Let's say you randomly select X squares from a battleship grid (let's say a 7x7 grid & no replacement), how many squares would you expect to be selected from a given column/row?

So, choose 14 squares, what's the expected value of # of squares selected from row A?

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    $\begingroup$ By symmetry the expected number selected in any row is the same. Thus if $X$ squares are selected, the expected number in Row A is $X/7$. $\endgroup$ – André Nicolas Sep 17 '15 at 19:11
  • $\begingroup$ $2$. Hint: expected number of squares selected from a square is $14/49$. $\endgroup$ – zhoraster Sep 17 '15 at 19:22

This sort of question can be answered using the linearity of expectation. Each square has a probability of $p=X/49$ of being selected. Thus the expected number of squares selected out of $1$ square is $p\cdot1+(1-p)\cdot0=p$, so by linearity of expectation the expected number of squares selected out of $7$ squares is $7X/49=X/7$.

  • $\begingroup$ Simple enough. I was hung up on the no replacement. So it doesn't matter if some of those squares end up being selected (say there are only 4 row A squares left on the last couple of selections)? $\endgroup$ – user271746 Sep 17 '15 at 19:18
  • $\begingroup$ @user271746: Alternatively: The expected number of squares selected on the entire grid is $X$. By linearity of expectation, this is $7$ times the expected number of squares per row, which is thus $X/7$. $\endgroup$ – joriki Sep 17 '15 at 19:20
  • $\begingroup$ @user271746: About your comment: I don't see how that could be expected to make a difference -- could you elaborate? $\endgroup$ – joriki Sep 17 '15 at 19:21
  • $\begingroup$ An example of why I think "without replacement" makes it more complex: On trial 1 you have a 7/49 chance of picking an A square. If you don't get an A, on trial 2 you have a 7/48 chance of picking an A square (but if you got an A on trial 1, only a 6/48 chance). I can't wrap my head around all the possible games (e.g., never getting any As, getting all 7 As on the first/last seven trials, and everything in between). Another way to think of it ...pick 14 cards from a deck, how many hearts do you expect to get? (maybe the without replacement doesn't matter) $\endgroup$ – user271746 Sep 17 '15 at 19:31
  • $\begingroup$ @user271746: I agree that in this way of thinking about it, thinks look very complex. What I don't understand is why and how any of this should invalidate the easier way of thinking about it that I suggested. $\endgroup$ – joriki Sep 17 '15 at 19:32

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