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A certain chessboard is infinite in size. There is a frog sitting in the center of every square. After a certain time, all the frogs jump such that

- They may jump to any possible square in the infinite chessboard
- They may jump and land at the same square again

Prove that its possible for all the frogs to jump simultaneously such that there are exactly two frogs per square after the jump.

Let there be n frogs sitting on n squares where $n\rightarrow \infty$

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Lets choose our king frog who’s sitting at a particular square, now

CASE I: King frog wants to stay at his original square, then its probability is $\frac { 1 }{ n } $. Then we are left with n-1 frogs who have n squares left. I am stuck at this point, I know the probability of remaining frogs to choose king’s square is $\frac { n-1 }{ n }$ but how to proceed after that?

CASE II: King frog doesn’t want to stay at his original square, then he has $n-1$ ways to go, also the probability of remaining of the n-1 frogs to land at king’s square is $\frac { n-1 }{ n } $. So we have

$$\lim _{ n\rightarrow \infty }{ \frac { n(n-1) }{ n } } $$

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Suppose you describe the squares of the board $B$ using integer coordinates:

$$

B

= \{(x,y) \mid x,y \in \mathbb{N}\}.

$$

Use the following map, which I will describe in steps: first map

$$

(x,0),(x,1) \mapsto (x,0) \quad \forall x \geq 0;

$$

this takes care of the first row, each square of which now has exactly $2$ frogs occupying it. Next map

$$

(x,2),(x,3) \mapsto (x,1) \quad \forall x \geq 0;

$$

this takes care of the second row, each square of which now has exactly $2$ frogs occupying it. In general, the map is

$$

(x,2i),(x,2i+1) \mapsto (x,i) \quad \forall x \geq 0, i \geq 0.

$$

I think you can see this will have exactly the property you want!

This problem has nothing to do with probability, but your idea of picking a king frog is useful. Starting with the king frog, enumerate both frogs and squares in a spiral (like the starting point for Ulam’s spiral). Then simply have frogs 1 and 2 jump to square 1, frogs 3 and 4 jump to square 2, and so forth.

Think about doing it for one layer of the chess board. It may be helpful to split the layer into two parts. I.e consider the split line being the natural numbers. Tell 0 to jump and 1 to move 1 left. Tell 2 to move 1 left and 3 to move 2 left…. tell $2n$ to move n left and $2n+1 $to move $n+1$ left.

Find a similar algorithm for $\{-1,-2,-3,\ldots\}$. And then just order that this rule should be followed on all levels of the chess board.

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