Dog probability

Here's a factorial calculator:

http://math.about.com/library/blcalcfactorial.htm

Can you explain how to calculate

Sorry, I can't but perhaps on the website they have a ehlp file that can explain it? They usually do.

If the dogs are all equal, the probability that a given dog out of ten will be first is 10%. The probability that it will be second is also 10%. The probability that it will either be first or second is 20%. If you are given that dog A came in first, the probability that dog B will come in second is 11.1% (1/9).

Can you show us your data for the dogs?

probability are not equal.we can put random probality to all dogs total=1
idea is we have probability of dogs to be first and to find probability to be second.

Let the probability that dog N will win be Pn. What is the probability that dog 1 will come in second given that dog 2 won the race? Since dog 2 is out of the question, the sum of the remaining probabilities is 1-P2, so the chance of dog 1 coming in second is P1/(1-P2). The overall probability that dog 2 will come in first (P2) and dog 1 will come in second (P1/(1-P2)) is the product of those probabilities or P1 P2 / (1-P2).

The probability that dog 1 will come in second regardless of which dog comes in first is the sum of the individual probabilities since they are mutually exclusive. This means the probability that dog 1 will be second regardless of which dog wins is:

P1P2 /(1-P2) + P1P3/(1-P3) + P1P4/(1-P4) ... + P1P10/(1-P10)

Your solution is probably how he needs to answer the question since it sounds like he doesn't have data for head-to-head results, but he needs to be careful about when to apply it. For example, if A, B, and C each roll a 3-sided die like these:
A: 9, 5, 1
B: 8, 4, 2
C: 7, 6, 3
The probability of each winning is:
A: 11/27
B: 7/27
C: 9/27
However:
if A wins, then B takes second 5/11 instead of 7/16
if B wins, then A takes second 1/7 instead of 11/20
if C wins, then A takes second 4/9 instead of 11/18