A scratchcard has ten windows. Nine are prize icons which include two
that match. One is a card-invalidating message. The object is to
uncover the two matching icons without uncovering the invalidating
message. What are the chances of winning? How did you figure it?

Hi, gyrocopter-ga:

For the sake of determining the chances of winning, we can ignore the
unmatched prize icons.  Scratching off any of these doesn't affect
winning or losing; it's irrelevant to the outcome one way or the
other.

So we consider only the order in which the two matching prizes and the
card-invalidating message would be chosen, if we devise our "strategy"
in advance.  Of course we will always have won or lost before the last
window is scratched off, but it still makes sense to consider a plan
for that "last" unscratched window as the last choice.

To win the knockout message must be third among the three windows that
matter, ie. the two matching prize icons and the card-invalidating
message.

Therefore the chance of winning is one in three.

To see this in a slightly different manner, there are 3! orders in
which three windows that matter can be chosen.  Two of these are
winners (picking the two matching prize icons first, but in either
order).  2/3! = 1/3.

regards, mathtalk-ga

Probability of picking 1st matching icon from all 10 uncovered = 1/10

Probability of then picking 2nd matching icon from the remaining 9 uncovered = 1/9

Therefore, probability of picking both matching icons = 1/10 x 1/9 = 0.01111

://www.google.co.uk/search?hl=en&ie=UTF-8&oe=UTF-8&q=%281%2F10%29*%281%2F9%29&meta=

This problem is a lot harder than the above comment would indicate.
For an example of a similar but easier problem, see page 9 of
http://www.ex.ac.uk/cimt/mepres/allgcse/as5act1.pdf