Probability of one horse finishing ahead of another In a 5 horse race.  Lets say the probability of a horse to win the race are as follows:
Horse A : 25%
Horse B : 4%
Horse C : 35%
Horse D : 10%
Horse E : 26%
How would you determine the probability of Horse E finishing the race in front of Horse C ? i.e. (Horse E could come 4th and Horse C 5th).
Is there a method for determining the probability given the information available ?

The given information is not enough to compute these probabilities in general, because it may be that for example Horse C is of a kind that it either wins or gets frustrated and finishes last. But it may also just be the best horse and if it doesn't win it may be very likely that it comes second. Which of these is the case is not captured by the data you have.
The problem can be solved making a simplifying assumption that in reality not may b0e true, even though it doesn't look wildly unrealistic either. The assumption I'm thinking of is that we assume that for all ranks the relative probability of any horse finishing on a lower rank assuming that certain given horses occupy the higher ranks is the same among the horses that don't yet have finished.
This for example implies that the probability for Horse A being second given that we know Horse C has won is $\frac{25}{25+4+10+26}=\frac{25}{100-35}=38.5\%$.
This assumption will determine the probabilities for all possible rankings, which can be fully computed using a fairly simple computer program (but complicated enough that I won't take the time to write it for you), going down from rank 1. One can then add all probabilities for cases in which E is ahead of C (or write the program so that it only computes those).
There may be a simpler way of doing this, but if nobody else explains it, here you are.
PS: The answer by Henry uses the same assumption, and it looks like $\frac{26}{61}$ is the result even without running a program.
