Odds of Winning a Scratch Card

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I have a scratch card that has 36 spaces that can be scratch. 9 of the spaces have a winning symbol and 27 spaces have an "X". I am allowed to scratch as many spaces until I either get the 9 winning symbols or 3 "X's". What are the odds that I will successfully scratch the 9 winning symbols before getting 3 "X's"?

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You may imagine that even if you win, you continue scratching until you have scratched a total of nine wins and two $x$'s. You may also temporarily assume that each space is numbered.

In other words, consider a win as getting exactly nine winning spaces and two X's. It should be clear that any win under the normal rules corresponds to a win using these rules and vice versa.

Consider our sample space as all ways of scratching eleven spaces where order matters. Our winning event in question is the set of ways of scratching eleven spaces where nine of the spaces are good where order matters.

What is the size of the sample space?

$\frac{36!}{(36-11)!} = \frac{36!}{25!} = P(36,11) = ~_{36}P_{11}=36^{\underline{11}}$ depending on your preferred notation. I'll continue by using $\frac{36!}{25!}$

What is the size of the event that you win?

Pick which of the winning spaces you used where order doesn't matter (One way of accomplishing this). Pick which of the X spaces you used where order doesn't matter ($\binom{27}{2}=~_{27}C_2=C(27,2)=\frac{27!}{2!25!}$ ways of doing this). Pick the order that the chosen spaces occur: ($11!$). Applying multiplication principle, there are $\binom{27}{2}11!$ ways in which you win.

The sample space is clear to see is unbiased, so we may use that the probability of the event occuring as the size of the event divided by the size of the sample space.


Note: Alternatively, we could have set the sample space as the number of ways in which we pick eleven spaces where order doesn't matter, which is also an unbiased sample space. In this case, we count the number of ways to win where order of selection doesn't matter.

This will give the same result, either way.

Both simplify as $\frac{\binom{27}{2}}{\binom{36}{11}}$