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64 people are standing on an 8x8 chessboard, facing randomly chosen neighbours. If two people are facing each other, they will give each other a high five. On average, how many high fives can we expect to see?

Source: puzzledquant.com

My approach: We divide the chessboard into $3$ groups, based on number of neighbours. If we place indicator variable on each square, for the middle $6$x$6$ = $36$ squares, we have $1/4$ * $1/4$ if the $2$ people face each other and this can happen in $4$ ways (for the $4$ neighbours) hence total we have $1/16$ * $4$ * $36$, similarly for corner squares we have $1/16$ * $2$ * $4$ and for the remaining we have $1/16$ * $3$ * $24$. However, the summation of these 3 don't match the answer of $335/36$.

Charlie
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1 Answers1

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You should not focus on squares but on edges that separate squares.

Let us call a square an $n$-square if the person on that square has $n$ neighbors.

For every edge that separates two fields there are $4$ possibilities:

  • the edge splits two $4$-squares
  • the edge splits a $4$-square and a $3$-square
  • the edge splits two $3$-squares
  • the edge splits a $3$-square and a $2$-square

Find out how many of each and apply linearity of expectation.

drhab
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  • Hi drhab, could you please show one of the possibilities, how to calculate, like say 1st one, I can think of in this way. $X_i$ will be 1 iff the 2 squares people look at a particular direction i.e. $1/4$ * $1/4$ i.e. for each number of such edges * $1/16$ would be the answer? Is this correct, if so we would not have $36$ in denominator in the answer right? Please tell me where I am going wrong. Thanks! – Charlie Oct 03 '23 at 13:10
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    Correct and I counted $60$ such edges. That gives term $\frac{60}{16}=\frac{135}{36}$. For the others I found the numbers $24$, $20$ and $8$. – drhab Oct 03 '23 at 14:08