How to calculate scratch card probability

Question

A scratch ticket has 15 cells.

How do you calculate the probability of scratching 6 winning images before scratching 9 losing images?

I'm trying to build a simple online game and struggling with the probabilities :(

Answer 1

I think you can formulate your problem in terms of the negative hypergeometric distribution, by asking the question as: What is the probability of having all six winning images revealed when we hit all nine losing images? Since then it is clear that the winning scenario (revealing all six winning images) happened first, regardless if after that we continue to reveal the remaining (losing) images. But if we have less than all six winning images revealed when revealing the nineth losing image we lose. In terms of the negative hypergeometric distribution with pmf

$$P(X=k)=\frac{{{k+r-1}\choose{k}}{{N-r-k}\choose{K-k}}}{N \choose K}$$

$N$ is the total number of images on the scratch ticket, $K$ is the total number of winning images, $r$ is the number of revealed losing images after which we stop and $k$ is the number of winning images we see when we have stopped.

Now your question becomes what is the value of

$$P(k=6,N=15,K=6,r=9)=\frac{{{6+9-1}\choose{6}}{{15-9-6}\choose{6-6}}}{15 \choose 6}=\frac{3003*1}{5005}=\frac{3}{5}$$

Answer 2

I assume that we know beforehand that the card will have exactly 6 winning images amongst the 15 on the card.

There are 6 winning images in the 15 images. When choosing the first image to reveal, there is a $\frac{6}{15}$ probability of it being a winning one.

There are now 5 winning images left in the remaining 14 images. When choosing the second image to reveal, there is a $\frac{5}{14}$ probability of it being a winning one.

There are now 4 winning images left in the remaining 13 images. etc.

The probability of going through all 6 reveals successfully is therefore $\frac{6}{15}\cdot \frac{5}{14}\cdot \frac{4}{13}\cdot \frac{3}{12}\cdot \frac{2}{11}\cdot \frac{1}{10}$.

Answer 3

Direct enumeration is another possibility. All possible drawings are obtained by the following Mathematica codes:

ticket = Join[Table[w, 6], Table[l, 9]];
ords = Permutations[ticket];
Length[ords] (* output 5005 *)

The count is $5005$.

Drawing 6 w's before the 9 l's requires that the first 14 scratches contain 6 w's. The number of drawings which satisfy this is

Length[ Select[ #[[1 ;; 14]] & /@ ords, Count[#, w]==6 &]]

and this gives $3003$. The required probability is $3003/5005$.

Answer 4

One more observation that I'm surprised no one has mentioned is that one scratches six winners before nine losers if and only if the last panel scratched is a loser. Since nine out of the fifteen are losers, the probability that one scratches six winners first is $9/15 = 3/5$.

Perhaps I'm not understanding the set-up correctly, if this reasoning doesn't apply.