one piece, then the second person takes one of the remaining pieces, then the first person again - and so on, until the pieces run out. Determine the maximum amount of cheese each player can guarantee for themselves.

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Let the total amount of cheese be 50 (pounds). Players $A$ (first) and $B$ are playing. It is clear that in the division, both would benefit from a greedy algorithm (each time the largest piece is chosen). In this case, $B$ will get the second and fourth largest pieces by weight.

Let's denote their total weight by $R$.

We will show how $A$ can guarantee himself 30 pounds.

First, $A$ divides the cheese into pieces weighing 30 and 20.

Let the weight of the smaller piece cut by $B$ later be $u$. If $u \leq 5$, then $A$ can obtain the set $L=\{30-u, 20-u, u, u\}$, where $0<20-d>c \geq d$.

The sum $R$ of the pieces in the 2nd and 4th positions is currently 20. After the cut, $R$ can only become larger if at least one of these pieces moves to a position with a higher number, being replaced by a larger piece. However, the piece $20-d$ cannot move "to the right": $a$ cannot be cut into two parts larger than $20-d$ (in both cases $20-d > a / 2$). Moreover, in case $L$, both parts $d$ cannot be moved to the 5th position. However, in case $M$, one of the two larger parts can be cut into two pieces larger than $d$. Let's consider two sub-cases.

1) The part $a=20$ is cut into pieces $s, 20-s$, where $d<5$), which means the 2nd largest part is 10. Therefore, $R<20$.

Now we will show how $B$ can guarantee himself 20 pounds (i.e., ensure $R \geq 20$).

Let the pieces after the first cut be $x$ and $y \geq x$. We will call pieces weighing at least 20 "large."

If $x \leq 10$, player $B$ cuts $y$ in half, and if $x \geq 20$, he cuts a piece of weight 20 from $y$. In both cases, two large pieces $a \geq b \geq 20$ are formed. If $A$ does not cut them, then $B$ will not cut them either, and he will be able to take one of them. If $A$ cuts one of them, then $B$ will cut the other into proportional parts. These four parts form two pairs $a_{1} \geq b_{1}$ and $a_{2} \geq b_{2}$, where $b_{1}+b_{2}=b$. The smaller pieces from the pairs are guaranteed to $B$, so $R \geq b \geq 20$.

If $10 \leq x \leq 20$, $B$ gets the set $\{20, x, y-20\}$, where all pieces are at least 10.

If $A$ then cuts the piece of 20, $B$ (cutting 10 from one of the "old pieces") gets the set $\{10+a \geq 10+b \geq 10 \geq 10-b \geq 10-a\}$, in which $R=20$.

Otherwise, before $B$, the set is $\{20, a, b, c\}$, where $a \geq b \geq c$. By cutting $b$ from 20, he transitions to the set $\{a, b, b, 20-b, c\}$. Since $a+b+c=30$, then $b+c \leq 20, a+b \geq 20$, so $c \leq 20-b \leq a$ and $R=b+(20-b)=20$.

## Answer

The first player gets 0.6, the second player gets 0.4 of the total piece.