Spinners from yesteryear: A challenging probability problem While browsing the Internet I found an old horse racing game where the results were determined by a spinner. The names of 6 different horses were listed an equal number of times on the spinner. Each time the spinner landed on a particular horse, you moved that horse forward 1 space. The first horse to move 5 spaces, won the race. 
It was obvious that each horse had an equal probability of winning. I wondered, though, what would happen if we weighted each horse differently, say p1,p2,...,p6, where p1+p2+...+p6 = 1. With this in mind, how would these weightings relate to their probability of winning the race? The problem is, to me at least, difficult. I simulated a few "races" as an example:
Race 1: p = [p1,p2,...,p6] = [50,10,10,10,10,10] --> Total wins for each horse from 100,000 simulations = [95928,831,776,839,825,801]
Race 2: p= [50,30,5,5,5,5] --> [78231,21636,29,43,30,31]
You get the idea. So the question is given an initial set of probabilities for each of the 6 horses, what are the final "winning" probabilities? 
Is there a formula (complicated or not) that describes this?
Thanks to everyone who gives it a shot!
 A: The following analysis gives a hideous formula that involves a nested sum of products. Perhaps someone else will be able to give a simplification.
Suppose one plays the above game with six horses, $h_0, h_1, \ldots, h_5$.  At each time step the probability of horse $h_i$ advancing one step is $p_i \geq 0$
with $\sum_{i=0}^{5}p_i = 1$. Let's first consider the case where we spin the wheel $m$ times and record how far each horse has advanced. In this case we don't stop when a horse has advanced $5$ steps.
The vector $(k_0, k_1, \ldots, k_5)$ of steps advanced by each horse follows a multinomial distribution. The probability mass function in this case is given
by 
$$P(k_0, k_1,\ldots, k_5) = \left\{ \begin{array}{cc}{m \choose k_0, k_1, k_2, k_3, k_4, k_5}p_0^{k_0}p_1^{k_1}p_2^{k_2}p_3^{k_3}p_4^{k_4}p_5^{k_5} & \sum_{i=0}^{5}k_i = m\\
0 & \text{otherwise}
\end{array} \right. $$
Now suppose instead we spin until $h_0$ advances $5$ steps. In this case the vector
$(k_1, k_2, k_3, k_4, k_5)$ of steps advanced by the other $5$ horses will follow a negative multinomial distribution. The probability mass function is given by
$$P(k_1,k_2,k_3,k_4,k_5) = \Gamma\left(5+\sum_{i=1}^5k_i\right)\frac{p_0^{5}}
{4!}\prod_{i=1}^{5}\frac{p_i^{k_i}}{k_i !}$$
Where we use the gamma function, to simplify notation. For natural numbers
$n$, $\Gamma(n) = (n-1)!$.
Now $h_0$ will have won the associated game if and only if 
$k_i < 5$ for $1 \leq i \leq 5$. Thus
$$P(h_0 \text{ wins}) = \frac{p_0^5}{4!}\sum_{ 
k_1 = 0}^4\sum_{k_2=0}^4\sum_{k_3=0}^4\sum_{k_4=0}^4\sum_{k_5=0}^4 \Gamma\left(5+\sum_{i=1}^5k_i\right)\prod_{i=1}^5\frac{p_i^{k_i}}{k_i!}$$
The same reasoning allows us to get the probabilities of the other horses winning, since our numbering of the horses was arbitrary. This result can be generalized to any number of horses and any number of steps needed to win.
