First of all lets assume that if horseD has twice the chances of horseF to finish 1st in a race of 7 horses, then it also has twice the chances to finish first when there are 100 horses or 2 horses or n horses. (It is not defined whether this is true or not. I'll continue assuming this.)
Let's take your example: "calculate the chance of HorseC finishing in 2nd place"
If horseA finishes 1st (0.35), then the probability of horseC finishing 2nd (1st amongst the others) is $\frac{0.15}{0.25+0.15+0.10+0.09+0.05+0.01}=\frac{0.15}{1-0.35}$. So this probability is $0.35\frac{0.15}{1-0.35}$.
Likewise if horseB finishes 1st, then the probability of horseC finishing 2nd is $0.25\frac{0.15}{1-0.25}=P_B\frac{P_C}{1-P_B}$.
The total probability of horse i to finish 2nd is the sum of these:
$P(i,2)=\sum_{i\neq j}{P_j\frac{P_i}{1-P_j}}$.
Can you continue in order to find a general formula for $P(i,n)$?
i suspect you are asking how to combine data in a coherent way to predict outcomes. i will suggest an approach to do so: we need a statistical model for observed data. For example, and following the horse racing application, suppose we observe the results of many horse races. In each race we assume two horses, horse A vs horse B, to keep it simple. Any of 5 jockeys can ride the horses. Maybe jockey #1 tends to ride horse A more often. Maybe jockey #2 is an unskilled jockey, etc. That will all come out in the data analysis. A possible model is the logistic regression:
$$ {\hat p}=\frac {1}{1+exp[-(\beta_0+\beta_1A_1+...+\beta_5A_5+\beta_6B_1+...+\beta_{10}B_{5})] }$$
where $\beta_0,...,\beta_{10}$ are values that will be estimated when fitting the model to a data set.
For each race we record the winning horse and the jockey of each horse. We do that by specifying:
$A_k=1$ if jockey $k$ rode horse A and $=0$ if not.
$B_k=1$ if jockey $k$ rode horse B and $=0$ if not.
If we code the data so that horse A wins & B loses is $=1$, and B wins & A loses is $=0$ then the result of the model is to predict the probability that A wins given jockey 3 on A and 1 on B. That would be the expression
$$ {\hat p}=\frac {1}{1+exp[-(\beta_0+\beta_3+\beta_6) ]}$$
Note that the $\beta's$ may be negative. In real horse race data, the impact of a jockey on the outcome is limited so all the A's and B's will come out near 0. But if they had a big effect, the coefficients would indicate that. The value of $\beta_0$ measures the ability of the horses apart from the jockeys. If, on the other hand, you are looking to fit your example into a theorem of Baye's type problem, you might let us know and start here:
http://en.wikipedia.org/wiki/Bayes%27_theorem
Best Answer
If the horses were priced fairly, no matter what bet you make you would expect to break even. But that's certainly not the case here. If you get your dollar back in addition to the listed payoffs, the probability of an outcome with payoff $r$ should be $1/(r+1)$. If you don't, it should be $1/r$. But neither the $1/r$'s nor the $1/(r+1)$'s add up to $1$.