Scratch card probabilities A scratch card is in the form of a 5 x 5 matrix. Scratching any square reveals a sum of money. The sums available are 4 x 2.50 dollars, 9 x 1.00 dollar, 12 x 0.50 dollar.
When the scratch card was made it was agreed to place the 2.50 values in the 4 corners. This is not known to the players. The remaining values are randomly distributed over the remaining 21 squares. The idea of the card is to scratch open squares until at least 4.00 dollars in total has been reached. The table below indicates the probability of uncovering at least 4.00 in total, by scratching respectively 1 or 2 or…….8 squares open. For example the probability of uncovering at least a total of 4.0 by scratching 3 squares is 0.287. The probability of needing to scratch 8 squares open to get to the 4.00 total would mean first scratching 7 squares with $0.50\square followed by a 2.50 or 1.00 or another 0.50 dollars to make up the 8 squares. Necessary and sufficient for this last case would seem to be 7 x 0.50 squares as the last square can take on any of the 3 values, giving a probability of
(12C7)/(25C7) = 0.00165. If however you take all 8 into account, this gives:
(12C8)/(25C8) + [(12C7)(9C1)]/(25C8) + [(12C7)(4C1)]/(25C8) = 0.009977 ??
What am I doing wrong here?
I also cannot confirm the 0.287, 0.230, 0.215, and 0.202 values below!
The probability for 7 squares is blank. I cannot format the table properly as I am unfamiliar the applicable text editor.
Number of squares   1 2 3   4   5   6   7   8
Probability 0   0.020   0.287   0.230   0.215   0.202       0.002
 A: There is a difference between these two events:


*

*uncovering a total of at least $4$ by scratching $N$ squares;

*needing to scratch $N$ squares to uncover a total of at least $4$.
The first event occurs, but the second does not occur,
if you scratch $N$ squares and the
first $N-1$ squares already have a total of $4$ or more.
The probabilities in the table are clearly
probabilities of the second kind of event,
so the order in which the squares are revealed matters.
When you write $\binom{12}{7} \binom91 / \binom{25}{8}$, for example,
that is the probability of scratching $7$ squares of value $0.5$ and
one square of value $1$ in any sequence.
But only in the case where the square of value $1$ is opened last
do you need $8$ squares to get a total of $4$ or more. In every other
case the total of $4$ is reached when the seventh square is opened.
Your calculation of the probability for $8$ squares therefore
overestimates the probability by including events that are not part
of the event you are supposed to measure.
In order to need to open $8$ squares,
it is necessary that all $7$ of the first $7$ opened squares
must have value $0.5$ each. The probability of that is 
$\binom{12}{7} / \binom{25}{7}$.
If that happens, then the eighth square will certainly bring the
total to $4$ or more, so a correct way to take into account
the eighth square is to multiply by $1$.
The final result (and the correct probability that you need
to open $8$ squares) is
$$
\frac{\binom{12}{7}}{\binom{25}{7}} \times 1 
= \frac{\binom{12}{7}}{\binom{25}{7}} \approx 0.0016476.
$$

If you must consider three separate cases depending on whether
the last square opened has value $0.5$, $1$, or $2.5$, then a
correct expression for the probability is
$$
\frac{\binom{12}{8}}{\binom{25}{8}}
+ \frac{\binom{12}{7}}{\binom{25}{7}} \times \frac{\binom91}{\binom{18}{1}}
+ \frac{\binom{12}{7}}{\binom{25}{7}} \times \frac{\binom41}{\binom{18}{1}},
$$
which happens to be exactly ${\binom{12}{7}}/{\binom{25}{7}}$;
in other words, we get the same result as before,
but make the calculation quite a bit more complicated than
this particular probability requires.
