0
$\begingroup$

Peter with a standard double six set lays 28 dominoes in one direction - what will be the last number on the end on then of the last domino?
Suppose he plays in both directions and discovers that he can use all the dominoes that he has available but Ann announces she hid one from before he started - which one will it be?

I have the answer 6 for the first part but anyone have a clue of the second part because in a pack there is 28 and if she hid one it makes it 29
this makes no sense to me
any one make sense of it?

  • 0
    The hidden one must be removed from the 28; you only have 27 dominoes left. – 2017-02-25
  • 0
    See my answer here: http://math.stackexchange.com/questions/2129527/domino-probability-problem/2129530#2129530 – 2017-02-25
  • 0
    oh that makes more sense – 2017-02-25
  • 0
    @rhysand You're welcome! And welcome to math.SE! :) – 2017-02-25

1 Answers 1

0

There is not enough information to give numeric values.

In general, consider the half-dominoes; there are $8$ of each value. Since each match point requires $2$ of the same value, a matched line of dominoes using the full set must inevitably have the same number at both ends, otherwise you would have an odd number of some half-domino value.

If a piece is removed and a matched line of the remaining dominoes is made, the answer depends on whether the removed piece was a "double" or not. If the removed piece is a "double", the ends of the matched line will be the same value as each other (but not necessarily the same value as the removed piece). If the removed piece was not a "double", the end values of the 27-piece matched line will match the two halves of the removed piece due to the parity argument used before.