I have a grid of 16 tiles face down. Half are good outcomes and half are bad outcomes. How would I calculate the probability of picking x number of Good outcomes before y number of bad outcomes are picked.


Once a tile is selected and it’s state is revealed (good or bad), it is left face up. That is to say there are no replacements. Also order dose matter because the x number of good outcomes must be picked before the y number of bad outcomes.



Calculate the discrete probability of x number of good outcomes occurring before y number of bad outcomes
You get $x$ *good* before $y$ *bad* if you get $x$ or more *good* out of $x+y-1$ attempts. Let's call the probability of this $P(x,y)$ and using hypergeometric probabilities we have $$P(x,y)= \sum\_{n=x}^8 \frac{{8 \choose n}{8 \choose y-1}}{{16 \choose n+y-1}}$$


There are some fairly obvious symmetric and anti-symmetric shortcuts using combinatorial arguments:


* $P(x,x)= \frac12$ since *good* and *bad* are equally likely
* $P(x,y)= 1 - P(9-x,9-y)$ if we arrange all $16$ and count from the other end
* $P(x,y)= P(9-y,9-x)$ counting from the other end and swapping *good* and *bad*
* $P(x,y)= 1 - P(y,x)$ since one must come before the other swapping *good* and *bad*


Here is a table of the values of $P(x,y)$


$$\begin{matrix} 
{P}&x: &\mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} \\
y: \\
\mathbf{1} && \frac{ 1 }{ 2 } & \frac{ 7 }{ 30 } & \frac{ 1 }{ 10 } & \frac{ 1 }{ 26 } & \frac{ 1 }{ 78 } & \frac{ 1 }{ 286 } & \frac{ 1 }{ 1430 } & \frac{ 1 }{ 12870 } \\
\mathbf{2} && \frac{ 23 }{ 30 } & \frac{ 1 }{ 2 } & \frac{ 37 }{ 130 } & \frac{ 11 }{ 78 } & \frac{ 17 }{ 286 } & \frac{ 29 }{ 1430 } & \frac{ 1 }{ 198 } & \frac{ 1 }{ 1430 } \\
\mathbf{3} && \frac{ 9 }{ 10 } & \frac{ 93 }{ 130 } & \frac{ 1 }{ 2 } & \frac{ 87 }{ 286 } & \frac{ 45 }{ 286 } & \frac{ 283 }{ 4290 } & \frac{ 29 }{ 1430 } & \frac{ 1 }{ 286 } \\
\mathbf{4} && \frac{ 25 }{ 26 } & \frac{ 67 }{ 78 } & \frac{ 199 }{ 286 } & \frac{ 1 }{ 2 } & \frac{ 797 }{ 2574 } & \frac{ 45 }{ 286 } & \frac{ 17 }{ 286 } & \frac{ 1 }{ 78 } \\
\mathbf{5} && \frac{ 77 }{ 78 } & \frac{ 269 }{ 286 } & \frac{ 241 }{ 286 } & \frac{ 1777 }{ 2574 } & \frac{ 1 }{ 2 } & \frac{ 87 }{ 286 } & \frac{ 11 }{ 78 } & \frac{ 1 }{ 26 } \\
\mathbf{6} && \frac{ 285 }{ 286 } & \frac{ 1401 }{ 1430 } & \frac{ 4007 }{ 4290 } & \frac{ 241 }{ 286 } & \frac{ 199 }{ 286 } & \frac{ 1 }{ 2 } & \frac{ 37 }{ 130 } & \frac{ 1 }{ 10 } \\
\mathbf{7} && \frac{ 1429 }{ 1430 } & \frac{ 197 }{ 198 } & \frac{ 1401 }{ 1430 } & \frac{ 269 }{ 286 } & \frac{ 67 }{ 78 } & \frac{ 93 }{ 130 } & \frac{ 1 }{ 2 } & \frac{ 7 }{ 30 } \\
\mathbf{8} && \frac{ 12869 }{ 12870 } & \frac{ 1429 }{ 1430 } & \frac{ 285 }{ 286 } & \frac{ 77 }{ 78 } & \frac{ 25 }{ 26 } & \frac{ 9 }{ 10 } & \frac{ 23 }{ 30 } & \frac{ 1 }{ 2 } 
 \end{matrix}$$


The fractions could easily be given a common denominator of ${16 \choose 8}=12870$


